Problem - Li Zhou (USA)
For a nonnegative integer m, let
\[A_m=\sum_{k=0}^\infty\left(\frac{1}{(6k+1)^{2m+1}}-\frac{1}{(6k+5)^{2m+1}}\right).\]Prove $A_0=\pi\sqrt{3}/6$ and, for $m\geq1$,
\[2A_m+\sum_{n=1}^m\frac{(-1)^n\pi^{2n}}{(2n)!}A_{m-n}=\frac{(-1)^m(4^m+1)\sqrt{3}}{2(2m)!}\left(\frac{\pi}{3}\right)^{2m+1}.\]Solution
Consider the function
\[f(z)=\frac{\sin\left(\frac{\pi}{6}(1-z)\right)}{\sin\left(\frac{\pi}{6}(1+z)\right)}\]with the goal of finding two different representations of its logarithmic derivative. We factor $f(z)$ as a Weierstrass product,
\[\begin{align*} f(z) &= \frac{\Gamma\left(\frac{1+z}{6}\right)\Gamma\left(1-\frac{1+z}{6}\right)}{\Gamma\left(\frac{1-z}{6}\right)\Gamma\left(1-\frac{1-z}{6}\right)},\;\text{reflection formula} \\ &= \frac{\Gamma\left(\frac{1+z}{6}\right)\Gamma\left(\frac{5-z}{6}\right)}{\Gamma\left(\frac{5+z}{6}\right)\Gamma\left(\frac{1-z}{6}\right)} \\ &= \frac{\left(\frac{1-z}{6}\right)}{\left(\frac{5-z}{6}\right)}\cdot\frac{\Gamma\left(\frac{1+z}{6}\right)\Gamma\left(\frac{11-z}{6}\right)}{\Gamma\left(\frac{5+z}{6}\right)\Gamma\left(\frac{7-z}{6}\right)} \\ &= \frac{1-z}{5-z}\cdot\frac{\frac{5+z}{6}\cdot\frac{7-z}{6}}{\frac{1+z}{6}\cdot\frac{11-z}{6}}\prod_{k=1}^\infty\frac{\frac{1}{1+\left(\frac{1+z}{6}\right)}\left(1+\frac{1}{k}\right)^\frac{1+z}{6}\frac{1}{1+\left(\frac{11-z}{6}\right)}\left(1+\frac{1}{k}\right)^\frac{11-z}{6}}{\frac{1}{1+\left(\frac{5+z}{6}\right)}\left(1+\frac{1}{k}\right)^\frac{5+z}{6}\frac{1}{1+\left(\frac{7-z}{6}\right)}\left(1+\frac{1}{k}\right)^\frac{7-z}{6}},\\ &\qquad\text{by Euler's product for }\Gamma(s)\\ &=\frac{1-z}{5-z}\cdot\frac{(5+z)(7-z)}{(1+z)(11-z)}\prod_{k=1}^\infty\frac{(6k+5+z)(6k+7-z)}{(6k+1+z)(6k+11-z)}\\ &= \frac{1-z}{5-z}\prod_{k=1}^\infty\frac{\left(1+\frac{z}{6k-1}\right)\left(1-\frac{z}{6k+1}\right)}{\left(1+\frac{z}{6k-5}\right)\left(1-\frac{z}{6k+5}\right)}\prod_{k=1}^\infty\frac{(6k-1)(6k+1)}{(6k-5)(6k+5)}.\tag{1}\end{align*}\]Now we wish to evaluate the rightmost product in the last line by induction, in particular, we show that
\[\prod_{k=1}^n\frac{(6k-1)(6k+1)}{(6k-5)(6k+5)}=\frac{30n+5}{6n+5}.\]Base case: It is easy to verify that $\prod_{k=1}^1\frac{(6k-1)(6k+1)}{(6k-5)(6k+5)}=\frac{35}{11}.$
Inductive step: Suppose it is true that $\prod_{k=1}^n\frac{(6k-1)(6k+1)}{(6k-5)(6k+5)}=\frac{30n+5}{6n+5}$, then
