Question

The sum of the series $\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+...\infty$ is $e^{-b^2}$. Find $b$
(Note : $b>0$)

Answer 1

The general term or $n$th term of the series
$T_n=\frac{2n}{(2n+1)!}$
Then dividing $2n$ by $(2n+1)$
$\therefore$ $T_n=\frac{(2n+1)-1}{(2n+1)!}$
$=\frac{(2n+1)}{(2n+1)!}-\frac{1}{(2n+1)!}$
$=\frac{(2n+1)}{(2n+1)\left(2n\right)!}-\frac{1}{(2n+1)!}$
$T_n=\frac{1}{\left(2n\right)!}-\frac{1}{(2n+1)!}$
$\therefore$ Sum of the series
$S=\sum _{n=1}^{\infty }{T}_n$
$=\sum _{n=1}^{\infty }\left(\frac{1}{\left(2n\right)!}-\frac{1}{(2n+1)!}\right)$

$=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}+...\infty$ $=1-1+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}+...\infty =e^{-1}$

Alternative Method :

$\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+...=\frac{3-1}{3!}+\frac{5-1}{5!}+\frac{7-1}{7!}+...$
$=\frac{3}{3!}-\frac{1}{3!}+\frac{5}{5!}-\frac{1}{5!}+\frac{7}{7!}-\frac{1}{7!}+...$

$=\frac{3}{3.2!}-\frac{1}{3!}+\frac{5}{5.4!}-\frac{1}{5!}+\frac{7}{7.6!}-\frac{1}{7!}+...$
$=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}+...$

$=1-1+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}+...\infty =e^{-1}$
Hence $b=1$