Question

Evaluate the following

$\displaystyle \sum_{n=0}^{\infty} \frac{ {n}^{3} }{n !}$
​

Answer 1

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

Before we start the question, let recall

$\boxed{\tt{ \: \: \: \displaystyle \sum_{n=0}^{\infty} \frac{1}{n !} = e \: \: \: }}$

Now,

Given expression is

$\rm :\longmapsto\:\displaystyle \sum_{n=0}^{\infty} \frac{ {n}^{3} }{n !}$

$\rm \:  =  \: \displaystyle \sum_{n=0}^{\infty} \frac{ {n}^{3} }{n(n - 1)!}$

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

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

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

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

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

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

$\rm \:  =  \: \displaystyle \sum_{n=0}^{\infty} \frac{n - 2+ 3}{(n - 2)!} + e$

$\rm \:  =  \: \displaystyle \sum_{n=0}^{\infty} \frac{n - 2}{(n - 2)!} + \displaystyle \sum_{n=0}^{\infty} \frac{3}{(n - 2) !} + e$

$\rm \:  =  \: \displaystyle \sum_{n=0}^{\infty} \frac{n - 2}{(n - 2)(n - 3)!} + \displaystyle \sum_{n=2}^{\infty} \frac{3}{(n - 2) !} + e$

$\rm \:  =  \: \displaystyle \sum_{n=0}^{\infty} \frac{1}{(n - 3)!} + 3e + e$

$\rm \:  =  \: \displaystyle \sum_{n=3}^{\infty} \frac{1}{(n - 3)!} + 3e + e$

$\rm \:  =  \: e+ 4e$

$\rm \:  =  \: 5e$

Hence

$\purple{\rm\implies \: \: \boxed{\tt{ \: \: \displaystyle \sum_{n=0}^{\infty} \frac{ {n}^{3} }{n !} \ = \: 5e \: \: }}}$

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MORE TO KNOW

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

$\rm :\longmapsto\:\boxed{\tt{ {e}^{ - x} \: = \: \displaystyle \sum_{n=0}^{\infty} \frac{ {( - 1)}^{n} {x}^{n} }{n !}}}$

$\rm :\longmapsto\:\boxed{\tt{ e = \displaystyle \sum_{n=0}^{\infty} \frac{1}{n !} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + - - - \infty }}$

$\rm :\longmapsto\:\boxed{\tt{ {e}^{ - 1} = \displaystyle \sum_{n=0}^{\infty} \frac{ {( - 1)}^{n} }{n !} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - - - \infty }}$

Answer 2

Answer:

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

Before we start the question, let recall

$\boxed{\tt{ \: \: \: \displaystyle \sum_{n=0}^{\infty} \frac{1}{n !} = e \: \: \: }}$

Now,

Given expression is

$\rm :\longmapsto\:\displaystyle \sum_{n=0}^{\infty} \frac{ {n}^{3} }{n !}$

$\rm \:  =  \: \displaystyle \sum_{n=0}^{\infty} \frac{ {n}^{3} }{n(n - 1)!}$

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

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

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

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

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

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

$\rm \:  =  \: \displaystyle \sum_{n=0}^{\infty} \frac{n - 2+ 3}{(n - 2)!} + e$

$\rm \:  =  \: \displaystyle \sum_{n=0}^{\infty} \frac{n - 2}{(n - 2)!} + \displaystyle \sum_{n=0}^{\infty} \frac{3}{(n - 2) !} + e$

$\rm \:  =  \: \displaystyle \sum_{n=0}^{\infty} \frac{n - 2}{(n - 2)(n - 3)!} + \displaystyle \sum_{n=2}^{\infty} \frac{3}{(n - 2) !} + e$

$\rm \:  =  \: \displaystyle \sum_{n=0}^{\infty} \frac{1}{(n - 3)!} + 3e + e$

$\rm \:  =  \: \displaystyle \sum_{n=3}^{\infty} \frac{1}{(n - 3)!} + 3e + e$

$\rm \:  =  \: e+ 4e$

$\rm \:  =  \: 5e$

Hence

$\purple{\rm\implies \: \: \boxed{\tt{ \: \: \displaystyle \sum_{n=0}^{\infty} \frac{ {n}^{3} }{n !} \ = \: 5e \: \: }}}$

$\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\pink{\rule{45pt}{7pt}}\blue{\rule{45pt}{7pt}}$