Question

$\frac{2}{1!}+\frac{4}{3!}+\frac{6}{5!}+\frac{8}{7!}+...to\infty =\frac{1}{\frac{2}{a!}+\frac{4}{5!}+\frac{6}{b!}+...to\infty }$ Find $b-a$

Answer 1

Given $\frac{2}{1!}+\frac{4}{3!}+\frac{6}{5!}+\frac{8}{7!}+...to\infty =\frac{1}{\frac{2}{a!}+\frac{4}{5!}+\frac{6}{b!}+...to\infty }$ .....(1)
Consider, $\frac{2}{1!}+\frac{4}{3!}+\frac{6}{5!}+\frac{8}{7!}+...to\infty$
$T_n=\frac{2n}{(2n-1)!}$
$T_n=\frac{2n-1+1}{(2n-1)!}$
$\Rightarrow T_n=\frac{1}{(2n-2)!}+\frac{1}{(2n-1)!}$
$\Rightarrow T_n=\frac{e+e^{-1}}{2}+\frac{e-e^{-1}}{2}$
$\Rightarrow T_n=e$
So, eqn (1), becomes
$e=\frac{1}{\frac{2}{a!}+\frac{4}{5!}+\frac{6}{b!}+...to\infty }$
$\Rightarrow \frac{2}{a!}+\frac{4}{5!}+\frac{6}{b!}+...to\infty =\frac{1}{e}$
$\Rightarrow \frac{2}{a!}+\frac{4}{5!}+\frac{6}{b!}+...to\infty =1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!-\frac{1}{7!}}+....\infty$
$\Rightarrow \frac{2}{a!}+\frac{4}{5!}+\frac{6}{b!}+...to\infty =\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+....\infty$
On comparing, we get
$a=3,b=7$
$\Rightarrow b-a=4$