<s>[INST] 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 
$$\sum_{u=2}^{\infty}
 {\frac{{{\left( -1 \right)}^{u}}}{4{{\left( u-1 \right)}^{2}}-1}}.$$
 [/INST] 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 , $$I=\frac{\pi}{4}-\frac{1}{2}.$$
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