Question

Evaluate the following

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

Answer 1

$\large\underline{\sf{Solution-}}$

Before start solving the problem, Lets recall

We know that,

$\rm :\longmapsto\: {e}^{x} = 1 + x + \dfrac{ {x}^{2} }{2!} + \dfrac{ {x}^{3} }{3!} + - - - -$

can be further expressed as

$\rm :\longmapsto\: {e}^{x} = \displaystyle\sum_{n=0}^{\infty} \frac{ {x}^{n} }{n!}$

So, on substituting x = 1, we get

$\rm :\longmapsto\: e = \displaystyle\sum_{n=0}^{\infty} \frac{1}{n!} = 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + - - - -$

Let's solve the problem now!!!

Given expression is

$\rm :\longmapsto\:\displaystyle\sum_{n=0}^{\infty} \frac{ {n}^{2} }{n!}$

which can be rewritten as

$\rm \: = \: \displaystyle\sum_{n=0}^{\infty} \frac{ {n}^{2} }{n(n - 1)!}$

$\rm \: = \: \displaystyle\sum_{n=0}^{\infty} \frac{ n }{(n - 1)!}$

can be rewritten as

$\rm \: = \: \displaystyle\sum_{n=0}^{\infty} \frac{ n - 1 + 1}{(n - 1)!}$

can be further split as

$\rm \: = \: \displaystyle\sum_{n=0}^{\infty} \frac{ n - 1}{(n - 1)!} + \displaystyle\sum_{n=0}^{\infty} \frac{1}{(n - 1)!}$

$\rm \: = \: \displaystyle\sum_{n=0}^{\infty} \frac{ n - 1}{(n - 1)(n - 2)!} +\bigg(1 + \dfrac{1}{1!} + \dfrac{1}{2!} + - - \bigg)$

$\rm \: = \: \displaystyle\sum_{n=0}^{\infty} \frac{1}{(n - 2)!} +e$

$\rm \: = \: \bigg(1 + \dfrac{1}{1!} + \dfrac{1}{2!} + - - \bigg) + e$

$\rm \: = \: e + e$

$\rm \: = \: 2e$

Hence,

$\\ \rm :\longmapsto\: \: \: \boxed{\tt{ \: \: \: \: \: \displaystyle\sum_{n=0}^{\infty} \frac{ {n}^{2} }{n!} = 2e \: \: \: \: \: }}$

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Explore more

$\rm :\longmapsto\: e = \displaystyle\sum_{n=0}^{\infty} \frac{1}{n!} = 1 + \dfrac{1}{1!} + \dfrac{1}{2!} + - - - -$

$\rm :\longmapsto\: {e}^{ - 1} = \displaystyle\sum_{n=0}^{\infty} \frac{ {( - 1)}^{n - 1} }{n!} = 1 - \dfrac{1}{1!} + \dfrac{1}{2!} - - - - -$

$\rm :\longmapsto\:\dfrac{e + {e}^{ - 1} }{2} = 1 + \dfrac{1}{2!} + \dfrac{1}{4!} + - - -$

$\rm :\longmapsto\:\dfrac{e - {e}^{ - 1} }{2} = 1 + \dfrac{1}{3!} + \dfrac{1}{5!} + - - -$