<s>[INST] 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.
 [/INST] You can get this by clever sum manipulations.  
Start from 
$$
\sum \frac1{(-4)^n}=-\frac14+\frac1{16}-\frac1{64} + \cdots = -\frac15
$$
Then look at 
$$
\sum \frac{n}{(-4)^n}=\begin{array}{lllll} 
-\frac14&+\frac1{16}&-\frac1{64} &+ \cdots \\ 
&+\frac1{16}&-\frac1{64}&- \cdots \\
&&-\frac1{64}&+ \cdots 
\end{array}\\
=\sum \frac1{(-4)^n} - \frac14\sum \frac1{(-4)^n} + \frac1{16}\sum \frac1{(-4)^n} \cdots = -\frac45\cdot \frac{1}{5} = -\frac{4}{25}
$$
Next do the same sort of vertical breakup, but with  $\sum \frac{n}{(-4)^n}$ in each row:
$$
\begin{array}{lllll} 
-\frac14&+\frac2{16}&-\frac3{64} &+\frac4{256} &-\cdots \\ 
&+\frac1{16}&-\frac2{64}&+\frac3{256} &-\cdots \\
&&-\frac1{64}&+\frac2{256} &- \cdots \\
&&&+\frac1{256} &-\cdots
\end{array} = \sum\frac{n(n+1)}{2}\frac1{(-4)^n} \\
=\sum \frac{n}{(-4)^n} - \frac14\sum \frac{n}{(-4)^n} + \frac1{16}\sum \frac{n}{(-4)^n} \cdots = -\frac45\cdot \frac{4}{25} = -\frac{16}{125}
$$
And the next step of the same trick, using rows of  $\sum\frac{n(n+1)}{2}\frac1{(-4)^n}$, will give
$$\sum\frac{n(n+1)(n+2)}{6}\frac1{(-4)^n} = -\frac45\cdot\frac{16}{125}=-\frac{64}{625}
$$
Finally, write 
$$
n^3 = 6 \frac{n(n+1)(n+2)}{6} -6 \frac{n(n+1)}{2} + n
$$
and you get the answer 
$$
-6\cdot\frac{64}{625} +6\cdot\frac{16}{125} -\frac{4}{25} =-0.0064
$$
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