I don't want English solution

Hint: Logarithmic differentiation on this .

First note that the sum is defined for all $a$ except the integral values, and Harmonic Numbers can take all real arguments except negative integers. Recall :

$\displaystyle H_x = \lim_{k\to \infty} \big( H_k - \sum_{n=1}^{k} \frac{1}{n+x} \big)$

Hence:

$\displaystyle H_x -H_y = \lim_{k\to \infty} \big( H_k - \sum_{n=1}^{k} \frac{1}{n+x} \big) - \lim_{k\to \infty} \big( H_k - \sum_{n=1}^{k} \frac{1}{n+y} \big)$

$\displaystyle = \lim_{k\to \infty} \big(\sum_{n=1}^{k} \frac{1}{n+y}-\sum_{n=1}^{k} \frac{1}{n+x} \big)$

$\displaystyle = \sum_{n=1}^{\infty} \bigg( \frac{1}{n+y} - \frac{1}{n+x} \bigg)$

Now rewrite our sum as:

$\displaystyle S = \frac{1}{2a} \sum_{n=1}^{\infty} \bigg(\frac{1}{n-a} - \frac{1}{n+a} \bigg) = \frac{1 }{2a} ( H_a - H_{-a})$

From the recurrence relation of the harmonic numbers: Replace $H_{-a}$ by $H_{1-a} - \frac{1}{1-a}$ :

$\displaystyle S= \frac{1 }{2a} (H_a - H_{1-a} + \frac{1}{1-a} )$

Apply the Reflection formula for the harmonic numbers:

$\displaystyle S = \frac{ 1 }{2a} (\frac{1}{a} - \frac{1}{1-a} - \pi \cot ( \pi a ) + \frac{1}{1-a} )$

$\displaystyle \boxed{ = \frac{1}{2a^2} - \frac{\pi \cot (\pi a) }{2a} }$

See this .

I doubt it's simpler. But it's at least different.

I can prove it using Harmonic Numbers, but it is not simpler than that theorem