Show that $\sum_ {k=1}^{\infty}\dfrac{\zeta(2k)-\zeta(3k)}{k}=\ln\left(\frac{2\cosh\left(\sqrt{3}\pi/2\right)}{3\pi}\right)$

Question

Question

$$\zeta(k)=1+\dfrac{1}{2^k}+\dfrac{1}{3^k}+\cdots+\dfrac{1}{n^k}+\cdots$$ Prove that : $$\sum_ {k=1}^{\infty}\dfrac{\zeta(2k)-\zeta(3k)}{k}=\ln\left(\frac{2\cosh\left(\sqrt{3}\pi/2\right)}{3\pi}\right).$$

We have $$\begin{align} \sum\limits_{k = 1}^\infty {\frac{{\zeta \left( {2k} \right) - \zeta \left( {3k} \right)}}{k}} &= \sum\limits_{k = 1}^\infty {\frac{1}{k}\left( {\sum\limits_{n = 2}^\infty {\left( {\frac{1}{{{n^{2k}}}} - \frac{1}{{{n^{3k}}}}} \right)} } \right)}\\&= \sum\limits_{n = 2}^\infty {\sum\limits_{k = 1}^\infty {\left( {\frac{1}{{k{n^{2k}}}} - \frac{1}{{k{n^{3k}}}}} \right)} }\\&= \sum\limits_{n = 2}^\infty {\left( { - \ln \left( {1 - \frac{1}{{{n^2}}}} \right) + \ln \left( {1 - \frac{1}{{{n^3}}}} \right)} \right)}\\& = - \ln \frac{3}{2} + \sum\limits_{n = 1}^\infty {\ln \left( {\frac{{{n^2} + n + 1}}{{{n^2} + n}}} \right)}\\&= - \ln \frac{3}{2} + \ln \left( {\prod\limits_{n = 1}^\infty {\frac{{\left( {n - {z_1}} \right)\left( {n - {z_2}} \right)}}{{n\left( {n + 1} \right)}}} } \right) \end{align}$$

Answer 1

Let $j:=\frac{-1+i\sqrt3}2.$ You proved that $$\sum_{k = 1}^\infty\frac{\zeta(2k)- \zeta(3k)}k=\ln\prod_{n =2}^\infty\frac{(n-j)(n+1+j)}{n(n + 1)}.$$ Now, $$\begin{align}\prod_{n =2}^\infty\frac{(n-j)(n+1+j)}{n(n + 1)}&=\frac1{(1-j)(1+j)(1+j/2)}\prod_{n =1}^\infty\left(1-\frac{j^2}{n^2}\right)\\ &=\frac{-2j}3\frac{\sin(\pi j)}{\pi j}\\ &=\frac{2\cosh\left(\pi\sqrt3/2\right)}{3\pi}. \end{align}$$ See also https://oeis.org/A073017.

Answer 2

Hoping that you do not mind, I shall use $(a,b)$ in place of $(z_1,z_2)$.

$$\prod\limits_{n = 1}^p \frac{(n - a)(n-b)} {n(n+1)}=\frac{\Gamma (p+1-a)\,\, \Gamma (p+1-b)}{\Gamma (1-a)\, \Gamma (1-b)\, \Gamma (p+1)\, \Gamma (p+2)}$$

Using Stirling approximation $$\log\Bigg(\frac{\Gamma (p+1-a)\,\, \Gamma (p+1-b)}{ \Gamma (p+1)\, \Gamma (p+2)}\Bigg)=$$ $$-(a+b+1) \log (p)+\frac{a^2-a+b^2-b-2}{2 p}+O\left(\frac{1}{p^2}\right)$$ But, $a+b=-1$; so what remains is $$\frac{1}{\Gamma (1-a)\, \Gamma (1-b)}=\frac{1}{\Gamma \left(\frac{3-i \sqrt{3}}{2}\right) \Gamma \left(\frac{3+i \sqrt{3}}{2}\right) }=\pi\, \text{sech}\left(\frac{\sqrt{3} \pi }{2}\right)$$

Answer 3

\begin{align} z_1 &= \frac{-1 + i\sqrt{3}}{2}, \quad z_2 = \frac{-1 - i\sqrt{3}}{2} \\ \prod_{n=1}^\infty \frac{(n - z_1)(n - z_2)}{n(n+1)} &= \lim_{N \to \infty} \prod_{n=1}^N \frac{(n - z_1)(n - z_2)}{n(n+1)} \\ &= \lim_{N \to \infty} \frac{\prod_{n=1}^N (n - z_1)\prod_{n=1}^N (n - z_2)}{N!(N+1)!} \\ &= \frac{1}{z_1 z_2} \lim_{N \to \infty} \frac{\prod_{n=0}^N (n - z_1)\prod_{n=0}^N (n - z_2)}{N!(N+1)!} \end{align} \begin{align} &= \frac{1}{z_1 z_2} \lim_{N \to \infty} \frac{N^{-({z_1} + {z_2})}(1/N^{ - {z_1}}N!\prod_{n=0}^N (n - z_1))(1/N^{ - {z_2}}N!\prod_{n=0}^N (n - z_2))}{N + 1} \\ &= \frac{1}{\Gamma(-z_1) \cdot \Gamma(-z_2)} \\ &= \frac{1}{\Gamma(\frac{1 + i\sqrt{3}}{2}) \cdot \Gamma(\frac{1 - i\sqrt{3}}{2})} \\ &= \frac{\sin(\pi \frac{1 + i\sqrt{3}}{2})}{\pi} \\ &= \frac{e^{i\pi \frac{1 + i\sqrt{3}}{2}} - e^{-i\pi \frac{1 + i\sqrt{3}}{2}}}{2i\pi} \\ &= \frac{e^{i\pi /2}e^{-\pi \sqrt{3}/2} - e^{-i\pi /2}e^{\pi \sqrt{3}/2}}{2i\pi} \\ &= \frac{i(e^{-\pi \sqrt{3}/2} + e^{\pi \sqrt{3}/2})}{2i\pi} \\ &= \frac{\cosh(\pi \sqrt{3}/2)}{\pi} \end{align} \begin{align} \boxed{\prod_{n=1}^\infty \frac{(n - z_1)(n - z_2)}{n(n+1)} = \frac{\cosh(\pi \sqrt{3}/2)}{\pi}} \\ \text{Finally, } \sum_{k=1}^\infty \frac{\zeta(2k) - \zeta(3k)}{k} &= -\ln\frac{3}{2} + \ln\left(\prod_{n=1}^\infty \frac{(n - z_1)(n - z_2)}{n(n+1)}\right) \\ &= -\ln\frac{3}{2} + \ln\frac{\cosh(\pi \sqrt{3}/2)}{\pi} \\ &= \ln\left(\frac{2\cosh(\pi \sqrt{3}/2)}{3\pi}\right) \end{align}