text:
\begin{document}

\begin{center}
{\bf\Large{The $q$-WZ Method for Infinite Series} }
\end{center}

\begin{center}
William Y. C. Chen$^1$ and Ernest X. W. Xia$^2$

Center for Combinatorics, LPMC-TJKLC\\
Nankai University\\
 Tianjin 300071, P. R. China

Email: $^1$chen@nankai.edu.cn,  $^2$xxwrml@mail.nankai.edu.cn

\end{center}

\noindent {\bf Abstract.} Motivated by the telescoping proofs of two
identities of Andrews and Warnaar, we find that  infinite
$q$-shifted factorials can be incorporated into the implementation
of the $q$-Zeilberger algorithm in the approach of  Chen, Hou and Mu
to prove nonterminating basic hypergeometric series identities. This
observation enables us to extend the $q$-WZ method to identities on
infinite series.
 As examples, we
will give the $q$-WZ pairs for some classical  identities such as
the $q$-Gauss sum, the $_6\phi_5$ sum,  Ramanujan's  $_1\psi_1$ sum
and Bailey's $_6\psi_6$ sum.

\noindent {\bf Keywords:} basic hypergeometric series, the
$q$-Gosper algorithm, the $q$-Zeilberger algorithm, the $q$-WZ
method.

\noindent {\bf AMS Classification:} 33D15, 33F10.

\section{Introduction}

Motivated by the telescoping proofs of two identities of Andrews and
Warnaar  \cite{GS06}, we find that infinite $q$-shifted factorials
can be incorporated into the implementation of the $q$-Zeilberger
algorithm as in the approach of Chen, Hou and Mu \cite{CHM2} to
prove nonterminating basic hypergeometric series identities. This
idea also leads to an extension of the $q$-WZ method to
nonterminating basic hypergeometric series identities.

 We will follow the
standard notation on $q$-series {\cite{GR2004}} and always assume
$|q|<1$. The $q$-shifted factorials $(a; q)_n$ and $(a;q)_\infty$
are defined as
\begin{eqnarray*}
(a;q)_n & = &\left\{
     \begin{array}{ll}
         1, \qquad & \mbox{if $n=0$}, \\ [8pt]
        (1-a) (1-aq) \cdots (1-aq^{n-1}), \qquad
              & \mbox{if $n\geq 1$},
     \end{array} \right. \\ [10pt]
     (a;q)_{-n}&=&\frac{1}{(aq^{-n};q)_n}, \qquad \mbox{if $n\geq
     1$},\\[8pt]
(a;q)_\infty & = & (1-a) (1-aq)  (1-aq^2) \cdots,\\[8pt]
 (a_1,a_2,\ldots,a_k;q)_n&=&(a_1;q)_n(a_2;q)_n\cdots (a_k;q)_n.
\end{eqnarray*}
An $_r\phi_{s}$ {\it basic hypergeometric  series} is defined by
\begin{align}\label{phi}
_r\phi_{s}\left[{a_1,a_2,\ldots,a_r \atop
b_1,b_2,\ldots,b_{s}};q,z\right]:=
\sum_{n=0}^\infty\frac{(a_1,a_2,\ldots,a_r;q)_n} {(q,b_1,\ldots,
b_{s};q)_n}\left[(-1)^nq^{n\choose 2}\right]^{1+s-r}z^n,
\end{align}
where $q\neq0$ when $r>s+1$. Further,  an $_r\psi_s$ {\it bilateral
basic hypergeometric  series} is defined by
\begin{align}\label{psi}
_r\psi_s\left[ {a_1,a_2,\ldots,a_r \atop b_1,b_2,\ldots,b_s};q,z
\right]:=\sum_{n=-\infty}^\infty\frac{(a_1,a_2,\ldots,a_r;q)_n}
{(b_1,b_2,\ldots,b_s;q)_n}\left[(-1)^nq^{n\choose
2}\right]^{s-r}z^n.
\end{align}
We assume that $q$, $z$ and the parameters are such that each
 term of the series is well-defined. We say that an $_r\phi_s$ series
terminates if one of its numerator parameters is of the form
$q^{-m}$ with $m=0,1,2,\ldots,$ and $q\neq 0$. Otherwise, we say
that the series $_r\phi_s$ is nonterminating \cite{GR2004}.

   For the ordinary  nonterminating hypergeometric identities,
Gessel \cite{Gessel} and Koornwinder \cite{Koor} provided computer
proofs of Gauss' summation formula and Saalsch\"utz' summation
formula by means of a combination of Zeilberger's algorithm and
asymptotic estimates. Vidunas
 \cite{Vidunas02} (see also Koepf \cite{Koepf03} and Koornwinder
\cite{Koorweb}) presented a method to evaluate $_2F_1\big({a,b \atop
c}\big|-1\big)$   when $c-a+b$ is an integer.
 Recently, Chen, Hou
and Mu  \cite{CHM2}  developed an approach to proving nonterminating
basic hypergeometric identities based on the $q$-Zeilberger
algorithm \cite{Koor0}. In this paper we will show how to apply the
$q$-WZ method to prove nonterminating basic hypergeometric summation
formulas by finding the  $q$-WZ pairs. We will present some examples
including
 the $q$-Gauss sum, the
$_6\phi_5$ sum, Ramanujan's $_1\psi_1$ sum and Bailey's
 $_6\psi_6$ sum  \cite{GR2004}.

\section{ The Andrews-Warnnar Identities}

In  this paper, we  give telescoping proofs of two identities on
partial theta functions of  Andrews and Warnnar \cite[Theorems 1.1
and  3.1]{GS06}:
    \begin{eqnarray}
\bigg(\sum_{n=0}^\infty(-1)^na^nq^{n\choose2
}\bigg)\bigg(\sum_{n=0}^\infty(-1)^nb^nq^{n\choose2
}\bigg)&=&(q,a,b;q)_\infty\sum_{n=0}^\infty\frac{(abq^{n-1};q)_n}
{(q,a,b;q)_n}q^n,
\label{Andrews-Warnaar-1}\\[8pt]
 1+\sum_{n=1}^\infty(-1)^nq^{n\choose
2}(a^n+b^n)&=&(a,b,q;q)_\infty \sum_{n=0}^\infty\frac{(ab/q;q)_{2n}}
{(q,a,b,ab;q)_n}q^n. \label{Andrews-Warnaar-2}
\end{eqnarray}
As will be seen, the telescoping proofs suggest that the  approach
developed by Chen, Hou and Mu \cite{CHM2} for proving nonterminating
basic hypergeometric identities can be extended so that infinite
$q$-shifted factorials are allowed in a $q$-hypergeometric term.
This idea immediately leads to an extension of the $q$-WZ method to
identities on infinite series.

 Note that the
formula (\ref{Andrews-Warnaar-2}) is a generalization of the
well-known Jacobi's triple product identity, when $b=q/a$, we get
the Jacobi's triple product identity
\begin{eqnarray}
\sum_{n=-\infty}^\infty(-1)^na^nq^{n\choose 2}=(a,q/a,q;q)_\infty,
\label{Jacobi's triple product identity}
\end{eqnarray}
 where $|q|<1$ and $a\neq 0$.

We now describe  how to prove the two identities
\eqref{Andrews-Warnaar-1} and \eqref{Andrews-Warnaar-2} by the
telescoping method. We consider \eqref{Andrews-Warnaar-1} first. Let
\[f(a)=\left(\sum_{n=0}^\infty(-1)^nq^{n\choose 2}a^n \right)
\left(\sum_{n=0}^\infty(-1)^nq^{n\choose 2}b^n \right).\]
 Note that the second factor does not contain the parameter $a$. It is
easily verified that
\begin{align}\label{R1}
f(a)=(1-a)f(aq)+aqf(aq^2).
\end{align}
So we are led  to show that the right hand side of
\eqref{Andrews-Warnaar-1} satisfies the same recurrence relation. Of
course, one should still  verify the boundary conditions. Let
\[g(a)=\sum_{n=0}^\infty D_n(a),\ \ \mbox{where}\ \ \ D_n(a)=(q,a,b;q)_\infty
\frac{(abq^{n-1};q)_nq^n} {(q,a,b;q)_n}.\] Then it is necessary to
show that
 \begin{equation} \label{fa} g(a) -(1-a)g(aq) -aqg(aq^2)=0.
\end{equation}
Here comes the key step of finding a telescoping relation for
$D_n(a)$.
 Note
that, for any $n\geq 0$, we have
\begin{align}\label{g1}
&D_n(a) -(1-a)D_n(aq)
-aqD_n(aq^2)\nonumber\\[8pt]
=&\frac{(abq^n;q)_n(a,b,q;q)_\infty q^n}{(q,a,b;q)_n} \bigg(
\frac{1-abq^{n-1}}{1-abq^{2n-1}}
-\frac{1-a}{1-aq^n}-\frac{aq(1-abq^{2n})}
{(1-aq^{n+1})(1-aq^n)(1-abq^n)} \bigg) \nonumber\\[8pt]
=&\frac{(abq^n;q)_n(a,b,q;q)_\infty q^n}{(q,a,b;q)_n} \bigg(
\frac{a(1-q^n)(1-bq^{n-1})}{(1-aq^n)(1-abq^{2n-1})}-
\frac{aq(1-abq^{2n})} {(1-aq^{n+1})(1-aq^n)(1-abq^n)}
\bigg) \nonumber\\[8pt]
=&z_{n+1}-z_n,
\end{align}
where
\[z_n=-\frac{(1-q^n)(1-bq^{n-1})(abq^n;q)_n(q,a,b;q)_\infty
aq^n}{(1-aq^n)(1-abq^{2n-1})(q,a,b;q)_n}.\] The above relation
reveals that the infinite $q$-shifted factorial $(a,b,q;q)_\infty$
can be incorporated into the telescoping relation and this step can
be automated by the $q$-Gosper algorithm. Moreover, one sees that
infinite $q$-shifted factorials can be incorporated into the
$q$-Zeilberger algorithm so that the recent approach of Chen, Hou
and Mu \cite{CHM2} can be extended. In particular, one can make the
$q$-WZ method work for infinite  basic hypergeometric  series.

