# Evaluate the series $\sum_{n=0}^{\infty} \frac{n^2-1}{n!}\frac{x^n}{n-1}$

I would like to show that the following sum converges $\forall x \in \mathbb{R}$ as well as calculate the sum:

$\sum_{n=0}^{\infty} \frac{n^2-1}{n!}\frac{x^n}{n-1}$

First for the coefficient:

$\frac{n^2-1}{n!(n-1)}=\frac{n+1}{n!}$

Then, what I did was to try and formulate this series to a series which I know:

$\sum_{n=0}^{\infty} \frac{n^2-1}{n!}\frac{x^n}{n-1}=\sum_{n=0}^{\infty} \left( \frac{n+1}{n!} \right)x^n=\cdots = \frac{1}{x} \sum_{n=0}^{\infty}\frac{(n+1)^2 x^{n+1}}{(n+1)!}$

I have ended up with this formula, which reminds me somehow the expansion of the exponential $e^x$

$\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x$

but I cannot see how the term $(n+1)^2$ affects the result.

Thanks.

• the sum is $$e^x(1+x)$$ prove this – Dr. Sonnhard Graubner Jan 20 '17 at 15:27
• Use $evaluate$ instead of $calculate$ !!!. – Felix Marin Jan 20 '17 at 21:33

$$\frac{n+1}{n!}=\frac n{n!}+\frac1{n!}=\frac1{(n-1)!}+\frac1{n!}$$

And then we get the well-known series expansion for $e^x$.

I think it might be easier to not incur the $(n+1)^2$, but to simply break up the sum into

$$\sum_{n=0}^\infty \frac{n+1}{n!} x^n \;\; =\;\; \sum_{n=0}^\infty \frac{n}{n!} x^n + \sum_{n=0}^\infty \frac{x^n}{n!}.$$

Clearly the second sum goes to $e^x$, but we can notice that

$$\sum_{n=0}^\infty \frac{nx^n}{n!} \;\; =\;\; 0 + x + \frac{2}{2!}x^2 + \frac{3}{3!}x^3 + \frac{4}{4!}x^4 + \ldots \;\; = \;\; x \left (1 + x + \frac{1}{2!}x^2 + \frac{1}{3!} x^3 + \ldots \right ) \;\; =\;\; xe^x.$$

Also you could compute $$\frac{d}{dx}(xe^x)=\frac{d}{dx}\sum_{n\geq 0} \frac{x^{n+1}}{n!}$$ More precisely: since the radius of convergence of the exponential's serie is $\infty$, we can do: $$e^x(1+x)=\frac{d}{dx}(xe^x)=\frac{d}{dx}\sum_{n\geq 0} \frac{x^{n+1}}{n!}=\sum_{n\geq 0} \frac{d}{dx}\frac{x^{n+1}}{n!}=\sum_{n\geq 0} \frac{n+1}{n!}x^{n}$$

• For a complete answer, it would be best to flesh such things out fully. – The Count Jan 20 '17 at 15:50