series of $\arctan \frac{2x^2}{n^2}$ .

Question

Find : $$\sum_{n=1}^{\infty}\arctan \frac{2x^2}{n^2}.$$ Where $x\in \mathbb{R}$.

Answer 1

Hint: Use the Taylor series of $\arctan$ to obtain $$ \sum_{n=1}^\infty \arctan \frac{t}{n^2} = \sum_{n=1}^\infty \sum_{m=0}^\infty \frac{(-1)^m (t/n^2)^{2m+1}}{2m+1} = \sum_{m=0}^\infty \frac{(-1)^m}{2m+1} t^{2m+1} \sum_{n=1}^\infty \frac{1}{n^{4m+2}} = \sum_{m=0}^\infty \frac{(-1)^m \zeta(4m+2)}{2m+1} t^{2m+1} = \sum_{m=0}^\infty \frac{B_{4m+2}(2\pi)^{4m+2}}{(4m+2)(4m+2)!} t^{2m+1}. $$ (I don't know if that actually leads to Lucian's formula, though...)

Answer 2

$\qquad\quad$ There is indeed a very beautiful closed form expression for this series, in terms of

$$k\pi+\dfrac\pi2-\arctan\bigg(\dfrac{\tan\pi x+\tanh\pi x}{\tan\pi x-\tanh\pi x}\bigg)$$

where k is $0$ on the interval $(0,A_0),~1$ on the interval $(A_0,A_1),~2$ on the interval $(A_1,A_2),~$ etc. where $A_k>0$ are the roots of the transcendental equation $\tan\pi x=\tanh\pi x$. The entire graphic is symmetrical with regards to the vertical axis Oy, since the series is even. Also, $A_k\simeq k+\frac54$.