Question

Sum the infinite series ((1)^2)/0! + ((2^2)/1! + ((3^2)/2! + ((4^2)/3! + ...

Answer 1

do you have any idea?

Answer 2

i know this part

Answer 3

\[\large \begin{align} \sum_{n=0}^{\infty} \frac{(n+1)^2}{n!}&=\sum_{n=0}^{\infty}\left(\frac{n^2+2n+1}{n!} \right)\\&=\sum_{n=0}^{\infty}\frac{n^2}{n!}+2\sum_{n=0}^{\infty}\frac{n}{n!}+\sum_{n=0}^{\infty}\frac{1}{n!}\\&=\sum_{n=0}^{\infty}\frac{n^2}{n(n-1)!} + 2\sum_{n=0}^{\infty}\frac{n}{n(n-1)!}+\underbrace{\sum_{n=0}^{\infty}\frac{1}{n!}}_{\huge =e} \\ &=\sum_{n=0}^{\infty}\frac{n}{(n-1)!}+2\sum_{n=0}^{\infty}\frac{1}{(n-1)!}+e \\ &=\sum_{n=0}^{\infty}\frac{n+2}{(n-1)!}+e \\
&=\sum_{n=-1}^{\infty}\frac{n+3}{n!}+e, \text{by index shifting} \\
&= \sum_{n=-1}^{\infty}\frac{n}{n!}+3\sum_{n=-1}^{\infty}\frac{1}{n!}+e\end{align} \]

Now, it's important to notice that the summation starts a a negative number. From combinatorics, the convention is that \(\large {n \choose r} = 0\) when \(r>n\), because the fraction has a negative factorial. So we can just re-index our sums to start at 0:
\[\large \begin{align} &= \sum_{n=0}^{\infty}\frac{n}{n!}+3\sum_{n=0}^{\infty}\frac{1}{n!}+e \\&= \underbrace{\sum_{n=0}^{\infty}\frac{n}{n!}}+3e+e\\
&=\underbrace{\frac{0}{0!}+\frac{1}{1!}+\frac{2}{2!}+\frac{3}{3!}+\frac{4}{4!}+\frac{5}{5!}+\ldots }+4e \\&=\underbrace{0+1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\ldots }_{\Large \text{series expansion for }e}+4e\\&=5e\end{align} \]