Question

The sum of the series $\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+...$ is equal to

• A. $2e^{-2}$
• B. $e^{-2}$
• C. $e^{-1}$
• D. $2e^{-1}$

Answer 1

The correct option is C $e^{-1}$

General term of given series is,

$T_n=\frac{2n}{(2n+1)!}=\frac{2n+1-1}{(2n+1)!}=\frac{1}{\left(2n\right)!}-\frac{1}{(2n+1)!}$

Thus required summation is,

$=\sum _{n=1}^{\infty }{T}_n=(\frac{1}{2!}-\frac{1}{3!})+(\frac{1}{4!}-\frac{1}{5!})+(\frac{1}{6!}-\frac{1}{7!})..........\infty$ ......(1)

Now we know $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...........\infty$

Putting $x=-1$ in above equation we get,

$e^{-1}=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-...........\infty$

Thus from (1) required summation of the given series is $e^{-1}$