 Now, let us return our attention to the proof of
 \eqref{Andrews-Warnaar-1}.
 Clearly, $z_0=0$. It is also easily seen that
  $\lim_{n\rightarrow¡¡+\infty } z_n=0$.
 Summing \eqref{g1} over all nonnegative integers, we obtain the
 recurrence relation \eqref{fa}. In order to show that $f(a)=g(a)$,
 we will use the recurrence relation of $f(a)-g(a)$ to reach this goal.

Let $H(a)=f(a)-g(a)$. From the recurrence relations for $f(a)$ and
$g(a)$, it follows that  $H(a)$ satisfies the  recurrence
 relation
\begin{align}\label{recurrence relation1}
H(a)=(1-a)H(aq)+aqH(aq^2).
\end{align}
Iterating the above relation yields that
\begin{align}{\label{Recurrence}}
H(a)=A_nH(aq^{n+1})+B_nH(aq^{n+2}),
\end{align}
 where $A_n$ and $B_n$ are given by
\[A_0=(1-a),\  B_0=aq,\   A_1=(1-a)(1-aq)+aq,\  B_{1}=(1-a)aq^2,\]
and
\[A_{n+1}=(1-aq^{n+1})A_n+aq^{n+1}A_{n-1},
\  B_{n+1}=aq^{n+2}A_n, \ \
 n\geq 1. \]
Hence we have
\[
A_{n+1}-A_{n}=-aq^{n+1}(A_n-A_{n-1}),
\]
which implies that
\begin{align*}
|A_{n+1}-A_{n}|  &=|(-1)^na^nq^{(n+1)!}||(A_1-A_0)|\\[8pt]
&\leq |a^nq^{(n+1)!}|\left(|A_1|+|A_0|\right).
\end{align*}
So, for fixed $a$ and $|q|<1$, the limit $\lim\limits_{n\rightarrow
+\infty}A_n$ exists. Since $B_{n+1}=aq^{n+2}A_n$,   the limit
$\lim\limits_{n\rightarrow +\infty}B_n$ also exists.  Again, by the
relation \eqref{Recurrence}, we find
 \[
 H(a)=H(0)\left(\lim_{n\rightarrow +\infty}A_n+\lim_{n\rightarrow
 +\infty}B_n\right).
 \]
 It remains to show that $H(0)=0$, that is,
 \begin{equation}\label{h0}
 \sum_{n=0}^\infty(-1)^nb^nq^{n\choose 2}
=(q,b;q)_\infty\sum_{n=0}^\infty\frac{q^n}{(q,b;q)_n}.
 \end{equation}
 Indeed, the above relation is a limiting case of Heine's
 transformation of $_2\phi_1$. For completeness, we give a derivation based on
 Euler's identities:
\begin{eqnarray}\label{identity-1}
(q,b;q)_\infty\sum_{n=0}^\infty\frac{q^n}{(q,b;q)_n} &=&
(q;q)_\infty\sum_{m=0}^\infty
\frac{q^m}{(q;q)_m}\sum_{n=0}^\infty\frac{q^{n\choose
2}(-bq^m)^n}{(q;q)_n}\nonumber\\[6pt]
&=&(q;q)_\infty\sum_{n=0}^\infty\frac{(-1)^nb^nq^{n\choose
2}}{(q;q)_n}\sum_{m=0}^\infty\frac{(q^{n+1})^m}{(q;q)_m}\nonumber
\\[6pt]
&=& \sum_{n=0}^\infty(-1)^nb^nq^{n\choose 2}.
\end{eqnarray}
Thus, we  have verified that $H(a)=0$.  This  completes the proof.
\qed

We remark that once the recurrence  relation \eqref{recurrence
relation1} is derived, one can also use the theorem of Chen, Hou and
Mu \cite[Theorem 3.1]{CHM2} to prove the existence of the limits of
$A_n$ and $B_n$.

We next present to a telescoping proof of \eqref{Andrews-Warnaar-2}.
Let
\[f(a)=1+\sum_{n=1}^\infty(-1)^nq^{n\choose 2}(a^n+b^n).\]
It is easily seen that
\begin{align}\label{R2}
(1+aq)f(a)-(1-a^2q)f(aq)-(aq+a^2q)f(aq^2)=(q-1)a.
\end{align}
 Let
\[g(a)=\sum_{n=0}^\infty D_n(a), \ \ \mbox{where}
\ \ \ D_n(a)=(q,a,b;q)_\infty
\frac{(ab/q;q)_{2n}q^n}{(q,a,b,ab;q)_n}.\] It will be shown that
\begin{align}\label{ga}
(1+aq)g(a)-(1-a^2q)g(aq)-(aq+a^2q)g(aq^2)=(q-1)a.
\end{align}
 Since
 \allowdisplaybreaks
\begin{align*}
&\frac{q^n-abq^{n-1}}{1-abq^{2n-1}}
-\frac{(1-a^2q)(1-ab)q^n}{(1+aq)(1-aq^n)
(1-abq^n)}\\[8pt]
&\qquad-\frac{(a^2q+aq)(1-abq^{2n})(1-abq)q^n}
{(1+aq)(1-aq^n)(1-aq^{n+1})
(1-abq^n)(1-abq^{n+1})}\\[8pt]
&\ =\frac{(1-abq^{2n})(-1+q+abq^{n+1}+a^2bq^{n+2}
-aq^{n+2}-q^{n+1}-a^2bq^{2n+2} +a^2bq^{2n+3})a}{
(1-aq^{n+1})(1-aq^n)(1-abq^{n})(1+aq)(1-abq^{n+1})}\\[8pt]
&\ -\frac{
(-1+q+abq^n+a^2bq^{n+1}-aq^{n+1}-q^n-a^2bq^{2n}+a^2bq^{2n+1})
a(1-bq^{n-1})(1-q^n)} {(1-aq^n)(1-abq^{2n-1})(1+aq)(1-abq^n)},
\end{align*}
 multiplying both sides by
$$\frac{(ab;q)_{2n}(a,b,q;q)_\infty}{(q,a,b,ab;q)_n},$$
we deduce that
\begin{align}\label{g2}
&D_n(a) -\frac{(1-a^2q)}{1+aq} D_n(aq)
-\frac{(a^2q+aq)}{1+aq}D_n(aq^2) =z_{n+1}-z_{n},
\end{align}
where
\begin{align*}
z_n=&\frac{(-1+q-aq^{n+1}-q^n+a^2bq^{n+1}+abq^n-a^2bq^{2n}
+a^2bq^{2n+1})a}{(1-aq^n)(1-abq^{2n-1})}\\[8pt]
 &\qquad\qquad\times\frac{(1-bq^{n-1})(1-q^n)(ab;q)_{2n}(q,a,b;q)_\infty}
{(1+aq)(1-abq^n)(ab;q)_n(q,a,b;q)_n}.
\end{align*}
Clearly, $z_0=0$  and $\lim_{n\rightarrow +\infty}z_{n+1}
=\frac{(q-1)a}{1+aq}$.
 Summing \eqref{g2} over all  nonnegative integers, we obtain the
 recurrence relation \eqref{ga}.

