calculus - 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.

Answer

One should instead notice that

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

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