Summation from n=1 ti infinity of (sin(1))^n.

How can I break this up? I think it will be a geometric series...

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- Nov 4th 2008, 10:26 AMveronicak5678Does the series converge?
Summation from n=1 ti infinity of (sin(1))^n.

How can I break this up? I think it will be a geometric series... - Nov 4th 2008, 10:34 AMPlato
$\displaystyle \left| r \right| < 1 \Rightarrow \quad \sum\limits_{n = 1}^\infty {r^n } = \frac{r}{{1 - r}}$

- Nov 4th 2008, 10:36 AMveronicak5678
I thought it was r ^ (n-1)? And what is my r and my a?

- Nov 4th 2008, 01:12 PMMathstud28
- Nov 4th 2008, 01:22 PMveronicak5678
Got it. Thank you!