Evaluation of $\sum_{n=1}^{\infty} (\frac{1}{4n-3}-\frac{1}{4n-1}) $ Prove that 
$$\sum_{n=1}^{\infty} (\frac{1}{4n-3}-\frac{1}{4n-1})=\frac{\pi}{4}$$
Could someone give me slight hint as how to convert above expression to $\int_{0}^{1} \tan^{-1}x.dx$
 A: Your series comes out to be
$$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\cdots$$
This can be written as $$\sum_{k=1}^\infty\frac{(-1)^k}{2k+1}$$
and is the well known Leibniz formula for $\pi$. Note that your series is on the hyperlinked Wikipedia page under the section "Convergence". The easiest way to show that this is equal to $\frac{\pi}{4}$ is to show that the $\arctan$ function has a Taylor Series
$$\arctan(x) = x-\frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$$
Now plug in $x=1$. The Taylor Series is valid here: the interval of convergence is $[-1,1]$ 
If you want a more complete proof the Wikipedia link should suffice, or see here. For another proof of the Taylor Series formula see the Math.SE post linked in the comments by @lab_bhattacharjee
A: By IBP,
\begin{eqnarray*}\int \tan^{-1}x.dx&=&x\times tan^{-1}x|-\int x\times \frac{1}{1+x^2}.dx \\
&=&x\times tan^{-1}x|-\int \frac{d(x^2+1)}{2dx}\times \frac{1}{1+x^2}.dx\\
&=&x\times tan^{-1}x|-\ln|x^2+1| |
  \end{eqnarray*}