  Let
$H(a)=f(a)-g(a)$. Then $H(a)$ satisfies the following recurrence
 relation
\begin{align}\label{recurrence relation2}
H(a)=\frac{1-a^2q}{1+aq}H(aq)+ \frac{aq+a^2q}{1+aq}H(aq^2).
\end{align}
By iteration, we obtain
\begin{align}{\label{Recurrence2}}
H(a)=A_nH(aq^{n+1})+B_nH(aq^{n+2}),
\end{align}
 where $A_n$ and $B_n$ are given by
\begin{align*}
&A_0=\frac{1-a^2q}{1+aq},\ \ \ A_1=\frac{(1-a^2q^3)(1-a^2q)}
{(1+aq)(1+aq^2)}+\frac{aq+a^2q}{1+aq},\\[8pt]
&B_0=\frac{aq+a^2q}{1+aq},\ \ \
B_1=\frac{(1-a^2q)(aq^2+a^2q^3)}{(1+aq)(1+aq^2)},
\end{align*}
and  for $n \geq 1$,
\begin{eqnarray}
 A_{n+1}& = & \frac{1-a^2q^{2n+3}}{1+aq^{n+2}}A_{n}
+\frac{aq^{n+1}+a^2q^{2n+1}}{1+aq^{n+1}}A_{n-1},\label{Formula-1} \\[6pt]
B_{n+1}& =& \frac{aq^{n+2}+a^2q^{2n+3}}{1+aq^{n+2}}A_n.
\label{Formula-2}
\end{eqnarray}
It follows from \eqref{Formula-1} that
\begin{align}{\label{relation1}}
A_{n+1}-\frac{-aq^{n+2}-a^2q^{2n+3}}{1+aq^{n+2}}A_n=
A_n-\frac{-aq^{n+1}-a^2q^{2n+1}}{1+aq^{n+1}}A_{n-1}. \ \ \ \ n\geq1.
\end{align}
For given  $a$ and $|q|<1$, we have
\[\lim_{n\rightarrow +\infty}\frac{-aq^{n+2}-a^2q^{2n+3}}
{1+aq^{n+2}}=0 .\] Now, for $\varepsilon_0=\frac{1}{2}$, there
exists an integer $N_0>0$, depending on  $a$ and $q$,
 such that for any integer $n\geq N_0$, we have
\begin{align}{\label{relation2}}
\left|\frac{-aq^{n+2}-a^2q^{2n+3}}{1+aq^{n+2}}
\right|<\varepsilon_0=\frac{1}{2}.
\end{align}
Therefore,  given  $a$ and $|q|<1$, for any integer $n\geq N_0$, we
have \allowdisplaybreaks
\begin{align*}
&|A_{n+1}-A_{N_0}|\\[8pt]
=&{\bigg|} A_{n+1} -\frac{-aq^{n+2}-a^2q^{2n+3}}{1+aq^{n+2}}A_n+
\frac{-aq^{n+2}-a^2q^{2n+3}}{1+aq^{n+2}}
(A_n-\frac{-aq^{n+1}-a^2q^{2n+1}}{1+aq^{n+1}}A_{n-1})\\[8pt]
+&\cdots \\[8pt]
+&\frac{(-aq^{n+2}-a^2q^{2n+3}) \cdots(-aq^{N_0+3}-a^2q^{2N_0+5})}
{(1+aq^{n+2})\cdots(1+aq^{N_0+3})}(A_{N_0+1}-
\frac{-aq^{N_0+2}-a^2q^{2N_0+3}}{1+aq^{N_0+2}}A_{N_0})\\[8pt]
+&\frac{(-aq^{n+2}-a^2q^{2n+3})
\cdots(-aq^{N_0+3}-a^2q^{2N_0+5})(-aq^{N_0+2}-a^2q^{2N_0+3})}
{(1+aq^{n+2})\cdots(1+aq^{N_0+3})(1+aq^{N_0+2})}A_{N_0}-A_{N_0}
 {\bigg|}\\[8pt]
 \leq & \left|A_{n+1} -\frac{-aq^{n+2}-a^2q^{2n+3}}
 {1+aq^{n+2}}A_n\right|
 +\left|\frac{-aq^{n+2}-a^2q^{2n+3}}{1+aq^{n+2}}\right|
\left|(A_n-\frac{-aq^{n+1}-a^2q^{2n+1}}{1+aq^{n+1}}
A_{n-1})\right|\\[8pt]
+&\cdots \\[8pt]
+&\left|\frac{(-aq^{n+2}-a^2q^{2n+3})
\cdots(-aq^{N_0+3}-a^2q^{2N_0+5})}
{(1+aq^{n+2})\cdots(1+aq^{N_0+3})}\right|\left|(A_{N_0+1}-
\frac{-aq^{N_0+2}-a^2q^{2N_0+3}}
{1+aq^{N_0+2}}A_{N_0})\right|\\[8pt]
+&\left|\frac{(-aq^{n+2}-a^2q^{2n+3})
\cdots(-aq^{N_0+3}-a^2q^{2N_0+5})(-aq^{N_0+2}-a^2q^{2N_0+3})}
{(1+aq^{n+2})\cdots(1+aq^{N_0+3})(1+aq^{N_0+2})}A_{N_0}\right|
+|A_{N_0}|\\[8pt]
\leq &(1+\frac{1}{2}+\frac{1}{2^2}+\cdots +\frac{1}{2^{n-N_0}})
\left|A_{N_0+1}-
\frac{-aq^{N_0+2}-a^2q^{2N_0+3}}{1+aq^{N_0+2}}A_{N_0}\right|
+(1+\frac{1}{2^{n+1-N_0}})|A_{N_0}|\\[8pt]
\leq & 3|A_{N_0+1}|+3\left|\frac{-aq^{N_0+2}-a^2q^{2N_0+3}}
{1+aq^{N_0+2}}A_{N_0}
\right|+\frac{3}{2}|A_{N_0}|\\[8pt]
 \leq & 3|A_{N_0+1}|+3|A_{N_0}|.
\end{align*}
Here we have used the relations \eqref{relation1} and
\eqref{relation2}. From the above inequality, we find
\[
|A_{n+1}|\leq 3|A_{N_0+1}|+4|A_{N_0}|, \ \ \ \ \ \ n\geq N_0-1.
\]
Thus, for  $n\geq N_0+1$, we have
\begin{align*}
|A_{n+1}-A_n|&=\left|\frac{-aq^{n+2}-a^2q^{2n+3}}{1+aq^{n+2}}A_n+
\frac{aq^{n+1}+a^2q^{2n+1}}{1+aq^{n+1}}A_{n-1}\right|\\[8pt]
&\leq \left(\left|\frac{-aq^{n+2}-a^2q^{2n+3}}{1+aq^{n+2}}\right| +
\left|\frac{-aq^{n+1}-a^2q^{2n+1}}{1+aq^{n+1}}\right|\right)
(3|A_{N_0+1}|+4|A_{N_0}|).
\end{align*}
It follows that the limit $\lim\limits_{n\rightarrow +\infty}A_n$
exists. Since
\[ B_{n+1}=\frac{aq^{n+2}+a^2q^{2n+3}}{1+aq^{n+2}}A_n,\]
 the limit $\lim\limits_{n\rightarrow +\infty}B_n$ also exists. By the
relation \eqref{Recurrence2}, we deduce that
 \[
 H(a)=H(0)(\lim_{n\rightarrow +\infty}A_n+\lim_{n\rightarrow
 +\infty}B_n).
 \]
The identity \eqref{identity-1} implies that  $f(0)=g(0)$. So we
have $H(a)=0$. This completes the proof. \qed

We also note that once the recurrence  relation \eqref{recurrence
relation2} is derived, one may assume that  $|a|<1$ and may use the
the  theorem in Chen, Hou and Mu \cite[Theorem 3.1]{CHM2}) to prove
the existence of the limits of $A_n$ and $B_n$. Moreover, we may
drop the assumption $|a|<1$ by analytic continuation.

\section{The $q$-WZ Pairs for Infinite Series}

Our approach to the $q$-WZ method for infinite series can  be
described as follows. The key step is to construct $q$-WZ pairs for
infinite sums.  Suppose that we aim to prove an identity of the
form:
\begin{align}\label{equation1}
\sum_{k=N_0}^\infty F_k(a_1,a_2,\ldots, a_t)=R(a_1,a_2,\ldots,a_t),
\end{align}
where $t$ is a positive integer, and the sum is either a unilateral
or bilateral basic hypergeometric series, namely, $N_0=0$ or
$N_0=-\infty$,  $R(a_1,a_2,\ldots,a_t)$ is either zero or a quotient
of two products
 of infinite $q$-shifted  factorials.

First, we set some parameters, say, $a_1,\ldots,a_p$, ($1\leq p\leq
t$) to $a_1q^n,\ldots, a_pq^n$, so that we get
\begin{align}\label{ Set parameters }
\sum_{k=N_0}^\infty F_k(a_1q^n,\ldots,
a_pq^n,a_{p+1},\ldots,a_t)=R(a_1q^n,\ldots,
a_pq^n,a_{p+1},\ldots,a_t).
\end{align}
If $R(a_1q^n,\ldots, a_pq^n,a_{p+1},\ldots,a_t)\neq 0$, Then set
\[
F(n,k)=\frac{F_k(a_1q^n,\ldots,
a_pq^n,a_{p+1},\ldots,a_t)}{R(a_1q^n,\ldots,
a_pq^n,a_{p+1},\ldots,a_t)}.
\]
Otherwise, we simply set
\[
F(n,k)=F_k(a_1q^n,\ldots, a_pq^n,a_{p+1},\ldots,a_t).
\]
Our goal is to show that
\begin{align}\label{Constant}
\sum_{k=N_0}^\infty F(n,k)= {\rm constant}, \ \ \ \ \ \
n=0,1,2,\ldots.
\end{align}
The constant can be determined by setting $n=0$ and setting
$a_1,a_2,\ldots, a_t$ to special values. We claim that the above
goal can be achieved by adopting the $q$-WZ method for infinite
sums.

