A calculus problem by Vishwesh Ramanathan

Using the series of $e^x$ . $e^x=1+x+\dfrac{x^2}{2!} + \dfrac{x^3}{3!} +\dfrac{x^4}{4!} + \cdots$ Multiplying x on the both sides. $xe^x = x+x^2+\dfrac{x^3}{2!} \dfrac{x^4}{3!}+\dfrac{x^5}{4!} +\cdots$ Diffrentiating both sides w.r.t $x$ and cancelling $dx$ on both sides. $xe^x+e^x=1+2x+\dfrac{3x^2}{2!}+\dfrac{4x^3}{3!}+\dfrac{5x^4}{4!}+\cdots$ Putting x=1: $e+e=1+2+\dfrac{3}{2!}+\dfrac{4}{3!}+\dfrac{5}{4!}+\cdots$ Rewritting RHS: $2e=1+\dfrac{2^2}{2!}+\dfrac{3^2}{3!}+\dfrac{4^2}{4!}+\cdots$ $5.45=\sum^{\infty}_{n=1} \dfrac{n^2}{n!}$

1^2/1! +2^2/2!+3^2/3!...
=1/0!+2/1!+3/2!+4/3!...
=(1/0!+0/0!)+(1/2!+2/2!)+(1/3!+3/3!)..
=1/0!+1/2!+1/3!.....+1/1!+2/2!+3/3!.. .
=e+1/0!+1/1!+1/2!... .
=e+e=2e.
=5.43656