How to evaluate $\sum\limits_{n=0}^\infty n^3x^n$, for $x\in[0,1)$.

Question

How can I solve the following sum?

$$\sum_{n=0}^\infty n^3x^n$$

For $x\in[0,1)$. Do I have to try values and see if its a geometric series or something like that?

Or do I have first find the differentiation of this?

Answer 1

It's easy to see your series' convergence radius is $\;1\;$ , just the same as the well-known standard geometric series. Thus your convergence interval is $\;(-1,1)\;$, and within this interval you can differentiate termwise the geometricseries:

$$f(x)=\sum_{n=1}^\infty x^n\implies f'(x)=\sum_{n=1}^\infty nx^{n-1}\implies f''(x)=\sum_{n=2}^\infty n(n-1)x^{n-2}=\sum_{n=1}^\infty(n^2-n) x^{n-2}\implies$$

$$\implies f'''(x)=\sum_{n=1}^\infty n(n-1)(n-2) x^{n-3}=\sum_{n=1}^\infty(n^3-3n^2+2n) x^{n-3}+$$

Thus, your series can be put as

$$\sum_{n=1}^\infty n^3 x^n=x^3f'''(x)+3x^2f''(x)+xf'(x)$$

And since $\;f(x)=\cfrac1{1-x}\;$ , then $\;f'(x)=\cfrac1{(1-x)^2}\;$ and etc....

Answer 2

\begin{align} n^3 = {} & \overbrace{n(n-1)(n-2)}^\text{3rd degree} {}+{} \overbrace{3n(n-1)}^\text{2nd degree} {}+ n \\[12pt] n^3 x^n = {} & \Big(n(n-1)(n-2)x^{n-3} \cdot x^3\Big) \\ & {} + \Big( 3n(n-1) x^{n-2}\cdot x^2 \Big) + \Big( n x^{n-1} \cdot x\Big) \\[12pt] = {} & x^3 \frac {d^3} {dx^3} x^n + 3x^2 \frac {d^2} {dx^2} x^n + x \frac d {dx} x^n \\[12pt] \sum_{n=0}^\infty n^3 x^n = {} & x^3 \frac {d^3} {dx^3} \sum_{n=0}^\infty x^n + 3x^2 \frac {d^2} {dx^2} \sum_{n=0}^\infty x^n + x\frac d {dx} \sum_{n=0}^\infty x^n \end{align} and this sum has a simple closed form.

Answer 3

One can aproach the series $\sum_{n=0}^∞ n^3x^n$ as the definite integral $\int_0^ {∞} x^3a^x dx $ for a between $0$ and $1$ and solving it we get the expression $6/ln^4(a)$.