how can i find $ \frac{1}{2\pi}\left ( \frac{\pi^{3}}{1!3}-\frac{\pi^{5}}{3!5}+\frac{\pi^{7}}{5!7}-... \right ) $

Question

Consider a sequence $ s_{n} = \frac{1}{2\pi}\left ( \frac{\pi^{3}}{1!3}-\frac{\pi^{5}}{3!5}+\frac{\pi^{7}}{5!7}-...+\frac{\left ( -1 \right )^{n-1}\pi^{2n+1}}{\left ( 2n-1 \right ) ! \left ( 2n+1 \right )} \right ) $

How can I attack this to find $ \lim_{n \to \infty}s_{n} \ $ ?

As the series $ \lim_{n \to \infty}s_{n} \ $ converges absolutely, so by simple manipulation I arrived to this :

$ \lim_{n \to \infty}s_{n} = \frac{1}{\pi}\left \{ \left ( \pi \right ) + \left ( \pi - \frac{\pi^{3}}{3!} \right ) + \left ( \pi - \frac{\pi^{3}}{3!} + \frac {\pi^{5}}{5!} \right )+...\right \} $

What to do next? Anyone please?

Answer 1

Consider $$f(x)=\frac{x^3}{1!3}-\frac{x^5}{3!5}+\frac{x^7}{5!7}-\cdots$$ (an infinite series). We want to find $f(\pi)/(2\pi)$. Then $$f'(x)=\frac{x^2}{1!}-\frac{x^4}{3!}+\frac{x^6}{5!}-\cdots =x\left(\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)$$ which is a series you should recognise. Now integrate to find $f(x)$ etc.

Answer 2

by Mclaurin series

$x\sin x=x^2-\frac{x^4}{3!}+..$

integrating both sides with limits $0$,$\pi$ we get the required sum.