Let us recall the boundary and limit conditions for the $q$-WZ
method. Let $f(n)$ denote the left hand side of \eqref{Constant},
i.e.,
\[ f(n)=\sum\limits_{k=N_0}^\infty F(n,k)\]
 and we aim to show that
\[ f(n)={\rm constant}\]
 for every nonnegative integer $n$. To this end, it suffices to
show that $f(n+1)-f(n)=0$ for every nonnegative integer $n$. This
can be done by finding  $G(n,k)$ such that
\begin{align}\label{WZPair}
F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k).
\end{align}
A pair of functions $(F(n,k),G(n,k))$ that satisfy \eqref{WZPair} is
called a {\it $q$-WZ pair}. Once a $q$-WZ pair is found, one can
check the boundary and limit conditions to ensure that $f(n)$ equals
the claimed constant. Here are the conditions:
\begin{itemize}
\item[(C1)] For each integer $n\geq 0$,
$\lim\limits_{k\rightarrow\pm \infty}G(n,k)=0$.

\item[(C2)] For each integer $k$, the limit
\begin{align}\label{limit}
f_k=\lim_{n\rightarrow \infty}F(n,k)
\end{align}
\ \ \ \ \ \ \ \ \ \ \  \ exists and is finite.

\item[(C3)] We have $\lim\limits_{L\rightarrow
\infty}\sum\limits_{n\geq0} G(n,-L)=0$.
\end{itemize}

The WZ method can be formally stated as follows.

\begin{thm}[Wilf and Zeilberger \cite{WZ2}]
 Assume that $(F(n,k),G(n,k))$ is a WZ pair \eqref{WZPair}.
 If {\rm (C1)} holds, then we have
\begin{align}\label{Con1}
\sum_{k}F(n,k)={\rm constant},\ \ \ \ \ \ \ n=0,1,2,\ldots.
\end{align}
If {\rm(C2)} and {\rm(C3)} hold, then we have the companion identity
\begin{align}\label{Identity}
\sum_{n=0}^\infty G(n,k)=\sum_{j\leq k-1}(f_j-F(0,j)),
\end{align}
 where $f_j$ is defined by \eqref{limit}.
\end{thm}

We now explain how to compute the desired $q$-WZ pair for the
identity \eqref{equation1}.
 In fact, it can be produced by applying the
 $q$-Gasper algorithm to
$F(n+1,k)-F(n,k)$. It should be noted that
  $F(n+1,k)-F(n,k)$ is  a  $q$-hypergeometric term with respect to
   $q^k$,  even if $F(n,k)$ contains
  infinite $q$-shifted factorials such as $(aq^n;q)_\infty.$
 Obviously,  $F(n+1,k)-F(n,k)$ is a  $q$-hypergeometric term when $R(a_1,\ldots,a_t)= 0$.
 Assume that $R(a_1,\ldots,a_t)\neq0$.  Let
 \allowdisplaybreaks
 \begin{align*}
 M_1&=\frac{R(a_1q^{n+1},\ldots,
a_pq^{n+1},a_{p+1},\ldots,a_l)}{R(a_1q^n,\ldots,
a_pq^n,a_{p+1},\ldots,a_t)},\\[8pt]
M_2 &=\frac{F_{k+1}(a_1q^{n+1},\ldots,
a_pq^{n+1},a_{p+1},\ldots,a_t)}{F_k(a_1q^{n+1},\ldots,
a_pq^{n+1},a_{p+1},\ldots,a_t)},\\[8pt]
M_3&=\frac{F_{k+1}(a_1q^n,\ldots,
a_pq^n,a_{p+1},\ldots,a_t)}{F_k(a_1q^{n+1},\ldots,
a_pq^{n+1},a_{p+1},\ldots,a_t)},\\[8pt]
M_4&=\frac{F_k(a_1q^n,\ldots,
a_pq^n,a_{p+1},\ldots,a_t)}{F_k(a_1q^{n+1},\ldots,
a_pq^{n+1},a_{p+1},\ldots,a_t)}.
 \end{align*}
 Since $M_1$ is a rational function in $q^n$ and is independent of
 $k$,
 $M_2, M_3, M_4$ are all rational functions in $q^k$. Observe that
\begin{equation}\label{fm}
 \frac{F(n+1,k+1)-F(n,k+1)}{F(n+1,k)-F(n,k)}=
\frac{M_2-M_1M_3}{1-M_1M_4}
\end{equation}
is a rational function in $q^k$,  i.e.,  $F(n+1,k)-F(n,k)$ is a
$q$-hypergeometric term with respect to $q^k$. It is necessary to
mention that even if $F(n,k)$ contains infinite $q$-shifted
factorials of the form $(aq^n;q)_\infty$, the quotient \eqref{fm}
 no longer contains the $q$-shifted
factorial $(aq^n;q)_\infty$ and it is still a rational function in
$q^k$. Consequently, we can employ the $q$-Gosper algorithm to
determine whether $G(n,k)$ exists.
 Nevertheless, it is also necessary to note that $G(n,k)$
contains infinite $q$-shifted factorials if $F(n,k)$ does.

There is another way to look at the above procedure.  Suppose that
$F(n,k)$ contains an infinite $q$-shifted factorial $(a;q)_\infty$,
where $a$ is a chosen parameter for the substitution $a\rightarrow
aq^n$. If we set $G'(n,k)=R(aq^n) G(n,k)$. Then the equation
\eqref{WZPair} becomes
\[ F(n+1, k)R(aq^{n})  - F(n,k) R(aq^n) = G'(n,k+1) - G'(n,k).\]
It is evident that the infinite $q$-shifted factorial
$(aq^n;q)_\infty$ will disappear in the above equation, and one can
use the $q$-Gosper algorithm to find a $q$-WZ pair if it exists.

We now take the $q$-binomial theorem  \cite[P. 354]{GR2004} as an
example to explain the above steps:
\begin{align}\label{q-binomial theorem}
\sum_{k=0}^\infty\frac{(a;q)_k}{(q;q)_k}z^k=
\frac{(az;q)_\infty}{(z;q)_\infty},\ \ \ \ \ |z|<1.
\end{align}
In this case,  we have \[ F_k(a)= \frac{(a;q)_k}{(q;q)_k}z^k, \quad
R(a)=\frac{(az;q)_\infty}{(z;q)_\infty}. \] We choose the parameter
$a$, and substitute $a$ with $aq^n$. Then we set
\[ F(n,k)={F_k(aq^n) \over R(aq^n)} ={ (aq^n;q)_k (z;q)_\infty  \over (q;q)_k
(azq^n;q)_\infty}z^k. \] In order to find $G(n,k)$ such that
 \eqref{WZPair} holds, it is easily checked that $F(n+1, k)-F(n,k)$ is a
$q$-hypergeometric term. By examining the $q$-Gosper algorithm, one
sees that it is capable to deal with the input $F(n+1,k)-F(n,k)$, or
we can set \[ G'(n,k)= R(aq^n) G(n,k)\] and find a solution of the
equation
\begin{equation}\label{New-WZ-Equation}
(1-azq^n)\frac{(aq^{n+1};q)_k}{(q;q)_k}z^k-\frac{(aq^n;q)_k}{(q;q)_k}z^k
= G'(n,k+1)-G'(n,k).
\end{equation}
Finally, we obtain the  $q$-WZ pair
\begin{align*}
F(n,k)&=\frac{(aq^n;q)_k(z;q)_\infty}
{(q;q)_k(azq^n;q)_\infty}z^k,\\[8pt]
G(n,k)&=-\frac{(aq^n;q)_k(z;q)_\infty(a-aq^k)}
{(q;q)_k(azq^n;q)_\infty(1-aq^n)}q^nz^k.
\end{align*}
If $|z|<1$,  it is easy to see that $F(n,k)$ and $G(n,k)$ satisfy
the conditions (C1), (C2) and (C3). By \eqref{Con1},
\[
\sum_{k=-\infty}^\infty F(n,k)=\sum_{k=0}^\infty F(n,k)={\rm
constant}, \ \ \ \ \ n=0,1,2,\ldots.
\]
Setting $z=0$ yields that the constant equals 1. Setting $n=0$,  we
have
\[
\sum_{k=0}^\infty F(0,k)={\rm constant}=1.
\]
 By
\eqref{Identity}, we get the companion identity of \eqref{q-binomial
theorem}
\begin{align*}
\sum_{j=0}^k\frac{(a;q)_j}{(q;q)_j}z^j=
(az;q)_\infty\sum_{j=0}^k\frac{z^j}{(q;q)_j}+
\frac{az^{k+1}(a;q)_{k+1}}{(q;q)_{k}}\sum_{n=0}^\infty
\frac{(az;q)_n(aq^{k+1};q)_n}{(a;q)_{n+1}}q^n.
\end{align*}

We now give a few more examples.

 \noindent {\bf
Example 3.1.} The $q$-Gauss sum \cite[P. 354]{GR2004}:
\begin{align}\label{Gauss}
\sum_{k=0}^\infty
\frac{(a,b;q)_k}{(q,c;q)_k}\left(\frac{c}{ab}\right)^k
=\frac{(c/a,c/b;q)_\infty} {(c,c/ab;q)_\infty},\ \ \ \ |c/ab|<1.
\end{align}

By computation, we obtain the following $q$-WZ pair
\begin{align*}
F(n,k)&=\frac{(b,aq^n;q)_k(c/ab,cq^n;q)_\infty}
{(q,cq^n;q)_k(c/a,cq^n/b;q)_\infty}
\left(\frac{c}{ab}\right)^k,\\[8pt]
G(n,k)&=-\frac{(a-aq^k)(b,aq^n;q)_k(c/ab,cq^n;q)_\infty}
{(1-aq^n)(q,cq^n;q)_k(c/a, cq^n/b;q)_\infty}
\left(\frac{c}{ab}\right)^kq^n.
\end{align*}
If $|c/ab|<1$, it is easy to verify that $F(n,k)$ and $G(n,k)$
satisfy the conditions (C1), (C2) and
 (C3).  By \eqref{Con1}, we have
\[
\sum_{k=-\infty}^\infty F(n,k)=\sum_{k=0}^\infty F(n,k)= {\rm
constant}, \ \ \ \ n=0,1,2,\ldots.
\]
Setting $c=0$ and $n=0$, we find that the constant equals 1, and  we
have
\[
\sum_{k=0}^\infty F(0,k)={\rm constant}=1.
\]
After simplification, we obtain the identity \eqref{Gauss}.