\[\begin{align*} \prod_{k=1}^{n+1}\frac{(6k-1)(6k+1)}{(6k-5)(6k+5)} &= \frac{(6(n+1)-1)(6(n+1)+1)}{(6(n+1)-5)(6(n+1)+5)}\prod_{k=1}^{n}\frac{(6k-1)(6k+1)}{(6k-5)(6k+5)} \\ &= \frac{(6(n+1)-1)(6(n+1)+1)}{(6(n+1)-5)(6(n+1)+5)}\cdot\frac{30n+5}{6n+5} \\ &= \frac{(6n+5)(6n+7)}{(6n+1)(6n+11)}\cdot\frac{5(6n+1)}{6n+5} \\ &= \frac{30(n+1)+5}{6(n+1)+5}.\end{align*}\]It follows that
\[\begin{align*} \prod_{k=1}^\infty\frac{(6k-1)(6k+1)}{(6k-5)(6k+5)} &= \lim_{n\to\infty}\frac{30n+5}{6n+5}\\ &= 5.\end{align*}\]So, continuing from $(1)$, we then have
\[\begin{align*} f(z) &= \frac{1-z}{1-\frac{z}{5}}\prod_{k=1}^\infty\frac{\left(1+\frac{z}{6k-1}\right)\left(1-\frac{z}{6k+1}\right)}{\left(1+\frac{z}{6k-5}\right)\left(1-\frac{z}{6k+5}\right)}\\ &= \prod_{k=-\infty}^\infty\left(1-\frac{z}{6k+1}\right)\left(1-\frac{z}{6k+5}\right)^{-1}.\tag{2}\end{align*}\]Since the bilateral product is conditionally convergent, we read $\prod_{k=-\infty}^\infty a_k$ as $\lim_{n\to\infty}\prod_{k=-n}^n a_k.$
We now find two expressions for the logarithmic derivative of $f(z)$, which will yield a generating function for $A_m$.
First,
\[\begin{align*} \frac{\mathrm d}{\mathrm dz}\log(f(z)) &=-\frac{\pi}{6}\cdot\frac{\cos\left(\frac{\pi}{6}(1-z)\right)\sin\left(\frac{\pi}{6}(1+z)\right)+\cos\left(\frac{\pi}{6}(1+z)\right)\sin\left(\frac{\pi}{6}(1-z)\right)}{\sin\left(\frac{\pi}{6}(1-z)\right)\sin\left(\frac{\pi}{6}(1+z)\right)}\\ &= -\frac{\pi}{6}\cdot\frac{\sqrt{3}/2}{\frac{1}{2}\cos\left(\frac{\pi z}{3}\right)-\frac{1}{4}}\\ &=\frac{\pi}{\sqrt{3}\left(1-2\cos\left(\frac{\pi z}{3}\right)\right)}.\end{align*}\]Second, let
\[\left|z\right|<1\]to take advantage of the absolute convergence and start from $(2)$,
\[\begin{align*} \frac{\mathrm d}{\mathrm dz}\log(f(z)) &= \frac{\mathrm d}{\mathrm dz}\sum_{n=-\infty}^\infty\left(\log\left(1-\frac{z}{6n+1}\right)-\log\left(1-\frac{z}{6n+5}\right)\right)\\ &= \sum_{n-\infty}^\infty\left(\frac{-\frac{1}{6n+1}}{1-\frac{z}{6n+1}}-\frac{-\frac{1}{6n+5}}{1-\frac{z}{6n+5}}\right),\\ &\qquad\text{swapping the sum and the derivative by absolute convergence}\\ &= -\frac{1}{z}\sum_{n=-\infty}^\infty\sum_{m=1}^\infty\left(\frac{z^m}{(6n+1)^m}-\frac{z^m}{(6n+5)^m}\right)\\ &= -\frac{1}{z}\sum_{m=1}^\infty\sum_{n=-\infty}^\infty\left(\frac{1}{(6n+1)^m}-\frac{1}{(6n+5)^m}\right)z^m,\\ &\qquad\text{swapping the order of summation is valid due to absolute convergence}\\ &=-\sum_{m=0}^\infty\sum_{n=-\infty}^\infty\left(\frac{1}{(6n+1)^{2m+1}}-\frac{1}{(6n+5)^{2m+1}}\right)z^{2m},\\ &\qquad(\log(f(z))'\text{ is even}\\ &=-2\sum_{m=0}^\infty\sum_{n=0}^\infty\left(\frac{1}{(6n+1)^{2m+1}}-\frac{1}{(6n+5)^{2m+1}}\right)z^{2m}\\ &=-2\sum_{m=0}^\infty A_m z^{2m}.\end{align*}\]Combining these two expressions for the logarithmic derivative of $f(z)$ yields
\[\begin{align*} \frac{\pi}{2\sqrt{3}\left(2\cos\left(\frac{\pi z}{3}\right)-1\right)}=\sum_{m=0}^\infty A_m z^{2m}.\tag{3}\end{align*}\]Setting $z=0$ in the above immediately gives $A_0=\frac{\pi\sqrt{3}}{6}.$
To verify
\[2A_m+\sum_{n=1}^m\frac{(-1)^n\pi^{2n}}{(2n)!}A_{m-n}=\frac{(-1)^m(4^m+1)\sqrt{3}}{2(2m)!}\left(\frac{\pi}{3}\right)^{2m+1}\]we sum over $m\geq1$ on each side of the alleged equality and see that they have the same generating functions. Starting on the left we have
\[\begin{align*} 2\sum_{m=1}^\infty A_m z^{2m} &= \frac{\pi}{\sqrt{3}\left(2\cos\left(\frac{\pi z}{3}\right)-1\right)}-\frac{\pi\sqrt{3}}{3}\tag{4}\end{align*}\]and
\[\begin{align*} \sum_{m=1}^\infty\sum_{n=1}^m\frac{(-1)^n\pi^{2n}}{(2n)!}A_{m-n}z^{2m} &= \sum_{m=0}^\infty\sum_{n=0}^m\frac{(-1)^{n+1}\pi^{2(n+1)}}{(2(n+1))!}A_{m-n}z^{2(m+1)}\\ &= \left(\sum_{m=0}^\infty A_m z^{2m}\right)\left(\sum_{n=0}^\infty\frac{(-1)^{n+1}\pi^{2(n+1)}}{(2(n+1))!}z^{2(n+1)}\right),\\ &\qquad\text{by the Cauchy product}\\ &=\frac{\pi\left(\cos(\pi z)-1\right)}{2\sqrt{3}\left(2\cos\left(\frac{\pi z}{3}\right)-1\right)}.\tag{5}\end{align*}\]Adding (4) and (5) gives $\frac{\pi\sqrt{3}}{6}\cos\left(\frac{\pi z}{3}\right)+\frac{\pi\sqrt{3}}{6}\cos\left(\frac{2\pi z}{3}\right)-\frac{\pi\sqrt{3}}{3}.$ Summing over the right side of the identity,
\[\begin{align*} \sum_{m=1}^\infty\frac{(-1)^m(4^m+1)\sqrt{3}}{2(2m)!}\left(\frac{\pi}{3}\right)^{2m+1}z^{2m} &= \frac{\pi\sqrt{3}}{6}\sum_{m=1}^\infty\left(\frac{(-1)^{m+1}}{(2m)!}\left(\frac{2\pi z}{3}\right)^{2m}+\frac{(-1)^{m+1}}{(2m)!}\left(\frac{\pi z}{3}\right)^{2m}\right)\\ &= \frac{\pi\sqrt{3}}{6}\cos\left(\frac{\pi z}{3}\right)+\frac{\pi\sqrt{3}}{6}\cos\left(\frac{2\pi z}{3}\right)-\frac{\pi\sqrt{3}}{3}.\end{align*}\]Since the generating functions for both sides match, the identity is true for $m\geq1$.