Prove that $\sum_{n=1}^{+\infty}\frac{(-1)^n n^3}{4^n}=-0.0064$ (Motivation) As homework, we have been asked to prove that the following series converges: $$\sum_{n=1}^{+\infty}\frac{(-1)^n n^3}{4^n}$$
I did it in two ways: 


*

*Using the alternating series test (Leibniz criterion), proving that $\frac{n^3}{4^n}$ is decreasing and also $\lim_\limits{n\to +\infty}{\frac{n^3}{4^n}}=0$.

*Using the n-th root test (Cauchy's criterion) and absolute convergence, proving that $\lim_\limits{n\to +\infty}{\frac{(\sqrt[n]n)^3}{4}}=\frac{1}{4}<1$.
However, Wolfram Alpha states another interesting result: That this series sums exactly to $-0.0064$. I would like to see how that result is obtained, so a proof for it.

As in other answers consider $$\sum_{n=1}^{+\infty}n^3 x^n$$ and rewrite $$n^3=n(n-1)(n-2)+a n(n-1)+bn$$ Expanding and identifying coefficients, you should get $a=3$ and $b=1$. So
$$\sum_{n=1}^{+\infty}n^3 x^n=\sum_{n=1}^{+\infty}n(n-1)(n-2) x^n+3\sum_{n=1}^{+\infty}n(n-1) x^n+\sum_{n=1}^{+\infty}n x^n$$
$$\sum_{n=1}^{+\infty}n^3 x^n=x^3\sum_{n=1}^{+\infty}n(n-1)(n-2) x^{n-3}+3x^2\sum_{n=1}^{+\infty}n(n-1) x^{n-2}+x\sum_{n=1}^{+\infty}n x^{n-1}$$
$$\sum_{n=1}^{+\infty}n^3 x^n=x^3\left(\sum_{n=1}^{+\infty} x^{n}\right)'''+3x^2\left(\sum_{n=1}^{+\infty} x^{n}\right)''+x\left(\sum_{n=1}^{+\infty} x^{n}\right)'$$ Now, for $|x|<1$ $$\sum_{n=1}^{+\infty} x^{n}=\frac x {1-x}\qquad \left(\sum_{n=1}^{+\infty} x^{n}\right)'=\frac{1}{(1-x)^2}$$ $$\left(\sum_{n=1}^{+\infty} x^{n}\right)''=\frac{2}{(1-x)^3}\qquad \left(\sum_{n=1}^{+\infty} x^{n}\right)'''=\frac{6}{(1-x)^4}$$ which make 
$$\sum_{n=1}^{+\infty}n^3 x^n=\frac{6x^3}{(1-x)^4}+\frac{6x^2}{(1-x)^3}+\frac{x}{(1-x)^2}={x(x^2+4x+1)\over (1-x)^4}$$ Using now $x=-\frac 14$ leads to $$\sum_{n=1}^{+\infty}\frac{(-1)^n n^3}{4^n}=-\frac{4}{625}=-0.0064$$
