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 
$$\sum_{u=2}^{\infty}
 {\frac{{{\left( -1 \right)}^{u}}}{4{{\left( u-1 \right)}^{2}}-1}}.$$
 A: Note that $$\frac{1}{4n^2-1}=\frac{1}{2(2n-1)}-\frac{1}{2(2n+1)}.$$ If I note $S$ the sum you want to compute, we find $$2S = \sum_{n=1}^{+\infty}\frac{(-1)^n}{2n+1}-\sum_{n=1}^{+\infty}\frac{(-1)^n}{2n-1}=\sum_{n=1}^{+\infty}\frac{(-1)^n}{2n+1}-\sum_{n=0}^{+\infty}\frac{(-1)^{n+1}}{2n+1}=1+2\sum_{n=1}^{+\infty}\frac{(-1)^n}{2n+1}.$$ Hence the only thing we need to compute is $$\sum_{n=1}^{+\infty}\frac{(-1)^n}{2n+1}.$$ But the series for the arctan is precisely $$\arctan(x) = \sum_{n=0}^{+\infty}\frac{(-1)^n}{2n+1}x^{2n+1}.$$ Thus $$\sum_{n=1}^{+\infty}\frac{(-1)^n}{2n+1} = \arctan(1)-1 = \frac{\pi}{4}-1.$$ Finally we have $$2S = 1+2(\pi/4-1)=\frac{\pi}{2}-1$$ and $$S=\frac{\pi}{4}-\frac{1}{2}.$$
A: You may note that $$\sum_{n\geq1}\frac{\left(-1\right)^{n+1}}{4n^{2}-1}=\frac{1}{2}\left(\sum_{n\in\mathbb{Z}}\frac{\left(-1\right)^{n+1}}{4n^{2}-1}-1\right)
 $$ and now we can use the well known summation formula$$\sum_{n\in\mathbb{Z}}\left(-1\right)^{n}f\left(n\right)=-\sum\left\{ \textrm{residue of }\pi\csc\left(\pi z\right)f(z)\textrm{ at }f\left(z\right)\textrm{ poles}\right\} 
 $$
 so it is sufficient to note that $f\left(z\right)=\frac{1}{4z^{2}-1}
 $ has poles at $z=\pm\frac{1}{2}
 $ and we have done. So $$\sum_{n\geq1}\frac{\left(-1\right)^{n+1}}{4n^{2}-1}=\color{red}{\frac{\pi}{4}-\frac{1}{2}}.$$
A: 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 
$$f(1)=\int_0^1x\arctan xdx=\frac{\pi}{4}-\frac{1}{2}. $$
A: Since
$$\frac{1}{4n^2-1}=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right) \tag{1}$$
we have:
$$ \sum_{n=1}^{+\infty}\frac{(-1)^n}{4n^2-1}=\frac{1}{2}\int_{0}^{1}\sum_{n\geq 1}(-1)^n(x^{2n-2}-x^{2n})\,dx =\color{red}{\frac{1}{2}\int_{0}^{1}\frac{x^2-1}{x^2+1}\,dx} \tag{2}$$
that is a straightforward integral to compute.
A: Another way to do it is converting the series into a double integral:
Let $$I=\int_{0}^{1}\int_{0}^{1} \frac{x^2}{1+x^2y^2}dydx.$$ Because the region of integration implies $$0<x,y<1$$ we can convert the integrand into a geometric series as such:
$$\frac{x^2}{1+x^2y^2}=\sum_{n=0}^{\infty}x^2(-x^2y^2)^n=\sum_{n=0}^{\infty}(-1)^nx^{2n+2}y^{2n}.$$ We replace the integrand with the series and integrate term by term to get that
$$I=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+3)(2n+1)}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n+1)(2n-1)}$$ if you change the starting point of the summation. 
Now we complete the proof by evaluating $$I=\int_{0}^{1}\int_{0}^{1} \frac{x^2}{1+x^2y^2}dydx.$$ directly.
Integrate with respect to $y$, we get $$I=\int_{0}^{1} x\arctan(x)dx.$$ Use the fact $\int_{0}^{1} \frac{1}{1+x^2y^2}dy=\frac{\arctan(x)}{x}$ to prove this. Now using integration by parts: $u=\arctan(x),du=\frac{dx}{1+x^2},v=\frac{x^2}{2},dv=xdx,$ we get $$I=\frac{\pi}{8}-\int_{0}^{1} \frac{x^2}{2(1+x^2)} dx.$$ To evaluate $$\int_{0}^{1} \frac{x^2}{2(1+x^2)} dx,$$ put $x=\tan(\theta),dx=\sec^2(\theta)d\theta$ to see that $$\int_{0}^{1} \frac{x^2}{2(1+x^2)} dx=\int_{0}^{\frac{\pi}{4}} \frac{\tan^2(\theta)}{2} d\theta=\int_{0}^{\frac{\pi}{4}} \frac{\sec^2(\theta)-1}{2} d\theta=\frac{1}{2}-\frac{\pi}{8}$$  Putting everything together , $$I=\frac{\pi}{4}-\frac{1}{2}.$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
\color{#f00}{\sum_{n = 1}^{\infty }{\pars{-1}^{n + 1} \over 4n^{2} - 1}} & =
\half\sum_{n = 1}^{\infty }{\pars{-1}^{n + 1} \over 2n - 1} -
\half\sum_{n = 1}^{\infty }{\pars{-1}^{n + 1} \over 2n + 1} =
\half\sum_{n = 0}^{\infty }{\pars{-1}^{n} \over 2n + 1} -
\half\sum_{n = 0}^{\infty }{\pars{-1}^{n} \over 2n + 3}
\\[5mm] & =
{1 \over 8}\bracks{%
2\sum_{n = 0}^{\infty }{\pars{-1}^{n} \over n + 1/2} -
2\sum_{n = 0}^{\infty }{\pars{-1}^{n} \over n + 3/2}}
\end{align}

With the identity $\ds{\pars{~\Psi:\ Digamma\ Function~}}$
$\ds{\Psi\pars{z + 1 \over 2} - \Psi\pars{z \over 2} = 2\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over k + z}}$:
\begin{align}
\color{#f00}{\sum_{n = 1}^{\infty }{\pars{-1}^{n + 1} \over 4n^{2} - 1}} & =
{1 \over 8}\bracks{%
\Psi\pars{3 \over 4} - \Psi\pars{1 \over 4} - \Psi\pars{5 \over 4} +
\Psi\pars{3 \over 4}}
\end{align}
With the $\ds{\Psi}$-Recurrence Formula: $\ds{\Psi\pars{5 \over 4} = \Psi\pars{1 \over 4} + 4}$. Then,
\begin{align}
\color{#f00}{\sum_{n = 1}^{\infty }{\pars{-1}^{n + 1} \over 4n^{2} - 1}} & =
{1 \over 4}\bracks{\Psi\pars{3 \over 4} - \Psi\pars{1 \over 4}} - \half
\end{align}
With Euler Reflection Formula,
$\ds{\Psi\pars{3 \over 4} - \Psi\pars{1 \over 4} =
\pi\cot\pars{\pi \over 4} = \pi}$

\begin{align}
\color{#f00}{\sum_{n = 1}^{\infty }{\pars{-1}^{n + 1} \over 4n^{2} - 1}} & =
\color{#f00}{{1 \over 4}\,\pi - \half} \approx 0.2854
\end{align}