 By
\eqref{Identity}, we obtain the companion identity of \eqref{Gauss}
\begin{align}\label{Gauss-c}
-\sum_{n=0}^\infty \frac{(a-aq^k)(b,aq^n;q)_k(c/ab,cq^n;q)_\infty}
{(1-aq^n)(q,cq^n;q)_k(c/a, cq^n/b;q)_\infty}
\left(\frac{c}{ab}\right)^kq^n =\sum_{j\leq k-1}(f_j-F(0,j)),
\end{align}
where
\[
f_j=\lim_{n\rightarrow \infty}F(n,j)=\frac{(b;q)_j(c/ab;q)_\infty}
{(q;q)_j(c/a;q)_\infty}\left(\frac{c}{ab}\right)^j,
\]
which can be restated as
\begin{align*}
\sum_{j=0}^k
\frac{(a,b;q)_j}{(q,c;q)_j}\left(\frac{c}{ab}\right)^j=&
\frac{(c/b;q)_\infty}{(c;q)_\infty}
\sum_{j=0}^k\frac{(b;q)_j}{(q;q)_j}
\left(\frac{c}{ab}\right)^j\\[8pt]
&+\frac{(a,b;q)_{k+1}c^{k+1}}{(q;q)_k(c;q)_{k+1}a^kb^{k+1}}
\sum_{n=0}^\infty\frac{(aq^{k+1},c/b;q)_n}
{(a;q)_{n+1}(cq^{k+1};q)_n}q^n.
\end{align*}

 \noindent{\bf {Example 3.2.}}
 The  $_6\phi_5$
 summation formula \cite[P. 356]{GR2004}:
\begin{align}\label{65}
\sum_{k=0}^\infty &
\frac{(1-aq^{2k})(a,b,c,d;q)_k}{(1-a)(q,aq/b,aq/c,aq/d;q)_k}
\left(\frac{aq}{bcd}\right)^k \nonumber\\[8pt]
&\qquad\qquad =\frac{(aq,aq/bc,aq/bd,aq/cd;q)_\infty}
{(aq/b,aq/c,aq/d,aq/bcd;q)_\infty},\ \ \ \ \ \ \ |aq/bcd|<1.
\end{align}

We get the following $q$-WZ pair:
\begin{align*}
F(n,k)=&\frac{(1-aq^{n+2k})(c,d, aq^n,bq^n;q)_k}
{(1-aq^n)(q,aq/b,aq^{n+1}/c,aq^{n+1}/d;q)_k}\\[8pt]
&\qquad \times\frac{(aq/b,aq/bcd,aq^{n+1}/c,aq^{n+1}/d;q)_\infty}
{(aq/bc,aq/bd,aq^{n+1},aq^{n+1}/cd;q)_\infty}
\left(\frac{aq}{bcd}\right)^k,\\[8pt]
G(n,k)=&\frac{(c,d;q)_k(a/b,a/bcd;q)_\infty}
{(q,a/b;q)_k(aq^n,aq^n/cd;q)_\infty}\\[8pt]
&\qquad\times \frac{(aq^n,bq^n;q)_k(aq^n/c,aq^n/d;q)_\infty}
{(aq^n/c,aq^n/d;q)_k(a/bd,a/bc;q)_\infty
}\\[8pt]
&\qquad\times \frac{(a-bc)(a-bd)(aq^n-cd)(1-q^k)}{(a-bcd)(bq^n-1)
(aq^{n+k}-c)(aq^{n+k}-d)} \left(\frac{aq}{bcd}\right)^kq^n.
\end{align*}
It is easily seen that $F(n,k)$ and $G(n,k)$ satisfy the conditions
(C1), (C2) and (C3). Therefore, by \eqref{Con1}, we have
$\sum\limits_{k=0}^\infty F(n,k)$ is a constant. Setting $n=0$ and
$a=0$, we find that the constant equals 1. Thus we have
\[ \sum_{k=0}^\infty F(0,k)={\rm constant}=1,\]
 which is nothing but \eqref{65}. Since
\[
f_k=\frac{(c,d;q)_k}{(q,aq/b;q)_k}\frac{(aq/b,aq/bcd;q)_\infty}
{(aq/bc,aq/bd;q)_\infty}\left(\frac{aq}{bcd}\right)^k
\]
and
\[
F(0,j)=\frac{(1-aq^{2j})(a,b,c,d;q)_j
(aq/b,aq/c,aq/d,aq/bcd;q)_\infty}
{(1-a)(q,aq/b,aq/c,aq/d;q)_j(aq,aq/bc,aq/bd,aq/cd;q)_\infty}
\left(\frac{aq}{bcd}\right)^j,
\]
 by \eqref{Identity},  we obtain the
companion identity
\begin{align*}
\sum_{j=0}^k&\frac{(1-aq^{2j})(a,b,c,d;q)_j }
{(1-a)(q,aq/b,aq/c,aq/d;q)_j}
\left(\frac{aq}{bcd}\right)^k\\[8pt]
&\qquad\qquad=\frac{(aq,aq/cd;q)_\infty}{(aq/c,aq/d;q)_\infty}
\sum_{j=0}^k\frac{(c,d;q)_j}
{(q,aq/b;q)_j}\left(\frac{aq}{bcd}\right)^j\\[8pt]
&\qquad\qquad+\frac{b(aq;q)_k(b,c,d;q)_{k+1}}
{(q,aq/b;q)_k(aq/c,aq/d;q)_{k+1}}
\left(\frac{aq}{bcd}\right)^{k+1}\\[8pt]
&\qquad\qquad\times\sum_{n=0}^\infty
\frac{(aq/cd;q)_n(aq^{k+1},bq^{k+1};q)_n}{(b;q)_{n+1}
(aq^{k+2}/c,aq^{k+2}/d;q)_{n}}q^n.
\end{align*}

\noindent{\bf Example 3.3.}  Ramanujan's $_1\psi_1$ sum \cite[P.
357]{GR2004}
\begin{align}\label{11}
_1\psi_1(a;b;q,z)=
\frac{(q,b/a,az,q/az;q)_\infty}{(b,q/a,z,b/az;q)_\infty}, \ \ \
|b/a|<|z|<1.
\end{align}

In this case, we find that
\begin{align*}
F(n,k)&=\frac{(aq^n;q)_k(z,b/az,bq^n,q^{1-n}/a;q)_\infty}
{(bq^n;q)_k(q,b/a,azq^n,q^{1-n}/az;q)_\infty}z^k,\\[8pt]
G(n,k)&=\frac{(z,b/az,bq^n,q^{-n}/a;q)_\infty(aq^n;q)_k(1-azq^n)}
{(q,b/a,azq^n,q^{-n}/az;q)_\infty(bq^n;q)_k(z-azq^n)}z^k.
\end{align*}
If $|b/a|<|z|<1$, utilizing the following identity
\begin{align}{\label{Trans}}
(a;q)_{-n}=\frac{(-q/a)^nq^{n\choose 2}}{(q/a;q)_n}, \ \ \ \
n=0,1,2,\ldots,
\end{align}
we can verify that $G(n,k)$ satisfies the condition (C1). It follows
that
\begin{align} \label{Ra1}
\sum_{k=-\infty}^\infty F(n,k)= {\rm constant}, \ \ \ \ \
n=0,1,2,\ldots.
\end{align}
Setting $n=0$,  $b=q$ and utilizing  the $q$-binomial theorem
\eqref{q-binomial theorem}, we see that the constant equals 1.
Setting $n=0$, we obtain the identity \eqref{11}. However, we note
  that the conditions for the companion identity do not hold in this case.

