exponential function - Efficient method to evaluate the following series: $sum_{n=1}^infty frac{n^2cdot (n+1)^2}{n!}$

How do I calculate the infinite series:

$$\frac{1^2\cdot 2^2}{1!}+\frac{2^2\cdot 3^2}{2!}+\dots \quad?$$
I tried to find the nth term $t_n$. $$t_n=\frac{n^2\cdot (n+1)^2}{n!}.$$
So, $$\sum_{n=1}^{\infty}t_n=\sum_{n=1}^{\infty}\frac{n^4}{n!}+2\sum_{n=1}^{\infty}\frac{n^3}{n!}+\sum_{n=1}^{\infty}\frac{n^2}{n!}$$
after expanding. But I do not know what to do next.

Thanks.

Answer

You are right, now you need to expand them separately and express each of them in form of $e$:

$$ \sum \limits_{n=1}^{\infty}\frac{n^2}{n!}= \sum \limits_{n=1}^{\infty} \frac{n+(n-1)n}{n!} = \sum \limits_{n=1}^{\infty} \frac 1{(n-1)!} +\frac 1{(n-2)!} = 2e $$

Similarly, we can show that,

$$\sum \limits_{n=1}^{\infty}\frac{n^3}{n!}= \sum \limits_{n=1}^{\infty} \frac{n+(n^2-1)n}{n!} = 5e$$

and,

$$\sum \limits_{n=1}^{\infty}\frac{n^4}{n!}= 15e$$