Define $$ f(x) = \sum_{n=1}^{\infty}\frac{\arctan(nx)}{n^2}$$ The function $f$ is continuous over $\mathbb{R}$, but non differentiable at $x=0$. I'm looking for a simple equivalent of $f$ there...
I tried comparing to an integral, or looking for an equivalent of $f'$, but didn't success.
If, for $x>0$, we use $$\tan ^{-1}(n x)=\frac \pi 2 -\tan ^{-1}\left(\frac{1}{n x}\right)$$ and $$\tan ^{-1}\left(\frac{1}{n x}\right)=\sum_{m=0}^\infty\frac {(-1)^m}{(2 m+1)\, x^{2 m+1}\,\, n^{2 m+1} }$$
$$f(x)=\frac{\pi ^3}{12}-\sum_{m=0}^\infty(-1)^m\,\,\frac { \zeta (2 m+3)}{(2 m+1)\, x^{2 m+1} }$$
Using $\sum_{m=0}^{20}$ for $x=2$, the absolute error is $2.13 \times 10^{-15}$.
Edit If we write
$$\tan ^{-1}\left(\frac{1}{n x}\right)=\sum_{m=0}^p +\sum_{m=p+1}^\infty $$ we can know in advance $p$ such that $$R_p=\frac{ \zeta (2 m+5)}{(2 m+3)\,x^{2 m+3}} \leq 10^{-k}$$
This is equivalent to $$2 p x^{2 p+3} \geq 10^k \quad \implies \quad p\sim\frac{1}{2 \log (x)}\,\,W\left(\frac{ \log (x)}{x^3}\, 10^k\right)$$ where $W(.)$ is Lambert function.
For $x=e$ and $k=20$, this gives as a real $p=19.6892$ while the exact solution is $19.6534$
$f$ is an odd function so that it suffices to investigate its behavior for $x > 0$. One can show that $$ \tag{*} - x \ln (x) - \frac x6 \le f(x) \le - x \ln (x) + 5 x $$ for $0 < x < 1/2$. It follows that $$ f(x) = - x \ln |x| + O(x) \quad \text{for } x \to 0 $$ and $$ \lim_{x \to 0} \frac{f(x)}{x} = +\infty \, . $$
In order to prove $(*)$ we need the following estimate for the $\arctan$ function $$ x - \frac{x^3}3 < \arctan(x) < x \quad \text{for } x > 0 $$ (see e.g. here), and an estimate for the harmonic numbers $$ \ln(N+1)<\sum_{n=1}^{N}\frac{1}{n}<1+\ln(N) \quad\text{for positive integers $N$} $$ (see e.g. here).
Now let $x \in (0, 1/2)$ and set $N = \lfloor 1/x \rfloor$. Note that $N \le \frac 1x < N+1$.
First we determine a lower bound for $f(x)$: $$ f(x) \ge \sum_{n=1}^N \frac{\arctan(nx)}{n^2} \ge \sum_{n=1}^N \frac{1}{n^2}\left( nx - \frac 13 n^3x^3 \right) \\ = x \sum_{n=1}^N \frac 1n - \frac 13 x^3 \frac{N(N+1)}{2} \\ \ge x \ln(N+1) - \frac 16 x^3 (N+1)^2 \\ \ge x \ln \frac 1x - \frac 16 x $$ and that proves the left inequality in $(*)$.
Now we determine an upper bound for $f(x)$: $$ f(x) \le \sum_{n=1}^N \frac{nx}{n^2} + \sum_{n=N+1}^\infty \frac{\pi/2}{n^2} \le x \sum_{n=1}^N \frac 1n + \frac{\pi}{2} \sum_{n=N+1}^\infty \frac{1}{n(n-1)} \\ \le x (1 + \ln(N)) + \frac{\pi}{2} \frac 1N \\ \le x \ln \frac 1x + x + \frac{\pi}{2} \frac{x}{1-x} \\ \le x \ln \frac 1x + \left( 1 + \pi \right) x $$ and that proves the right inequality in $(*)$.
