Find limit of modified geometric sum $\sum_{k=1}^{\infty} k^3 q^k,$ where $\left| q\right| < 1$ I've been trying to find the limit of the infinite series
$\displaystyle\sum_{k=1}^{\infty} k^3 q^k, \quad \left| q\right| < 1$
I already determined that it converges using the ratio test, but I am at a loss when it comes to actually determine the limit.
I know how to calculate the limit of
$\displaystyle\sum_{k=1}^{\infty} k q^k, \quad \left| q\right| < 1$
This is simple by derivation with respect to q applied to
$1 + q + q^2 + \ldots q^n = \frac{1-q^{n+1}}{1-q}$
and then multiplying with $q$ and taking the limit. But this gets way too complicated when doing it three times. Is there a better way?
 A: Note that
$$ \sum_{k=0}^\infty k^nq^k = q\sum_{k=0}^\infty k^{n-1} \cdot kq^{k-1}  =  q\Big(\sum_{k=0}^\infty k^{n-1}  q^k\Big)' $$
So you can use recursion to calculate it.
\begin{align} \sum_{k=0}^\infty q^k &= \frac{1}{1-q} \\
\sum_{k=0}^\infty kq^k &= q\Big(\sum_{k=0}^\infty q^k\Big)' = q\Big(\frac{1}{1-q}\Big)' = \frac{q}{(1-q)^2} \\ 
\sum_{k=0}^\infty k^2q^k &= q\Big(\sum_{k=0}^\infty kq^k\Big)' = q\Big(\frac{q}{(1-q)^2}\Big)' = \frac{q^2+q}{(1-q)^3} \\
\sum_{k=0}^\infty k^3q^k &= q\Big(\sum_{k=0}^\infty k^2q^k\Big)' = q\Big(\frac{q^2+q}{(1-q)^3}\Big)' = \frac{q^3 + 4q^2 +q}{(1-q)^4}  \end{align}
A: Hint:
$$k^3=k(k-1)(k-2)+3k(k-1)+k.$$
On the other hand,
\begin{align}
\sum_{k=1}^{\infty} k q^k&= q\sum_{k=1}^{\infty} kq^{k-1}=q\biggl(\sum_{k=0}^{\infty} q^{k}\biggr)\vphantom{\Bigr)}^{\!'}\\
\sum_{k=1}^{\infty} k(k-1) q^k&= q^2\sum_{k=1}^{\infty} k(k-1)q^{k-2}=q^2\biggl(\sum_{k=0}^{\infty} q^{k}\biggr)\vphantom{\Bigr)}^{\!''}\\
\sum_{k=1}^{\infty} k(k-1)(k-2) q^k&=\dots\dots
\end{align}
A: There are technical issues with using differentiation; it is a highly non-trivial fact (compared to the sum you want to evaluate) that you can differentiate a (convergent) power series term-wise. You can look up Taylor expansion (with remainder term) and its proof, but for now observe that there is a simple elementary solution:

$q · \sum_{k=0}^n k^3 q^k = \sum_{k=0}^n k^3 q^{k+1} = \sum_{k=1}^{n+1} (k-1)^3 q^k = \sum_{k=0}^{n+1} (k-1)^3 q^k$.
Thus $\sum_{k=0}^n k^3 q^k - q · \sum_{k=0}^n k^3 q^k = \sum_{k=0}^n (k^3-(k-1)^3) q^k - n^3 q^{n+1}$.
Thus $(1-q) · \sum_{k=0}^n k^3 q^k = \sum_{k=0}^n (3k^2-3k+1) q^k - n^3 q^{n+1}$.

As you can see, we have reduced the degree of the polynomial part of each term in the series. I leave it to you to repeat the process, which would let you find the closed-form for $\sum_{k=0}^n k^3 q^k$, from which it is then easy to find $\lim_{n→∞} \sum_{k=0}^n k^3 q^k$ as desired.
