How can I evaluate the infinite series $\sum_{n=0}^\infty\frac{ n^2}{n!} $? Can someone help me to evaluate

$$\sum_{n=0}^\infty\frac{n^2}{n!}?$$

It can be written as $$\sum_{n=1}^\infty\frac{n}{(n-1)!},$$ but I am unable to analyze this.
 A: You got a very good start. Now rewrite the numerator $n$ as $(n-1)+1$. Our sum is equal to
$$\sum_1^\infty \frac{n-1}{(n-1)!}+\sum_1^\infty \frac{1}{(n-1)!}.$$
By the method you already used, the first sum is $\sum_2^\infty \frac{1}{(n-2)!}$. This sum is $e$. More simply, the second sum is $e$.
A: Factorials in denominator are always appealing. So start writing $$S=\sum_{n=0}^\infty\frac{ n^2}{n!}x^n=\sum_{n=0}^\infty\frac{ n(n-1)+n}{n!}x^n=x^2\sum_{n=0}^\infty\frac{ n(n-1)}{n!}x^{n-2}+x\sum_{n=0}^\infty\frac{ n}{n!}x^{n-1}$$  Recognize that the first sum is the second derivative of $\sum_{n=0}^\infty\frac{ x^n}{n!}=e^x$ and that the second sum is the first derivative of the same quantity. So $$S=x^2 e^x+x e^x=x(x+1)e^x$$ Now, replace $x$ by $1$ and you are done.
A: Hint We can (after reindexing and checking convergence) write the sum as the value of the series
$$\sum_{k = 0}^{\infty} \frac{n + 1}{n!} x^n$$
at $x = 1$.

 Additional hint The appearance of $(n + 1) x^n$ in the summand suggests that we rewrite this as $$\frac{d}{dx} \sum_{k = 0}^{\infty} \frac{1}{n!} x^{n + 1} = \frac{d}{dx}\left(x \sum_{k = 0}^{\infty} \frac{1}{n!} x^n\right).$$

A: $$
\sum_{n=0}^\infty\frac{n^2}{n!} = \sum_{n=1}^\infty\frac{n}{(n-1)!}=\sum_{n=0}^\infty\frac{n+1}{n!}=\frac{d}{dx}xe^x|_{x=1}=2e
$$
