What is the sum of this infinite series?

[anemone]

Evaluate the infinite series $1-\dfrac{2^3}{1!}+\dfrac{3^3}{2!}-\dfrac{4^3}{3!}+\cdots$.

 

[Opalg]

Evaluate the infinite series $1-\dfrac{2^3}{1!}+\dfrac{3^3}{2!}-\dfrac{4^3}{3!}+\cdots$.

[sp]
First, you need to check the identity $(n+1)^3 = n(n-1)(n-2) + 6n(n-1) + 7n + 1.$ Then $$ \begin{aligned} \sum_{n=0}^\infty \frac{(-1)^n(n+1)^3}{n!} &= \sum_{n=0}^\infty \frac{(-1)^n \bigl(n(n-1)(n-2) + 6n(n-1) + 7n + 1\bigr)}{n!} \\ &= \sum_{n=0}^\infty (-1)^n \biggl( \frac1{(n-3)!} + 6\frac1{(n-2)!} + 7\frac1{(n-1)!} + \frac1{n!} \biggr). \end{aligned}$$ The last line there looks suspicious, because for small values of $n$ there are some factorials of negative numbers in the denominators. But if you check what happens when $n$ is small, you see that those terms can be ignored. For example, in the first term \(\displaystyle \sum_{n=0}^\infty (-1)^n \frac1{(n-3)!},\) the summation actually starts at $n=3$ rather than $n=0.$ If you replace the summation index $n$ by $n-3$ then that sum becomes \(\displaystyle \sum_{n=0}^\infty (-1)^{n+3} \frac1{n!} = -\sum_{n=0}^\infty \frac{(-1)^n}{n!}.\) Therefore $$\sum_{n=0}^\infty \frac{(-1)^n(n+1)^3}{n!} = \sum_{n=0}^\infty \biggl( -\frac{(-1)^n}{n!} + 6\frac{(-1)^n}{n!} - 7\frac{(-1)^n}{n!} + \frac{(-1)^n}{n!} \biggr) = - \sum_{n=0}^\infty\frac{(-1)^n}{n!} = -\frac1e.$$[/sp]

 

[anemone]

Very well done, Opalg and thanks for participating!:)