$1 + \frac{2^{3}}{2!} + \frac{3^{3}}{3!} + \frac{4^{3}}{4!} + \ldots\infty$ equals

Question

Dr. Abubacker · Accepted Answer

(a)
Here, $T_{n} = \frac{n^{3}}{n!} = \frac{n^{2} - 1}{(n - 1)!} + \frac{1}{(n - 1)!}$
$= \frac{n + 1}{(n - 2)!} + \frac{1}{(n - 1)!}$
$= \frac{1}{(n - 3)!} + \frac{3}{(n - 2)!} + \frac{1}{(n - 1)!}$
$\Rightarrow \sum_{}^{}{T_{n} = \sum_{}^{}\left( \frac{1}{(n - 3)!} + \frac{3}{(n - 2)!} + \frac{1}{(n - 1)!} \right)}$
$= e + 3e + e = 5e$

Q. $1 + \frac{2^{3}}{2!} + \frac{3^{3}}{3!} + \frac{4^{3}}{4!} + \ldots\infty$ equals

Explanation

Reviewed by Dr. Abubacker

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