Evaluation of this series $\sum\limits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}}{4{{n}^{2}}-1}}=??$ I start first to use $u=n+1$ then 
$$\sum_{u=2}^{\infty}
 {\frac{{{\left( -1 \right)}^{u}}}{4{{\left( u-1 \right)}^{2}}-1}}.$$

Note $\sum_{u=2}^{\infty}
 {\frac{{{\left( -1 \right)}^{u}}}{4{{\left( u-1 \right)}^{2}}-1}}=\sum_{n=1}^{\infty}
 {\frac{{{\left( -1 \right)}^{n+1}}}{4{n^{2}}-1}}$. Let
$$ f(x)=\sum_{n=1}^{\infty}
 {\frac{{{\left( -1 \right)}^{n+1}}}{4n^2-1}}x^{2n+1}. $$
Then $f(1)=\sum_{n=1}^{\infty}
 {\frac{{{\left( -1 \right)}^{n+1}}}{4n^2-1}}, f(0)=0$ and
\begin{eqnarray}
f'(x)&=&\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}x^{2n}
\end{eqnarray}
and $f'(0)=0$.
So
$$ (\frac{f'(x)}{x})'=\sum_{n=1}^{\infty}(-1)^{n+1}x^{2n-2}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n}=\frac{1}{1+x^2} $$
Thus
$$ \frac{f'(x)}{x}=\int\frac{1}{1+x^2}dx=\arctan x+C. $$
and
$$ f'(x)=x\arctan x+Cx.$$
But $f'(0)=0$ from which we have $C=0$.
So 
$$f(1)=\int_0^1x\arctan xdx=\frac{\pi}{4}-\frac{1}{2}. $$
