Calculating the limit based on the cauchy product

Question

I'm attempting to find the limit of $$ \sum^\infty_{k=0} k^2 q^k$$ using the result of the cauchy product $$ \sum^\infty_{k=0} k q^k * \sum^\infty_{k=0} q^k$$ and I have calculated the cauchy product to be $$ \sum^\infty_{n=0} \sum^n_{k=0} kq^n$$ and then I've arranged that in the form $$\sum^\infty_{n=0} \sum^n_{k=0} kq^n = \sum^\infty_{n=0}\left(q^n \sum^n_{k=0}k \right) = \sum^\infty_{n=0}\left(q^n \frac{n(n+1)}{2} \right)$$ however I'm not sure if

1. This is actually correct and
2. This actually helps with figuring out the the original problem.

Answer 1

1. Yes, I think this is correct.
2. Yes, it helps: Assuming you know or can prove that these sums actually converge, let's define $$s_d :=\sum_{k=0}^{\infty} k^dq^k$$ Then what you've shown so far is $s_0s_1=\frac{s_2+s_1}{2}$ (just separating your last sum into two) and thus $s_2=2s_0s_1-s_1$. All the basic techniques to calculate sums like this (that I know of) are using expressions of $s_d$ in terms of $s_{d-1}, s_{d-2}, ...$. Now you can proceed to reduce $s_1$ to an expression in terms of $s_0$, e.g. by doing the same Cauchy product trick as above to $s_0\times s_0$.

An interesting ALTERNATIVE SOLUTION (if you can take term-by-term differentiation of absolutely convergent series for granted, or are happy to take it as a given just for the sake of the result): $$\frac{ds_d}{dq}=\sum_{k=1}^{\infty} k^{d+1}q^{k-1}=\frac{1}{q}s_{d+1}$$ by taking the derivative inside the sum and differentiating term by term (allowed in the interior of the region of convergence because the series converges absolutely). So if you know $s_d$, you can differentiate it directly and express $s_{d+1}$ in terms of it.