# Series/Sequences Questions

• Mar 2nd 2010, 07:26 PM
carnage
Series/Sequences Questions
The first question is to find $\Sigma^{b}_{n = 1}\frac{\sin n}{2^n}$ and the second being to find $\Sigma^{\infty}_{n = 1}\frac{\sin n}{2^n}$.

The thing is I'm not sure how to even begin these, so if someone could help me out that would be great. Thanks.
• Mar 2nd 2010, 07:41 PM
Prove It
Quote:

Originally Posted by carnage
The first question is to find $\Sigma^{b}_{n = 1}\frac{\sin n}{2^n}$ and the second being to find $\Sigma^{\infty}_{n = 1}\frac{\sin n}{2^n}$.

The thing is I'm not sure how to even begin these, so if someone could help me out that would be great. Thanks.

Are you asked to evaluate them or to show that they are convergent?

Showing convergence is easy, just use the comparison test with

$\sum_{n = 1}^{\infty}\frac{1}{2^n}$ (this is a geometric series).
• Mar 2nd 2010, 08:17 PM
carnage
Quote:

Originally Posted by Prove It
Are you asked to evaluate them or to show that they are convergent?

Showing convergence is easy, just use the comparison test with

$\sum_{n = 1}^{\infty}\frac{1}{2^n}$ (this is a geometric series).

I believe we're supposed to evaluate them.
• Mar 2nd 2010, 09:24 PM
chisigma
Is...

$\sin n = \frac{e^{in} - e^{-in}}{2i}$ (1)

... so that...

$\sum_{n=1}^{\infty} \frac{\sin n}{2^{n}} = \frac{1}{2i}\cdot \{\sum_{n=1}^{\infty} (\frac{e^{i}}{2})^{n} - \sum_{n=1}^{\infty} (\frac{e^{-i}}{2})^{n}\}=$

$= \frac{1}{2i}\cdot \{\frac{e^{i}}{2-e^{i}} - \frac{e^{-i}}{2-e^{-i}} \}$ (2)

The numerical evaluation of (2) is tedious but not difficult and it is left to the reader (Wink) ...

Kind regards

$\chi$ $\sigma$