Q. The sum of the series $\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \ldots$ is equal to
A) $\frac{(e^{2} - 1)}{2}$
B) $\frac{(e - 1)^{2}}{2e}$
C) $\frac{(e^{2} - 1)}{2e}$
D) $\frac{(e^{2} - 2)}{e}$
Explanation
(b)
We know that,
$e + e^{- 1} = 2 + \frac{2}{2!} + \frac{2}{4!} + ...\infty$
$\Rightarrow \frac{e^{2} + 1 - 2e}{e} = 2\left( \frac{1}{2!} + \frac{1}{4!} + ...\infty \right)$
$\Rightarrow \frac{(e - 1)^{2}}{2e} = \frac{1}{2!} + \frac{1}{4!} + ...\infty$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.