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Question: Sum of the infinite series \(1 + 2 + \frac{1}{2!} + \frac{2}{3!} + \frac{1}{4!} + \frac{2}{5!} + ......

Sum of the infinite series 1+2+12!+23!+14!+25!+.....1 + 2 + \frac{1}{2!} + \frac{2}{3!} + \frac{1}{4!} + \frac{2}{5!} + ..... is.

A

e2e^{2}

B

e+e−1e + e^{- 1}

C

e−e−12\frac{e - e^{- 1}}{2}

D

3e−e−12\frac{3e - e^{- 1}}{2}

Answer

3e−e−12\frac{3e - e^{- 1}}{2}

Explanation

Solution

Sum of series

x21.2+x43.4+x65.6+....∞\frac{x^{2}}{1.2} + \frac{x^{4}}{3.4} + \frac{x^{6}}{5.6} + ....\infty

2[x21.2+x43.4+x65.6+..∞]2\left\lbrack \frac{x^{2}}{1.2} + \frac{x^{4}}{3.4} + \frac{x^{6}}{5.6} + ..\infty \right\rbrack

= 2log⁡ex−log⁡e(x+1)−log⁡e(x−1)2\log_{e}x - \log_{e}(x + 1) - \log_{e}(x - 1).