I'v got roughly half way through this question:
For (fixed) x which is an element of the real numbers, consider the series
$\sum_{n=1}^\infty \frac{x^{n-1}}{2^nn} $
For which x does this series converge? For which x is the series conditionally convergent?
So far I've managed to deduce that for x=-1, the series is in oscillating harmonic form i.e.
$\sum_{n=1}^\infty \frac{x^{n-1}}{2^nn} = \sum_{n=1}^\infty \frac{1}{2^nn} * (-1)^{n-1} $
here $a_n$ = $\frac{1}{2^nn}$ where as n approaches infinity, $\frac{1}{2^nn}$ approaches 0. Therefore the series converges according to the alternating series test.
For all other values between -1 and 1, the series must be convergent as $a_n$ is decreasing.
Therefore I concluded that the series is absolutely convergent in the interval [-1,1].
Thats what I got so far but I'm quite uncomfortable with this area. Would anyone mind correcting/helping?