Q. The value of $\frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \ldots$ is
A) $e^{\frac{1}{2}}$
B) $e^{- 1}$
C) $e$
D) $e^{- \frac{1}{3}}$
Explanation
(b)
Here, $T_{n} = \frac{2n}{(2n + 1)!} = \frac{1}{2n!} - \frac{1}{(2n + 1)!}$
$\sum_{}^{}T_{n} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + ...$
$= 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + ...$
$= e^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.