The discussion centers around proving the identity $$\sum_{n=0}^\infty \frac{n^2}{n!}=2e$$. Participants explore various approaches to this mathematical series, focusing on its derivation and implications.
While there is general agreement on the identity being discussed, multiple competing views and approaches to the proof remain, and the discussion does not reach a consensus on a single method.
Participants do not clarify the assumptions or specific methods used in their proofs, leaving some steps and reasoning potentially unresolved.
Readers interested in mathematical series, proofs in combinatorics, or those studying exponential functions may find this discussion relevant.
lfdahl said:Prove that
$$\sum_{n=0}^\infty \frac{n^2}{n!}=2e.$$
kaliprasad said:$\sum_{n=0}^{\infty}\frac{n^2}{n!}= \sum_{n=0}^{\infty}\frac{n(n-1)+n}{n!}= \sum_{n=0}^{\infty}\frac{n(n-1)+n}{n!}$
$=\sum_{n=1}^{\infty}(\frac{n(n-1)}{n!} +\frac{n}{n!})$
$=\sum_{n=2}^{\infty}(\frac{n(n-1)}{n!}) +\sum_{n=1}^{\infty}\frac{n}{n!}$ as 1st 2 terms in 1st sum are zero
$=\sum_{n=2}^{\infty}\frac{1}{(n-2)!} +\sum_{n=1}^{\infty}\frac{1}{(n-1)!}$
$=\sum_{n=0}^{\infty}\frac{1}{n!} +\sum_{n=0}^{\infty}\frac{1}{n!}$
$=e + e = 2e$
I like Serena said:Nice solution kaliprasad!
Just to provide another one:
Let $f(x)=\sum\limits_{n=1}^\infty \frac{n^2 x^{n-1}}{n!}$.
Note that the requested sum is equal to $f(1)$, which intentionally starts at $n=1$, possible since the first term is $0$ anyway.
Then $F(x)=\int_0^x f(x) = \sum \frac{n x^n}{n!} = x\sum \frac{x^{n-1}}{(n-1)!} = xe^x$.
Therefore $f(x)=F'(x)=e^x+xe^x$.
Thus the requested sum is $f(1)=2e. \ \blacksquare$