Well, the first thing to always try is see if you can get it into a form you recognize. For instance, (let me replace "k" with "z" because I typically use k as an index in sums and I don't want to confuse myself)
S(z) = \sum_{n=1}^\infty \frac{n}{z^n} = \sum_{n=1}^\infty n (z^{-1})^n.
Now, for a moment, let me write w = z^{-1}, so that the sum is
\tilde{S}(w) = \sum_{n=1}^\infty n w^n.
Is the sum starting to look a bit like a more familiar sum to you? It almost looks like a geometric sum, right? If it weren't for that factor of n, it would be.
So, the question you ask at this point is whether or not there is a way to involve the geometric sum somehow.
It may not be quite obvious to you how to do this, but notice two things: I can start the sum at n = 0, because the n=0 term is 0, and if I pull a factor of w out of the sum,
\tilde{S}(w) = w \sum_{n=0}^\infty n w^{n-1}.
Now comes another common trick when dealing with sums: notice that nw^{n-1} = \frac{d}{dw} w^n. That means,
\tilde{S}(w) = w \sum_{n=0}^\infty \frac{d}{dw}\left[w^n\right] = w \frac{d}{dw}\left[\sum_{n=0}^\infty w^n\right],
where I've used the linearity of the derivative to pull it outside of summation (the sum of the derivatives is the derivative of the sum). Note that you can't always do this - there are cases where the convergence of the infinite sum is not nice enough that you can exchange the derivative and the sum. In this case however, since the sum is uniformly (absolutely?) convergent (for a domain I will specify momentarily) we can do that. (I may be off on which kind of convergence is sufficient).
Anywho, now you can see that the sum inside the derivative is just the geometric series sum, so you can use that result to find \tilde{S}(w), and then set w = 1/z again to get S(z).
Recall, however, that the geometric series only gives a finite result if |w| < 1, which implies that S(z) is only finite if |z| > 1.
