Solveeit Logo
Login

Question

Question: \[\frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + .........\infty =\]...

23!+45!+67!+.........∞=\frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + .........\infty =

A

e

B

2 e

C

e2e^{2}

D

1/e

Answer

1/e

Explanation

Solution

HereTn=(2n+1)−1(2n+1)!=1(2n)!−1(2n+1)!T_{n} = \frac{(2n + 1) - 1}{(2n + 1)!} = \frac{1}{(2n)!} - \frac{1}{(2n + 1)!}

⇒S=∑n=1∞Tn=(12!+14!+16!+.......∞)−(13!+15!+17!+.......∞)⇒S=(e+e−12−1)−(e−e−12−1)⇒e−1=1e\Rightarrow S = \sum_{n = 1}^{\infty}T_{n} = \left( \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + .......\infty \right) - \left( \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + .......\infty \right) \Rightarrow S = \left( \frac{e + e^{- 1}}{2} - 1 \right) - \left( \frac{e - e^{- 1}}{2} - 1 \right) \Rightarrow e^{- 1} = \frac{1}{e}