\noindent{\bf Example 3.4.} Bailey's $_6\psi_6$ summation formula
\cite[P. 357]{GR2004}:
\begin{align}\label{66}
\sum_{k=-\infty}^\infty&\frac{(1-aq^{2k})(b,c,d,e;q)_k}
{(1-a)(aq/b,aq/c,aq/d,aq/e;q)_k}
\left(\frac{a^2q}{bcde}\right)^k \nonumber\\[8pt]
&\qquad\qquad
=\frac{(aq,aq/bc,aq/bd,aq/be,aq/cd,aq/ce,aq/de,q,q/a;q)_\infty}
{(aq/b,aq/c,aq/d,aq/e,q/b,q/c,q/d,q/e,a^2q/bcde;q)_\infty}.
\end{align}

We obtain the following $q$-WZ pair:
\begin{align*}
F(n,k)=&\frac{(1-aq^{n+2k})(d,e,bq^n,cq^n;q)_k (aq/b,aq/c;q)_\infty}
{(1-aq^n)(aq/b,aq/c,aq^{n+1}/d,aq^{n+1}/e;q)_k
(aq/bd,aq/be;q)_\infty}\\[8pt]
& \qquad\times \frac{(q/d,q/e,a^2q/bcde,aq^{n+1}/d,
aq^{n+1}/e,q^{1-n}/b,q^{1-n}/c;q )_\infty}
{(q,aq/cd,aq/ce,aq^{n+1},aq^{n+1}/de,
q^{1-n}/a,aq^{1-n}/bc;q)_\infty} \left(\frac{a^2q}
{bcde}\right)^k,\\[8pt]
G(n,k)=&\frac{(d,e,bq^n,cq^n;q)_k(a/b,a/c,1/e,a^2/bcde,
1/d;q)_\infty}{(a/b,a/c,aq^n/d,aq^n/e;q)_k
(q,a/bd,a/be,a/cd,a/ce;q)_\infty }\\[8pt]
&\qquad\times \frac{(aq^n/d,aq^n/e,q^{-n}/b ,q^{-n}/c;q)_\infty
(-1+aq^n)} {
(aq^n,aq^n/de,aq^{-n}/bc,q^{-n}/a;q)_\infty(1-bq^n) (1-cq^n)}\\[8pt]
&\qquad\times\frac{(a-bd)(a-be)(a-cd)(a-ce)(aq^n-de)q^n}
{(aq^{n+k}-d)(aq^{n+k}-e)(a-ad)(1-e)(a^2-bcde)} \left(\frac{a^2q}
{bcde}\right)^k.
\end{align*}
Since $|a^2q/bcde|<1$, from  the identity \eqref{Trans} it follows
that $G(n,k)$ satisfies the condition (C1). By \eqref{Con1}, we find
\begin{align} \label{Ra2}
\sum_{k=-\infty}^\infty F(n,k)= {\rm constant}, \ \ \ \ \
n=0,1,2,\ldots.
\end{align}
In order to determine the constant, we set $n=0$ and $b=a$. From the
 $_6\phi_5$  summation formula  \eqref{65}, we see that
 the constant equals
\begin{eqnarray*}
\sum_{k=-\infty}^\infty F(0,k) & =
&\sum_{k=0}^\infty\frac{(1-aq^{2k})(a,c,d,e;q)_k}
{(1-a)(aq/c,aq/d,aq/e;q)_k
}\\[8pt]
& & \quad \times \frac{(aq,aq/cd,aq/ce,aq/de;q)_\infty}
{(aq/c,aq/d,aq/e,aq/cde;q)_\infty}\left(\frac{aq}{cde}\right)^k=1,
\end{eqnarray*}
 which can be restated as \eqref{66}. Nevertheless, we note that
the conditions for the companion identity do not hold in this case.

\vspace{1cm}
 \noindent{\bf Acknowledgments.} The authors would like to thank George E. Andrews, Q.H.
  Hou and Doron Zeilberger for valuable comments.  This work was supported by  the 973
Project, the PCSIRT Project of the Ministry of Education, the
Ministry of Science and Technology, and the National Science
Foundation of China.

config: arxiv


TITLE: Riemann $\zeta(3)$ convergence with Cauchy
QUESTION [3 upvotes]: I'm an undergraduate freshman math student, and we were asked to prove that the sequence $a{_n} =\sum_{k=1}^{n} \frac{1}{k^3}$ converges (obviously, we weren't asked to calculate its limit.) Our teacher hinted to prove that it's a Cauchy sequence. We don't know much, only Cauchy, several sentences about sequences and limits and monotonic sequences and such (basic first few months of undergraduate freshman). I'm stuck. any hints / ideas?
Here's my attempt:
Let $\varepsilon > 0$. We need to find N, such that for all $m > n > N$, $a_{m}-a_{n} < \varepsilon$. $a_{m}-a_{n} = \sum_{k = n+1}^{m} \frac{1}{k^3}$.
$\sum_{k = n+1}^{m} \frac{1}{k^3} < \frac{m-n}{(n+1)^3}$.
But this leads nowhere.
Note: We don't have to prove it by Cauchy, any solution (from the little we have learnt) will do.

REPLY [0 votes]: The integral test shows us that it converges and that
$$\sum_{n=1}^\infty\frac1{n^3}<1+\int_1^\infty\frac1{x^3}\ dx=\frac32$$
And we also know that it is greater than the first term, since the partial sums are monotonically increasing, so
$$1<\sum_{n=1}^\infty\frac1{n^3}<\frac32$$


TITLE: On the number of antichains of a poset
QUESTION [1 upvotes]: I am not an expert in combinatorics, but I need to count (or to approximate) the number of antichains of a poset.
The idea relies on approximating this number by embedding my poset into another one, that I call "rectangular" but actually I don't know if there is already a standard definition.
Let's say that my poset has depth $d$ (t.i. the maximal lenght of a chain) and spread $a$ (t.i., the max of cardinalities among all maximal antichains). Then I embedd this poset into another one, that has $d$ independent chains of lenght $a$, not considering the $0$ and $1$ elements. I claim that the second one contains more antichains than the first one, and that this number is $d \cdot 2^a$.
Is this idea correct? May I have a better upper bound of the number of antichains in a general poset?
Thank you all in advance for any answer!

REPLY [3 votes]: You appear to switch the usage of $d$ and $a$ when you identify your poset with a subset of the other; in what follows, I am using $d$ and $a$ as you defined them, i.e., $d$ is the maximum chain length and $a$ is the maximum antichain size.
Elaborating on Dave Pritchard's point 1: by Dilworth's Theorem, there exist $a$ disjoint chains whose union (as sets) is the (ground set of the) poset.  Suppose these chains have lengths $c_1, c_2, \ldots, c_a$.  Every antichain contains at most one vertex from each of these chains, so the number of antichains is at most $(c_1 + 1)(c_2 + 1)\cdots(c_a + 1) \leq (n/a + 1)^a$, where $n$ is the number of elements of the poset.  We can also use $n \leq da$ to get your upper bound $(d + 1)^a$, which is sharp for a disjoint union of $a$ chains of length $d$.


TITLE: Three points on a circle
QUESTION [7 upvotes]: If three points are randomly chosen on the boundary of a circle, what is the probability that there exists a diameter of the circle such that all three points lie on the same side of it?
I have a solution, but I'm very curious to see how others do it (and whether we get the same answer)!
Thanks.

REPLY [3 votes]: Here's my answer:
Suppose points $A$ and $B$ are placed first. The measure $\theta$ of the minor arc that has $A$ and $B$ as endpoints is uniformly distributed over the interval $(0,\pi)$. Given a value for $\theta$, though, we can know that the probability that point $C$ will be placed so that all three points are on the same side of a diameter is equal to $1-\frac{\theta}{2\pi}$. Averaging over the interval $(0,\pi)$ gives
$$
\frac{1}{\pi} \int_0^\pi (1-\frac{\theta}{2\pi})d\theta=\frac{1}{\pi}\left[\theta-\frac{1}{4\pi}\theta^2\right]_0^\pi=\frac{3}{4}
$$


TITLE: To finde the center of $D_4$
QUESTION [2 upvotes]: is there a nice/smart way to find the center of $D_4$? rather then going through every element?

REPLY [1 votes]: By its represantation, $$D_4=<a,b> | bab^{-1}=a^{-1} , a^4=e=b^2$$
Thus, $ba^2b^{-1}=a^{-2}=a^2$ so $a^2\in Z(D_4)$
From that we see that $Z(G)$ has at least $2$ elements but can not be $4$ as $G/Z(G)$ become cylic and can not be $8$ as it is nonabelian.
$$Z(D_4)=\{e,a^2\}$$


text:
\section{Technical Preliminaries}

In this section, we summarize the preliminary material on
controlled and weakly controlled vector fields that we will
need, and define precisely the ingredients of Theorem \ref{main}.
The reader is referred to [2, Chapter 2] and [14, Chapter 2.5]
for a detailed treatment of controlled vector fields.  However,
our definitions differ slightly from these sources.  The main
distinction is that our notion of a quasi-distance function is
more flexible than the corresponding notions in \cite{G-al} and
\cite{dPW}.  This is necessary to make Theorem \ref{isotopy} true
``in the controlled category.''

