The sum of the series $\frac{1^{2}}{3!} + \frac{2^{2}}{4!} + \frac{3^{2}}{5!} + \ldots to\ \infty$ equals

Question

Dr. Abubacker · Accepted Answer

(c)
Here, $T_{n} = \frac{n^{2}}{(n + 2)!} = \frac{n^{2} - 4 + 4}{(n + 2)!}$
$= \frac{(n - 2)}{(n + 1)!} + \frac{4}{(n + 2)!}$
$\Rightarrow T_{n} = \frac{1}{n!} - \frac{3}{(n + 1)!} + \frac{4}{(n + 2)!}$
$	herefore S = \Sigma T_{n} = \sum_{n = 1}^{\infty}\frac{1}{n!} - 3\sum_{n = 1}^{\infty}{\frac{1}{(n + 1)!} + 4\ \sum_{n = 1}^{\infty}\frac{1}{(n + 2)!}}\$
$= (e - 1) - 3(e - 2) + 4\left( e - 2 - \frac{1}{2} ight)$
$= 2e - 5$

Q. The sum of the series $\frac{1^{2}}{3!} + \frac{2^{2}}{4!} + \frac{3^{2}}{5!} + \ldots to\ \infty$ equals

Explanation

Reviewed by Dr. Abubacker

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