# Proof of $\sum_{n=1}^\infty \frac{1}{n^4 \binom{2n}{n}}=\frac{17\pi^4}{3240}$

Recently, I was able to prove that

$$\sum_{n=1}^\infty \frac{1}{n \binom{2n}{n}}= \frac{\pi}{3\sqrt{3}}$$

$$\sum_{n=1}^\infty \frac{1}{n^2 \binom{2n}{n}}= \frac{\pi^2}{18}$$

But does anybody know how to prove that $\displaystyle\sum_{n=1}^\infty \frac{1}{n^4 \binom{2n}{n}}= \frac{17}{36}\zeta(4)$?

I obtained the formula

$$\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^3 \binom{2n}{n}}= 4\int_0^x \frac{(\arcsin t)^2}{t}dt$$

I am trying to divide by $x$ and integrate both sides, but so far I have been unsuccessful.

Here is my solution:

Step 1. Reduction to an integral representation

Let $S$ denote the summation in question. Then by the successive application of integration by parts, we obtain

\begin{align*} S &= 8 \int_{0}^{1} \frac{1}{x} \int_{0}^{x/2} \frac{\arcsin^{2} t}{t} \, dt dx = -8\int_{0}^{1} \frac{\arcsin^{2} (x/2)}{x} \log x \, dx \\ &= 8 \int_{0}^{1} \frac{\arcsin (x/2)}{\sqrt{1 - (x/2)^{2}}} \log^{2} x \, \frac{dx}{2}. \end{align*}

Thus with the substitution $x = 2 \sin\theta$, we have

$$S = 8 \int_{0}^{\frac{\pi}{6}} \theta \log^{2} (2 \sin\theta) \, d\theta.$$

To evaluate this integral, note that for $0 < \theta < \frac{\pi}{6}$ we have

$$e^{i\theta} \cdot 2 \sin \theta = i \cdot (1 - e^{2i\theta}).$$

Taking logarithm (with the branch cut $(-\infty, 0]$ as usual) to both sides, it follows that

$$i\theta + \log (2\sin\theta) = \frac{i\pi}{2} + \log(1 - e^{2i\theta}).$$

Cubing both sides and integrating on $\left( 0, \frac{\pi}{6} \right)$ and taking imaginary parts only,

$$S = \frac{2}{3} \left( \frac{\pi}{6} \right)^{4} + \frac{8}{3} \Im \int_{0}^{\frac{\pi}{6}} \left( \frac{i\pi}{2} + \log(1 - e^{2i\theta}) \right)^{3} \, d\theta. \tag{1}$$

Step 2. Some complex-analysis techniques

Now we focus on the integral in the imaginary part:

$$I := \int_{0}^{\frac{\pi}{6}} \left( \frac{i\pi}{2} + \log(1 - e^{2i\theta}) \right)^{3} \, d\theta \tag{2}.$$

Once we evaluate the imaginary part of $I$, the identity $(1)$ immediately gives us the answer. We first make the substitution $z = 1 - e^{2i\theta}$ and $\omega = e^{-i\pi/3}$ to obtain

$$I = \int_{0}^{\omega} \left( \frac{i\pi}{2} + \log z \right)^{3} \frac{dz}{2i(z-1)}.$$

Here, the path of integration is a circular arc joining from $0$ to $\omega$ centered at $1$ (green-colored path in the figure below).

But since the integrand is analytic for $0 < \Re z < 1$, we may change the path of integration as $z = \omega t$ for $0 \leq t \leq 1$ (blue-colored path in the figure above). This gives

$$I = \frac{1}{2i} \int_{0}^{1} \left( \frac{i\pi}{6} + \log t \right)^{3} \frac{\omega \, dt}{\omega t - 1}.$$

Plugging $t = e^{-x}$, $I$ reduces to

\begin{align*} I &= \frac{1}{2} \int_{0}^{\infty} \left( \frac{\pi}{6} + ix \right)^{3} \frac{\omega e^{-x}}{1 - \omega e^{-x}} \, dx = \frac{1}{2} \sum_{n=1}^{\infty} \omega^{n} \int_{0}^{\infty} \left( \frac{\pi}{6} + ix \right)^{3} e^{-nx} \, dx \\ &= \frac{1}{2} \sum_{n=1}^{\infty} \omega^{n} \left( -\frac{6 i}{n^4}-\frac{\pi }{n^3}+\frac{i \pi ^2}{12 n^2}+\frac{\pi ^3}{216 n} \right). \end{align*}