\subsection{Controlled Vector Fields}

Let $X$ be a real $C^\infty$ manifold with a Whitney
stratification $\cS$.

\begin{defn}\label{proj}
Let $S \in \cS$ be a stratum, and let $U_S$ be an open
neighborhood of $S$.  A tubular projection $\Pi_S : U_S \to S$
is a smooth submersion restricting to the identity on $S$.
\end{defn}

\begin{defn}\label{qdprelim}
Let $M$ be a smooth manifold, and let $E \to M$ be a vector
bundle with zero section $Z$.  A quasi-norm on $E$ is a smooth
function $\rho : E \setminus Z \to \R_+$ such that
$\rho(\lambda \, e) = \lambda \, \rho(e)$ for every
$e \in E \setminus Z$ and $\lambda \in \R_+$.
\end{defn}

\begin{defn}\label{qd}
Let $S \in \cS$ be a stratum, and let $U_S$ be an open
neighborhood of $S$.  \\ A quasi-distance function
$\rho : U_S \setminus S \to \R_+$ is a smooth function
satisfying the following condition.  There exist a vector
bundle $p : E \to S$ with zero section $Z$, an open neighborhood
$U' \subset E$ of $Z$, and a diffeomorphism $\phi : U' \to U$,
such that $\phi|^{}_Z = p|^{}_Z$ and $\rho \circ \phi$ is the
restriction to $U' \setminus Z$ of a quasi-norm on $E$.
\end{defn}

\begin{defn}\label{cdat}
Control data on $(X, \cS)$ is a collection $\{ U_S, \Pi_S, \rho_S
\}_{S \in \cS}$, where $U_S \supset S$ is an open neighborhood,
$\Pi_S : U_S \to S$ is a tubular projection, and $\rho_S : U_S
\setminus S \to \R_+$ is a quasi-distance function, subject to the
compatibility conditions $\Pi_S \circ \Pi_T = \Pi_S$ and $\rho_S
\circ \Pi_T = \rho_S$, both being equalities between maps of $U_S
\cap U_T$.
\end{defn}

\begin{defn}\label{compat}
Let $\{ U_S, \Pi_S, \rho_S \}$ be control data on $(X, \cS)$,
let $\cU \subset X$ be an open subset, let $A$ be a set,
and let $f : \cU \to A$ be a map of sets.  We say that
$\{ U_S, \Pi_S, \rho_S \}$ is $f$-compatible on $\cU$ if, for
every $S \in \cS$, there is a neighborhood $U'_S$ of $S \cap \cU$
such that $f \circ \Pi_S = f$ on $U'$.
\end{defn}

\begin{defn}\label{chom}
Let $(X, \cS)$, $(\hat X, \hat \cS)$ be two Whitney stratified
manifolds, with control data $\{ U_S, \Pi_S, \rho_S \}$ and
$\{ U_{\hat S}, \Pi_{\hat S}, \rho_{\hat S} \}$.  A controlled
homeomorphism $\phi : X \to \hat X$ is a homeomorphism which takes
strata diffeomorphicly onto strata, establishing a bijection
$S \mapsto \hat S$, and satisfies the following condition.  For
every $S \in S$, there is a neighborhood $U'_S \subset U_S$ of $S$
such that $\phi \circ \Pi_S = \Pi_{\hat S} \circ \phi$ and
$\phi \circ \rho_S = \rho_{\hat S} \circ \phi$ on $U'_S$.
\end{defn}

\begin{defn}\label{cvf}
Suppose we are given control data $\{ U_S, \Pi_S, \rho_S\}$ on
$(X, \cS)$.  A controlled vector field $V$ on $X$ compatible with
$\{ U_S, \Pi_S, \rho_S\}$ is a collection $\{ V_S \}_{S \in \cS}$
of smooth vector fields on the individual strata, satisfying the
following condition.  For every $S \in \cS$, there exists a
neighborhood $U'_S \subset U_S$ of $S$ such that:

(a) $(\Pi_S)_* V_x = V_{\Pi_S (x)}$ for every $x \in U'_S$;

(b) $V_x \, \rho_S = 0$ for every $x \in U'_S \setminus S$.
\end{defn}

Integrating controlled vector fields is a basic technique for
constructing controlled homeomorphisms, going back to the work
of Thom \cite{Th} and Mather \cite{Ma}.  The following variant of
[15, Lemma 4.11] and [13, Theorem 1.1] will serve as our basic
tool for constructing controlled vector fields.

\begin{lemma}\label{ecvf}
Let $\{ U_S, \Pi_S, \rho_S\}$ be control data on $(X, \cS)$.  Let
$S$ be a stratum, let $U \subset S$ be open in $S$, and let $V$ be
a smooth vector field on $U$.  Then there exist an open $\cU \subset
X$ with $\cU \cap S = U$ and a controlled vector field $\tilde V$ on
$\cU$, compatible with $\{ U_S, \Pi_S, \rho_S\}$, such that
$\tilde V|^{}_U = V$.  Furthermore, the vector field $\tilde V$ can
be chosen to be continuous as a section of $T\cU$. \hfill$\Box$
\end{lemma}

\subsection{Weakly Controlled Vector Fields}

Controlled vector fields are not suitable for discussing the
ascending and descending sets.  Indeed, the trajectory of a
controlled vector field can not approach a point on a smaller
stratum as time tends to infinity.  We will therefore need the
following definition.

\begin{defn}\label{tvf}
Suppose we are given control data $\{ U_S, \Pi_S, \rho_S\}$ on
$(X, \cS)$.  A weakly controlled vector field $V$ on $X$ is a
collection $\{ V_S \}_{S \in \cS}$ of smooth vector fields on the
individual strata, satisfying the following condition.  For every
$S \in \cS$, there exists a neighborhood $U'_S \subset U_S$ of $S$
and a number $k > 0$ such that:

(a) $(\Pi_S)_* V_x = V_{\Pi_S (x)}$ for every $x \in U'_S$;

(b) $|V_x \, \rho_S| < k \cdot \rho_S(x)$ for every
$x \in U'_S \setminus S$.
\end{defn}

\begin{prop}\label{tvf2thom}
Weakly controlled vector fields integrate to stratum preserving
homeomorphisms.  More precisely, let $V$ be a weakly controlled
vector field on $X$.  Then for every $x \in X$, there is a
neighborhood $U_x$ of $x$ and number $t_0 > 0$ such that, for
every $t \in [-t_0, t_0]$, the time-$t$ flow of $V$ is defined
on $U_x$ and gives a stratum preserving homeomorphism
$\psi_{V,t} : U_x \to \psi_{V,t} (U_x)$, which is smooth on each
stratum.
\end{prop}

\noindent
{\bf Proof:}  The main thing to check is that a trajectory of
$V$ lying in a stratum $T$ can not approach a point on a smaller
stratum $S \subset \bar T$ in finite time.  This follows from
condition (b) in Definition \ref{tvf}.  See [14, Proposition 2.5.1]
for more details.
\hfill$\Box$

\vspace{.1in}

The next two definitions clarify the statement of Theorem \ref{main}.

\begin{defn}\label{gradlike}
Let $f : X \to \R$ be a $C^\infty$ function.  An $\cS$-preserving
$\nabla f$-like vector field $V$ on an open subset $\cU \subset X$
is a weakly controlled vector field (compatible with some control
data on $(X, \cS)$) satisfying:

(a)  $V_p = 0$ for all $p \in \Sigma_f \cap \cU$;

(b)  $V_x \, g > 0$ for all $x \in \cU \setminus \Sigma_f$.

\noindent
A $\nabla f$-like vector field on an arbitrary subset
$A \subset X$ (e.g., on a closed ball) is the restriction to $A$
of a $\nabla f$-like vector field on some open $\cU \supset A$.
\end{defn}

\begin{defn}\label{andmfds}
Let $f : X \to \R$ be a $C^\infty$ function, let $V$ be a
$\nabla f$-like vector field on some $A \subset X$, and let
$p \in \Sigma_f \cap A$.  We define $M^-_V (p)$ to be the set
of all $x \in A$ such that the trajectory $\psi_{V,t} (x)$ is
contained in $A$ for all $t \geq 0$ and we have
$\displaystyle \lim_{t \to \infty} \psi_{V,t} (x) = p$.  The
ascending set $M^+_V (p)$ is defined similarly.
\end{defn}

\subsection{The Flow Topology}

In this section, we discuss the notion of the flow topology on
the set of weakly controlled vector fields.  It will give us a
degree of flexibility, making some of our constructions less
tied to the choice of control data.

Let $\cV(X, \cS)$ be the set of all weakly controlled vector
fields on $(X,\cS)$, compatible with all possible control data,
and let $\cV(X)$ be the union of the $\cV(X, \cS)$ over all
Whitney stratifications $\cS$ of $X$.  Fix a Riemannian metric
$g$ on $X$.  Let $V \in \cV(X)$, let $K \subset X$ be a compact
subset, let $t > 0$ be a number such that $\psi_{V,s}(x)$
is defined for all $x \in K$ and all $s \in [-t, t]$, and let
$\epsilon > 0$ be any positive number.  Define
$$\cU(V, K, t, \epsilon) = \{ V' \in \cV \; | \;
\forall x \in K, s \in [-t, t] : \mbox{dist}_g
(\psi_{V,s}(x), \, \psi_{V',s}(x)) < \epsilon \},$$
where we set $\mbox{dist}_g (\psi_{V,s}(x), \, \psi_{V',s}(x)) =
+ \infty$ if $\psi_{V',s}(x)$ is undefined.

\begin{defn}\label{flow-t}
The flow topology on $\cV(X)$ is the weakest topology in which
all the $\cU(V, K, t, \epsilon)$ are open sets.
\end{defn}

It is easy to see that the flow topology is independent of the
metric $g$.

\begin{prop}\label{approx}
(i)  Let $\cS$ and $\hat \cS$ be two Whitney stratifications of $X$
such that $\hat \cS$ is a refinement of $\cS$.  Fix a vector field
$\hat V \in \cV(X, \hat \cS)$ and control data $\{ U_S, \Pi_S,
\rho_S \}$ on $(X, \cS)$.  Then there exists a sequence $\{ V_i
\in \cV(X, \cS) \}_{i \in \N}$ of vector fields compatible with
$\{ U_S, \Pi_S, \rho_S \}$ such that $V_i \to \hat V$ in the flow
topology.

