A nice relationship between $\zeta$, $\pi$ and $e$

I just happened to see this equation today, any suggestions on how to prove it? $$\sum_{n=1}^\infty{\frac{\zeta(2n)}{n(2n+1)4^n}}=\log{\frac{\pi}{e}}$$

• Welcome to MSE. Have you attempted to prove it? If so, please include this in your post so that we can see you've made some effort. – TheSimpliFire Dec 1 '18 at 18:56
• Mmm, delicious... – Frpzzd Dec 1 '18 at 19:03
• Can you say where you saw this? – Zacky Dec 1 '18 at 19:12
• Okay, found the source: facebook.com/photo.php?fbid=1609500362528719. The second sum looks cute too: $$\sum_{n=1}^\infty \left(\frac{\zeta(2n)}{n}-\frac{\zeta(2n+1)}{4^n}\right)\frac{1}{2n+1} =\gamma +\ln\left(\frac{\pi}{e}\right)$$ – Zacky Dec 1 '18 at 19:36
• @TheSimpliFire Thanks for the welcome. To be completely honest I have little experience in dealing with the zeta function and it's properties. I played around with the idea of using the Wallis product (as Frpzzd did in his answer) but ultimately gave up. I'll keep your suggestion in mind. – EvanHehehe Dec 1 '18 at 21:10

An approach that does not require complex analysis:

$$\sum_{n=1}^\infty \frac{\zeta(2n)}{n(2n+1)4^n} =\sum_{n=1}^\infty \frac{\zeta(2n)}{n4^n}-2\sum_{n=1}^\infty \frac{\zeta(2n)}{(2n+1)4^n}=S_1-2S_2$$

To calculate $$S_1$$: \begin{align} S_1=\sum_{n=1}^\infty \frac{\zeta(2n)}{n4^n} &=\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{n(4k^2)^n}\\ &=\sum_{k=1}^\infty \sum_{n=1}^\infty \frac{1}{n(4k^2)^n}\\ &=-\sum_{k=1}^\infty \ln\bigg(1-\frac{1}{4k^2}\bigg)\\ &=-\ln \prod_{k=1}^\infty \frac{(2k+1)(2k-1)}{(2k)^2}\\ &=\ln(\pi/2) \end{align} using the Wallis Product.

You may calculate $$S_2$$ similarly.

• Thanks for this answer. I had the Wallis product in my mind but wasn't able to transform the equation into anything coherent. – EvanHehehe Dec 1 '18 at 21:14
• @EvanHehehe No problem, I'm glad to help. – Frpzzd Dec 1 '18 at 21:15

The left-hand side is $$\sum_{n\ge 1}\frac{1}{\Gamma (2n)n(2n+1)4^n}\int_0^\infty\frac{x^{2n-1} dx}{e^x-1}=\int_0^\infty\frac{dx}{e^x-1}\sum_{n\ge 1}\frac{(x/2)^{2n-1}}{(2n+1)!}\\=\int_0^\infty\frac{dx}{e^x-1}\frac{\sinh\tfrac{x}{2}-\tfrac{x}{2}}{(\tfrac{x}{2})^2}.$$The rest is an exercise in complex analysis.

The identities \begin{align}\sum_{k=1}^{\infty}\frac{\zeta\left(2k\right)}{k}z^{2k}&=\ln\left(\frac{\pi z}{\sin\left(\pi z\right)}\right)&&(\lvert z \rvert < 1)\\ \sum_{k=1}^{\infty}\frac{\zeta\left(2k\right)}{(2k+1)2^{2k}}&=\frac{1}{2}-\frac {1}{2}\ln 2\end{align} (DLMF 25.8.8, 25.8.9) follow from the more basic identity \begin{align} \sum_{k=2}^{\infty}\frac{\zeta\left(k\right)}{k}z^{k}&=-\gamma z+\ln\Gamma\left (1-z\right)&&(\lvert z \rvert < 1) \end{align} (DLMF 25.8.7), which in turn follows from the product formula for the Gamma function (DLMF 5.8.2): $$\frac{1}{\Gamma\left(z\right)}=ze^{\gamma z}\prod_{k=1}^{\infty}\left(1+\frac{% z}{k}\right)\mathrm{e}^{-z/k}\text{.}$$

• you meant looking at $\sum_{k=2}^{\infty}\frac{\zeta\left(k\right)}{k+1}z^{k+1}= ?$ – reuns Dec 1 '18 at 20:07

Since \begin{align*} \zeta(2n) &= \frac{(-1)^{n+1} B_{2n} (2\pi)^{2n}}{2(2n)!}, \end{align*} for integers $$n > 0$$, the given sum $$S$$ is \begin{align*} S &= \sum_{n=1}^\infty \frac{\zeta(2n)}{n(2n+1)4^n} = \sum_{n=1}^\infty (-1)^{n+1} \frac{B_{2n}}{(2n)(2n+1)}\frac{\pi^{2n}}{(2n)!} \end{align*} But \begin{align*} f(z) &= \sum_{n=1}^\infty \frac{B_n}{n(n+1)} \frac{z^{n+1}}{n!} \end{align*} has $$\operatorname{Im} f(\pi i) = \pi S$$ and \begin{align*} f''(z) &= \frac{1}{z}\sum_{n=1}^\infty B_n \frac{z^n}{n!} = \frac{1}{z}\left(-1 + \frac{z}{e^z - 1}\right) = -\frac{1}{z} + \frac{1}{e^z - 1} \end{align*} by (a) definition of the Bernoulli numbers $$B_n$$. A bit of careful integration gives the required sum.