Evaluation of $\sum_{n=1}^{\infty} (\frac{1}{4n-3}-\frac{1}{4n-1}) $

Question

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 $\int_{0}^{1} \tan^{-1}x.dx$

Answer 1

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

Answer 2

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*}