(ii) In the situation of part (i), assume that $\hat \cS$ has a unique
point stratum $\{ p \}$, and that $f : X \to \R$ is a smooth function
whose only stratified critical point with respect to $\hat \cS$ is $p$.
Assume also that $V$ is $\nabla f$-like.  Then all the $V_i$ can be
chosen to be $\nabla f$-like too.
\end{prop}

\noindent
{\bf Proof:}  This is an exercise using Lemma \ref{ecvf} and
partitions of unity.  For part (i), the first step is to show that
there is a sequence $\{ \hat V_i \in \cV(X, \hat \cS)\}$ of continuous
vector fields compatible with the same control data as $V$, such that
$\hat V_i \to \hat V$ in the flow topology.  The second step is to
show that each $\hat V_i$ can be approximated in the $C^0$ topology
by continuous vector fields from $\cV(X, \cS)$ compatible with
$\{ U_S, \Pi_S, \rho_S \}$.  Then it remains to note that the $C^0$
topology on continuous weakly controlled vector fields is stronger
than the flow topology.  Part (ii) is similar.
\hfill$\Box$


TITLE: Product of a discrete and absolutely continuous random variable, part deux
QUESTION [0 upvotes]: I have the independent random variables $U\sim N(0,1)$, $V\sim N(0,1)$, $W\sim b(1,1/2)$
I define $X=WU + (1-W)(V+1)$. I need to determine that $X$ is absolutely continuous, and determine a density function of $X$. I have the same problem as with my earlier question: I really don't know how the joint distribution is defined for a discrete and an absolutely continuous variable.
Bonus question: I also need to determine if $UW$ is a discrete  variable. I'm thinking yes, since ${UW=0}\supseteq{W=0}$, and $P({W=0})=1/2$.

REPLY [3 votes]: $$f_X(x)=\tfrac12f_U(x)+\tfrac12f_{V+1}(x)=\tfrac12f_U(x)+\tfrac12f_{V}(x-1)$$
Note that $UW$ is neither purely discrete nor purely continuous since its distribution has an atom of mass $\frac12$ at $0$ and the rest of the distribution is a densitable measure of mass $\frac12$.


TITLE: 3 dimensional gaussian integral with vector length as a factor
QUESTION [3 upvotes]: I'm currently trying to calculate a weird gaussian integral:
$$ \int\limits_{\mathbb{R}^3} \mathrm{d}^3\vec x \ |\vec x| \exp( -a \cdot x^2) $$
where $a > 0$ of course.
I can easily evaluate it if there is only $x_1, x_2$ or $x_3$ or products of those in front of the exponential function. But I have now idea how to seperate $|\vec x| = \sqrt{ x_1 + x_2 + x_3}$.
any ideas?

REPLY [1 votes]: Since your integral only depends on $|\vec x|$, you can compute it in spherical coordinates $(r=|\vec x|,\theta,\phi)$ because the integration over $\theta$ and $\phi$ is simply a constant.
You get :
$$\int\limits_{\mathbb{R}^3} \mathrm{d}^3\vec x \ |\vec x| \exp( -a \cdot x^2)=\int_{r=0}^{+\infty}\int_{\theta=0}^\pi\int_{\phi=0}^{2\pi}r\exp( -a \cdot r^2)\cdot r^2\sin\theta \mathrm{d} \phi\mathrm{d} \theta\mathrm{d}r$$
The first two integrals gives $4\pi$ and :
$$\int\limits_{\mathbb{R}^3} \mathrm{d}^3\vec x \ |\vec x| \exp( -a \cdot x^2)=4\pi\int_{0}^{+\infty}r^3\exp( -a \cdot r^2)\mathrm{d}r$$
which is a fairly simple gaussian integral ($\frac{2\pi}{a^2}$ says Maple)


TITLE: Relation between function of B.V and A.C. function?
QUESTION [3 upvotes]: Define the $B.V.$ norm $$\|f\|_{B.V}=V(f)+|f(a)|$$ on the closed interval $[a,b]$.
  How to prove the set of absolutely continuous functions in $[a,b]$ is a closed subspace of $BV[a,b]$?

I have shown every absolutely continuous function is of bounded variation. I am stuck in proving it actually forms a closed subspace.
Could someone kindly help? Thanks very much!

REPLY [2 votes]: Assume that $f_n\to f$ in the BV norm, and $f_n$ are absolutely continuous functions, while $f$ is just a BV function. We wish to show that $f$ is absolutely continuous, i.e. $\forall \varepsilon>0$, $\exists \delta>0$ such that for all $a\leq a_1\leq b_1 \leq a_2\leq b_2\leq \dots\leq a_k\leq b_k\leq b$, $k\in \mathbb  N$ with $\sum (b_i-a_i)<\delta$ we have $\sum |f(b_i)-f(a_i)|<\varepsilon$.
By assumption $V(f_n-f)\to 0$, and $V(f_n-f_m)\to 0$ as $m,n\to \infty$, since $f_n$ is a Cauchy sequence. In particular, we can choose large $N$ such that $V(f_n-f_m)<\varepsilon$ and $V(f_n-f)<\varepsilon$ for all $n,m\geq N$. Since $f_N$ is absolutely continuous, we can choose $\delta$ small enough such that $\sum |f_N(b_i)-f_N(a_i)|<\varepsilon$. Then for $n\geq N$ we have
$$\sum |f_n(b_i)-f_n(a_i)| \leq \sum |f_N(b_i)-f_N(a_i)|+ \sum |f_n(b_i)-f_n(a_i) -f_N(b_i)+f_N(a_i)|$$
$$\leq \varepsilon +V(f_n-f_N)\leq 2\varepsilon$$
Therefore, similarly we obtain
$$\sum |f(b_i)-f(a_i)| \leq \sum |f_n(b_i)-f_n(a_i)| + V(f_n-f)$$
$$\leq 2\varepsilon+ \varepsilon= 3\varepsilon $$
and this completes the proof that $f$ is absolutely continuous.


TITLE: How to solve this lim?
QUESTION [0 upvotes]: $\lim _{n \rightarrow+\infty}(n-3)(3 \ln (n-1)-\ln (n+2)-2 \ln (n+1))$
I tried to get all in one ln but then I get this:
$\ln \left(\lim _{n\rightarrow \infty}\left(\frac{(n-1)^{3}}{(n+2)(n+1)^{2}}\right)^{(n-3)}\right)$
$\ln \left(\lim _{n \rightarrow \infty}\left(\frac{\left(n^{3}-3 n^{2}+3 n-1\right)^{(n-3)}}{\left(n^{3}+4 n^{2}+5 n+2\right)^{(n-3)}}\right)\right)$
Have no idea how to get out of this..

REPLY [1 votes]: The expression under limit can be written as $$(n-3)\log\frac{(n-1)^3}{(n+2)(n+1)^2}=(n-3)\log\frac{n^3-3n^2+\cdots}{n^3+4n^2+\cdots} $$ and this gets expressed further as $$ (n-3)\log\left(1-\frac{7n^2+\dots}{n^3+4n^2+\dots}\right)$$ and using the limit $\lim\limits _{x\to 0} \dfrac {\log(1+x)}{x}=1$ the limit of the expression in question is same as that of expression $$-(n-3)\cdot\frac{7n^2+\dots}{n^3+4n^2+\dots}$$ which is $-7$.


TITLE: Doubt regarding properties of open sets
QUESTION [0 upvotes]: I am studying Metric Spaces from the book by Satish Shirali and H. L. Vasudeva. The chapter on Topology of Metric Spaces talks about a certain property about the union of open sets. It states:

The union of any finite, countable or uncountable family of open sets is open but the intersection of an infinite number of open sets need not be open.

What exactly is meant by the phrase "finite, countable or uncountable" ?

REPLY [1 votes]: That's just a long way of saying that the union of any family $\mathcal F$ of open sets is still an open set. Note that $\mathcal F$ may be finite, countable or uncountable.


TITLE: Computation of Digits in Tetration
QUESTION [1 upvotes]: According to Wikipedia, $^44$ has $8.1 \cdot 10^{153}$ digits. How can I calculate the number of digits for an arbitrarily large tetration, such as $^{11}11$?
Thank you!

REPLY [1 votes]: Do you know about logarithms? The number of digits is one more than the floor of the base-10 logarithm of the number. For large numbers, the "one more" and "floor" don't affect things much (relatively). By applying the power law/rule/identity, we get:
$$\log_{10}\left(4^{4^{4^4}}\right)=\left(4^{4^4}\right)\log_{10}4\approx0.602*\left(4^{4^4}\right)\approx8.1*10^{153}$$
$$\log_{10}\left({}^{11}11\right)=\left({}^{10}11\right)\log_{10}11\approx1.041*\left({}^{10}11\right)\gtrapprox\left({}^{10}11\right)\gtrapprox\left({}^{10}10\right)$$
But these numbers are so much larger than ${}^34$ that we can't write them using regular exponent notation. ${}^311$ already has something like $297121486766$ digits (rather than $154$), and things just get worse from there.