Given:
$$\sum_{k=1}^{\infty} \frac{kx^{k+1}}{3^k}$$
Im guessing its equivalent to:
$$\sum_{k=1}^{\infty} k\left(\frac{-1}{3}\right)^k x^{k+1}$$
But I am not sure on how to advance past this step.
How would I find the function it represents?
Also do I need to find its interval of convergence and find the function over this interval or it doesn't necessarily matter?
Hint It is of the form $x\sum k(\alpha x)^k$, and the derivative of $\sum x^k$ is $\sum k x^{k-1}$.
First you need to write it as a proper Taylor series, which means the exponent should just be $k$, so change indexes:
$$ k' = k + 1 \rightarrow k = k' - 1 \\ \sum_1^\infty \frac{kx^{k + 1}}{3^k} = \sum_2^\infty \frac{(k - 1)x^k}{3^{k - 1}} = \frac{1}{3}\sum_2^\infty \left(k - 1\right)\left(\frac{x}{3}\right)^k $$
As per Pedro's hint, this does almost look like a geometric series (i.e. $\frac{1}{1 - x} = \sum x^i$). The fact that the first two terms are missing suggests that this might be a second derivative (but it just might be a first derivative...integrate and see where it goes). I would integrate this power series and see if you can come up with an easier function to deal with. As for the radius of convergence, the ratio test should show you that this series converges when $|x| < 3$.
Also, don't fret if some terms are missing, you can always just add them in then subtract them off. For instance you could write:
$$ \frac{1}{3}\sum_2^\infty \left(k - 1\right)\left(\frac{x}{3}\right)^k = \frac{1}{3}\left(\left(\sum_0^\infty \left(k - 1\right)\left(\frac{x}{3}\right)^k\right) + 1\right) $$
It's $+1$ because you need to subtract $(0 - 1) = -1$ and then the $k = 1$ term already gives $0$ in the sum, so no need to subtract off $0$.
