<s>[INST] Solving the infinite series $1-\frac{2^3}{1!}+\frac{3^3}{2!}-\frac{4^3}{3!}+\cdots$ I have the following question:

Evaluate the infinite series:
   $$S=1-\frac{2^3}{1!}+\frac{3^3}{2!}-\frac{4^3}{3!}+\cdots$$
  (a) $\displaystyle\frac1e$   (b) $\displaystyle\frac{-1}e$ (c) $\displaystyle\frac{2}e$ (d) $\displaystyle\frac{-2}e$

Now in the book they have given a strange explanation using $$(n+1)^3=[n(n-1)(n-2)+6n(n-1)+7n+1].$$ I don't understand how did they get this. So I have two doubts:


*

*How did they get this "trick"?

*If you would have got this question in a competitive exam, how would you have solved it (in the minimum amount of time), either directly or with numerical methods?

 [/INST] Since $(n+1)^3 = \color{red}{1}\cdot n(n-1)(n-2)+\color{red}{6}\cdot n(n-1) +\color{red}{7}\cdot n+\color{red}{1}$ we have:
$$\begin{eqnarray*}\sum_{n=0}^{+\infty}\frac{(-1)^n (n+1)^3}{n!} &=& \frac{13}{2}+\sum_{n\geq 3}\frac{(-1)^n (n+1)^3}{n!}\\&=&\frac{13}{2}+\color{red}{1}\cdot\sum_{n\geq 3}\frac{(-1)^n}{(n-3)!}+\color{red}{6}\cdot\sum_{n\geq 3}\frac{(-1)^n}{(n-2)!}+\color{red}{7}\cdot\sum_{n\geq 3}\frac{(-1)^n}{(n-1)!}+\color{red}{1}\cdot\sum_{n\geq 3}\frac{(-1)^n}{n!}\\&=&\frac{13}{2}-\sum_{n\geq 0}\frac{(-1)^n}{n!}+6\cdot\sum_{n\geq 1}\frac{(-1)^n}{n!}-7\cdot\sum_{n\geq 2}\frac{(-1)^n}{n!}+\sum_{n\geq 3}\frac{(-1)^n}{n!}\\&=&\sum_{n\geq 0}(-1+6-7+1)\frac{(-1)^n}{n!}\\&=&-\sum_{n\geq 0}\frac{(-1)^n}{n!}=\color{red}{-\frac{1}{e}}.\end{eqnarray*}$$
We simply get the initial decomposition by induction. Obviously $(n+1)^3-n(n-1)(n-2)$ is a quadratic polynomial in $n$, whose leading term is $6n^2$. So we substract $6n(n-1)$ and we are left with a linear polynomial and so on. Another chance is to tackle our problem this way:
$$\begin{eqnarray*} \sum_{n\geq 0}\frac{(-1)^n(n+1)^3}{n!}&=&\sum_{n\geq 1}\frac{(-1)^n n(n+1)^2}{n!}+\sum_{n\geq 0}\frac{(-1)^n(n+1)^2}{n!}\\&=&\sum_{n\geq 0}\frac{(-1)^n(n+1)^2}{n!}-\sum_{n\geq 0}\frac{(-1)^n(n+2)^2}{n!}\\&=&-\sum_{n\geq 0}\frac{(-1)^n(2n+3)}{n!}\\&=&-3\sum_{n\geq 0}\frac{(-1)^n}{n!}+2\sum_{n\geq 0}\frac{(-1)^n}{n!}=\frac{2-3}{e}.\end{eqnarray*}$$
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