Taking the imaginary part,

\begin{align*} \Im I &= -3 \sum_{n=1}^{\infty} \frac{\cos (n\pi/3)}{n^4} + \frac{\pi}{2} \sum_{n=1}^{\infty} \frac{\sin (n\pi/3)}{n^3} + \frac{\pi^2}{24} \sum_{n=1}^{\infty} \frac{\cos (n\pi/3)}{n^2} - \frac{\pi^3}{432} \sum_{n=1}^{\infty} \frac{\sin (n\pi/3)}{n}. \tag{3} \end{align*}

Step 3. Evaluation of a series

Note that for $0 < \theta < \pi$, we have

$$\sum_{n=1}^{\infty} \frac{\sin n\theta}{n} = \Im \sum_{n=1}^{\infty} \frac{e^{in\theta}}{n} = - \Im \log(1 - e^{i\theta}) = \frac{\pi-\theta}{2}.$$

Integrating both sides, we obtain

$$\sum_{n=1}^{\infty} \frac{1-\cos n\theta}{n^{2}} = \frac{\theta (2 \pi -\theta )}{4} \quad \Longrightarrow \quad \sum_{n=1}^{\infty} \frac{\cos n\theta}{n^{2}} = \frac{\theta ^2}{4}-\frac{\pi \theta }{2}+\frac{\pi ^2}{6}.$$

Repeating this procedure, we obtain

$$\sum_{n=1}^{\infty} \frac{\sin n\theta}{n^{3}} = \frac{\theta ^3}{12}-\frac{\pi \theta ^2}{4}+\frac{\pi ^2 \theta }{6}$$

and

$$\sum_{n=1}^{\infty} \frac{\cos n\theta}{n^{4}} = -\frac{\theta ^4}{48}+\frac{\pi \theta ^3}{12}-\frac{\pi ^2 \theta ^2}{12}+\frac{\pi^4}{90}.$$

Plugging $\theta = \frac{\pi}{3}$, we have

\begin{align*} \sum_{n=1}^{\infty} \frac{\sin (n \pi / 3)}{n^{2}} &= \frac{\pi}{3} \\ \sum_{n=1}^{\infty} \frac{\cos (n \pi / 3)}{n^{2}} &= \frac{\pi^3}{36} \\ \sum_{n=1}^{\infty} \frac{\sin (n \pi / 3)}{n^{3}} &= \frac{5 \pi^3}{162} \\ \sum_{n=1}^{\infty} \frac{\cos (n \pi / 3)}{n^{4}} &= \frac{91 \pi ^4}{19440} \end{align*}

Plugging these to $(3)$, we have

$$\Im I = \frac{23 \pi^4}{12960} \quad \Longrightarrow \quad S = \frac{17 \pi^4}{3240} = \frac{17}{36}\zeta(4)$$

as desired.

• (+1) For the hard and precise work! With the final nice rediscovery of the Bernoulli polynomials for $\theta=2\pi x$. I'll add that when $\sin$ is replaced by $\cos$ you'll get the Clausen functions. – Raymond Manzoni May 26 '13 at 19:37
• @sos440: An impressive solution! I had no success with dilogarithms in my first try and thought hard to find some simplifying combinatorial argument. Would upvote once again if I could. – Start wearing purple May 26 '13 at 20:15
• @RaymondManzoni, Thank you! I didn't know that this family of series are related to the Bernoulli polynomial. What an amazing relationship! – Sangchul Lee May 27 '13 at 2:37
• @O.L., Thank you! Motivated by this posting, I also tried polylogarithmic solution at first, which only ended with a complete mess. :( – Sangchul Lee May 27 '13 at 2:39
• @sos440: Thank you so much sos440! I was able to arrive at that logsine integral but I had no idea to evaluate it. Thank you! – Shobhit Bhatnagar May 27 '13 at 3:19