Series/Sequences Questions

**carnage** · March 2nd 2010, 06:26 PM

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.

**Prove It** · March 2nd 2010, 06:41 PM

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).

**carnage** · March 2nd 2010, 07:17 PM

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.

**chisigma** · March 2nd 2010, 08:24 PM

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

...

Kind regards

$\chi$ $\sigma$

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