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.
Given that $P(x)$ is a polynomial of degree $n$, we have $$P(x)=\sum_{k=0}^n\binom xk\Delta^kP(0)$$ which is Newton's series. Therefore \begin{align} \sum_{m\ge0}\frac{P(m)}{m!}&=\sum_{m\ge0}\sum_{k=0}^n\frac{\Delta^kP(0)}{k!(m-k)!}\\ &=\sum_{k=0}^n\frac{\Delta^kP(0)}{k!}\sum_{m\ge k}\frac1{(m-k)!}\\ &=e\sum_{k=0}^n\frac{\Delta^kP(0)}{k!} \end{align}
The forms of the power series, you reached in the last step of your work, are known as Bell numbers. Bell numbers are defined as:
$$ B_{m} = \frac{1}{e}\,\sum_{n=0}^{\infty} \frac{n^m}{n!} $$.
So your result can be written in terms of Bell numbers:
$$ e( B_{4} + 2\,B_{3} + B_{2})\,, $$
where $ B_{4}=15 \,, B_{3}=5\,, B_{2}=2\,. $ See here.
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,
$$\sum \limits_{n=1}^{\infty}\frac{n^4}{n!}= 15e$$
