Question

How would I approach this problem?
\[\sum{\frac{n^2*(-1)^n}{n!}}\]
I'm trying to find the value of the series.

Answer 1

series yeah!

Answer 2

A hint given is that
\[e^{-1} = \sum{\frac{(-1)^n}{n!}}\]

Answer 3

You have to use on of the tests to figure out whether its convergent or divergent. I think ratio test would work for this.

Answer 4

I know the series converges, but the problem asks to find the value of the series.

Answer 5

Here's what I tried:
\[\begin{align*}\sum_{n=0}^\infty \frac{n^2(-1)^n}{n!}&=\sum_{n=0}^\infty \frac{n(-1)^n}{(n-1)!}\\
&=\sum_{k=-1}^\infty \frac{(k+1)(-1)^{k+1}}{k!}&\text{substitution, }k=n-1\\
&=\sum_{k=0}^\infty \frac{(k+1)(-1)^{k+1}}{k!}\\
&=\sum_{k=0}^\infty \frac{k(-1)^{k+1}}{k!}+\sum_{k=-1}^\infty \frac{(-1)^{k+1}}{k!}\\
&=-\sum_{k=0}^\infty \frac{k(-1)^{k}}{k!}-\sum_{k=-1}^\infty \frac{(-1)^{k}}{k!}\\
&=-\color{red}{\sum_{k=0}^\infty \frac{k(-1)^{k}}{k!}}-\frac{1}{e}
\end{align*}\]
Checking on wolfram, you'd find that the red sum comes to \(-\dfrac{1}{e}\), but I'm assuming that's not a valid reason for the assignment. Well, continuing with this sort of decomposition, we come to some problems:
\[\begin{align*}
\sum\cdots &=-\sum_{k=0}^\infty \frac{k(-1)^{k}}{k!}-\frac{1}{e}\\
&=-\sum_{k=0}^\infty \frac{(-1)^{k}}{(k-1)!}-\frac{1}{e}\\
&=-\sum_{m=-1}^\infty \frac{(-1)^{m+1}}{m!}-\frac{1}{e}&\text{substituting, }m=k-1\\
&=\sum_{\color{red}{m=-1}}^\infty \frac{(-1)^{m}}{m!}-\frac{1}{e}
\end{align*}\]
That first \(m=-1\) term throws everything off...

Answer 6

Well, as it turns out, the sum from \(m=-1\) to \(\infty\) works out the same as from \(m=0\) to \(\infty\). Hmm...

Answer 7

Yeah, for some reason, \(\dfrac{(-1)^{-1}}{(-1)!}=0\)... I guess the reasoning works out.