$\sum_{k=0}^\infty \frac{1}{(4k+1)(4k+3)}$

Question

$$\sum_{k=0}^\infty \frac{1}{(4k+1)(4k+3)}=\frac{1}{2}\sum_{k=0}^\infty \left(\frac{1}{4k+1}-\frac{1}{4k+3}\right)$$ $$=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\,...\right)=\frac{1}{2}\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}$$ What happens next? How can we find a value for this? I know that this converges since it is equivalent to $\frac{1}{k^2}$ at infinity but what about computing the sum? This should equal to $\pi/4$.

Answer 1

Let:

$$S(x) = \frac 1 2 \sum_{k = 0}^\infty \frac {(-1)^k} {2 k + 1} x^{2 k + 1}$$

Clearly we want $S(1)$.

You find the derivative $S'$ by differentiating the series termwise. (why?) This results in a common infinite series that is not too hard to express in closed form. By integration, you can obtain your answer, using standard facts about trigonometric functions.

Answer 2

The sum equals $\pi/4$, as in https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80

Answer 3

The sum value obtained in the bracket is the expansion of $tan^-1(x)$, which can be obtained by taylor series of the function about $x=1$, which gives the value