How I could evaluate this :$ \sum_{n=1}^{\infty}({-1})^{n+1}n(tan^{-1}s-s+\frac{s^3}{3}+....({-1})^{n+1} \frac{s^{2n+1}}{2n+1}) $? let $s$ be a complex variable which $Re(s)>0$. Evaluate :  $ \sum_{n=1}^{\infty}({-1})^{n+1}n(tan^{-1}s-s+\frac{s^3}{3}+....({-1})^{n+1} 
\frac{s^{2n+1}}{2n+1}) $
I would be interest for any replies or any comments 
 A: For $n \in \mathbb{Z}_{+}$, let $a_n$ be the expression
$\displaystyle\;\tan^{-1}s-s+\frac{s^3}{3} + \cdots + (-1)^{n+1}$. It has an integral representation:
$$\begin{align}
a_n &= \int_0^s \left[\frac{1}{1+t^2} - \left( 1 + ( -t^2) + \cdots + (-t^2)^n\right)\right]dt\\
&= \int_0^s \left[\frac{1}{1+t^2} - \frac{1-(-t^2)^{n+1}}{1+t^2}\right]dt\\
&= \int_0^s \frac{(-t^2)^{n+1}}{1+t^2} dt
\end{align}
$$
As a result, we can rewrite the series at hand as
$$\sum_{n=1}^\infty (-1)^{n+1} n a_n
= \sum_{n=1}^\infty \int_0^s \frac{n t^{2(n+1)}}{1+t^2} dt
= \sum_{n=1}^\infty \int_0^1 \frac{n s (st)^{2(n+1)}}{1+(st)^2} dt
\tag{*1}
$$
Notice for $|s| < 1$, the sequences of functions 
$\displaystyle\;\sum_{n=1}^N \frac{n s (st)^{2(n+1)}}{1+(st)^2}\;$
converges uniformly and absolutely to
$\displaystyle\;\frac{s(st)^4}{(1+(st)^2)(1-(st)^2)^2}\;$
as $N \to \infty$ for $t$ over $[0,1]$, we can switch the order of summation and taking limit in $(*1)$ and
get
$$\begin{align}
\sum_{n=1}^\infty (-1)^{n+1} n a_n 
&= \int_0^1 \frac{(st)^4}{(1+(st)^2)(1-(st)^2)^2} sdt
= \int_0^s \frac{t^4}{(1+t^2)(1-t^2)^2} dt\\
&= \frac14 \left[ \log\frac{1-s}{1+s} + \tan^{-1} s + \frac{s}{(1-s^2)}\right]\
\end{align}
$$
A: For the time being I can give you the following answer. Rearrange your series a
$$ S = \tan^{-1}(s)\sum_{n=1}^{\infty}(-1)^{n+1}n- \sum_{n=1}^{\infty}(-1)^{n+1}n \sum_{m=0}^{n}\frac{(-1)^m\,s^{2m+1}}{(2m+1)} $$

$$\implies S = \frac{\tan^{-1}(s)}{4}- \sum_{n=1}^{\infty}(-1)^{n+1}n \sum_{m=0}^{n}\frac{(-1)^m\,s^{2m+1}}{(2m+1)}  $$

Note: We can find a closed form for the first series as
$$ \sum_{n=1}^{\infty}(-1)^{n+1}n=\frac{1}{4} $$
by considering the series 
$$ \sum_{n=1}^{\infty} x^n = \frac{x}{1-x}.$$
