1. ## Series/Sequences Questions

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

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

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

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

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

3. 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

$\displaystyle \sum_{n = 1}^{\infty}\frac{1}{2^n}$ (this is a geometric series).
I believe we're supposed to evaluate them.

4. Is...

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

... so that...

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$