Find limit of modified geometric sum $\sum_{k=1}^{\infty} k^3 q^k,$ where $\left| q\right| < 1$

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}

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}

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.