Prove the sum identity ∑n2n=2e.

[lfdahl]

Prove that

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

 

[kaliprasad]

Prove that

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

Spoiler

$\sum_{n=0}^{\infty}\frac{n^2}{n!}= \sum_{n=0}^{\infty}\frac{n(n-1)+n}{n!}= \sum_{n=0}^{\infty}\frac{n(n-1)+n}{n!}$
$=\sum_{n=1}^{\infty}(\frac{n(n-1)}{n!} +\frac{n}{n!})$
$=\sum_{n=2}^{\infty}(\frac{n(n-1)}{n!}) +\sum_{n=1}^{\infty}\frac{n}{n!}$ as 1st 2 terms in 1st sum are zero
$=\sum_{n=2}^{\infty}\frac{1}{(n-2)!} +\sum_{n=1}^{\infty}\frac{1}{(n-1)!}$
$=\sum_{n=0}^{\infty}\frac{1}{n!} +\sum_{n=0}^{\infty}\frac{1}{n!}$
$=e + e = 2e$

 

[lfdahl]

Spoiler

$\sum_{n=0}^{\infty}\frac{n^2}{n!}= \sum_{n=0}^{\infty}\frac{n(n-1)+n}{n!}= \sum_{n=0}^{\infty}\frac{n(n-1)+n}{n!}$
$=\sum_{n=1}^{\infty}(\frac{n(n-1)}{n!} +\frac{n}{n!})$
$=\sum_{n=2}^{\infty}(\frac{n(n-1)}{n!}) +\sum_{n=1}^{\infty}\frac{n}{n!}$ as 1st 2 terms in 1st sum are zero
$=\sum_{n=2}^{\infty}\frac{1}{(n-2)!} +\sum_{n=1}^{\infty}\frac{1}{(n-1)!}$
$=\sum_{n=0}^{\infty}\frac{1}{n!} +\sum_{n=0}^{\infty}\frac{1}{n!}$
$=e + e = 2e$

Well done, kaliprasad! Thankyou very much for your participation and a fine solution!(Yes)

 

[I like Serena]

Nice solution kaliprasad!

Just to provide another one:

Spoiler

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 $f(1)=2e. \ \blacksquare$

 

[lfdahl]

Nice solution kaliprasad!

Just to provide another one:

Spoiler

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 $f(1)=2e. \ \blacksquare$

Another very fine solution, I like Serena! Thankyou very much for your elegant approach!(Handshake)