The sum of the series frac{1^2}{1 cdot 2!} + | vedclass

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Question

(C) The $n^{th}$ term of the series is $T_n = \frac{\sum_{k=1}^{n} k^2}{n(n+1)!}$.Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we g

Vedclass · Accepted Answer

$\frac{3e - 1}{6}$

Vedclass · Answer

(C) The $n^{th}$ term of the series is $T_n = \frac{\sum_{k=1}^{n} k^2}{n(n+1)!}$.Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:$T_n = \frac{n(n+1)(2n+1)}{6n(n+1)!} = \frac{2n+1}{6(n!)} = \frac{1}{6} \left( \frac{2n}{n!} + \frac{1}{n!} \right) = \frac{1}{6} \left( \frac{2}{(n-1)!} + \frac{1}{n!} \right)$.The sum $S = \sum_{n=1}^{\infty} T_n = \frac{1}{6} \left( 2 \sum_{n=1}^{\infty} \frac{1}{(n-1)!} + \sum_{n=1}^{\infty} \frac{1}{n!} \right)$.Since $\sum_{n=1}^{\infty} \frac{1}{(n-1)!} = e$ and $\sum_{n=1}^{\infty} \frac{1}{n!} = e - 1$,$S = \frac{1}{6} [2(e) + (e - 1)] = \frac{3e - 1}{6}$.

The sum of the series $\frac{1^2}{1 \cdot 2!} + \frac{1^2 + 2^2}{2 \cdot 3!} + \frac{1^2 + 2^2 + 3^2}{3 \cdot 4!} + \dots + \frac{1^2 + 2^2 + \dots + n^2}{n(n + 1)!} + \dots \infty$ is equal to:

Similar Questions

$\frac{1 + \frac{2^2}{2!} + \frac{2^4}{3!} + \frac{2^6}{4!} + \dots \infty}{1 + \frac{1}{2!} + \frac{2}{3!} + \frac{2^2}{4!} + \dots \infty} = $

The sum of the series $\frac{1}{2 !} + \frac{1+2}{3 !} + \frac{1+2+3}{4 !} + \ldots$ is equal to :

$1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dots \infty = \,$

If $S_n$ denotes the sum of the products of the first $n$ natural numbers taken two at a time,then $\sum\limits_{n = 0}^\infty \frac{S_n}{(n + 1)!} = $

$1 + \frac{a - b}{a} + \frac{1}{2!} \left( \frac{a - b}{a} \right)^2 + \frac{1}{3!} \left( \frac{a - b}{a} \right)^3 + \dots \infty = $

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