Find the sum of a complicated telescoping series

Question

I think with this problem I am supposed to simplify the interior here or regroup but I am unsure of how to approach it:

$$\sum _{n=1}^{\infty }\:\left(sin\left(\frac{\left(n+1\right)\pi }{2n+1}\right)-sin\left(\frac{n\pi }{2n-1}\right)\right)$$

This appears to be a telescoping series where the interior terms cancel but I'm not sure how to begin this the right way. Do I start by trying to simplify the inside or trying to break it up into two separate sums? I just want to know where to start on this one.

Answer 1

Andronicus almost but not quite answer re-done... $$ \sum _{n=1}^{N }\:\left(\sin\left(\frac{\left(n+1\right)\pi }{2n+1}\right)-\sin\left(\frac{n\pi }{2n-1}\right)\right)= \\ \sum _{n=1}^{N }\:\sin\left(\frac{\left(n+1\right)\pi }{2n+1}\right)-\sum _{n=1}^{N }\:\sin\left(\frac{n\pi }{2n-1}\right) = \\ \sum _{n=2}^{N+1 }\:\sin\left(\frac{n\pi }{2n-1}\right)-\sum _{n=1}^{N }\:\sin\left(\frac{n\pi }{2n-1}\right)= \\ \sin\left(\frac{(N+1)\pi }{2N+1}\right) -\sin{\pi} $$ and then $$ \lim_{N\to\infty}\left[\sin\left(\frac{(N+1)\pi }{2N+1}\right) -\sin{\pi}\right] =\sin\frac{\pi}{2}-\sin \pi = 1 - 0 = 1. $$

Answer 2

You could make it more obvious writind $$a_n=\sin \left(\frac{n+1}{2 n+1}\pi\right)-\sin \left(\frac{ n}{2 n-1}\pi\right)=\cos \left(\frac{\pi }{4 n+2}\right)-\cos \left(\frac{\pi }{4 n-2}\right)$$ which is clearly telescoping. So $$S_p=\sum_{n=1}^p a_n=\cos \left(\frac{\pi }{2 (2 p+1)}\right)$$

For large values of $p$, using Taylor expansion $$S_p=1-\frac{\pi ^2 }{32 }\frac{(p-1)}{ p^3}+O\left(\frac{1}{p^4}\right)$$

Try it for $p=10$. You have $$S_{10}=\cos \left(\frac{\pi }{42}\right)\approx 0.997204$$ while the above truncated expansion gives $$S_{10}\sim 1-\frac{9 \pi ^2}{32000}\approx 0.997224$$