Question

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

Answer 1

Numerator of R.H.S.= $1+\frac{2^2}{2!}+\frac{2^4}{3!}+\frac{2^6}{4!}+...$
=$\frac{1}{2^2}(2^2+\frac{2^4}{2!}+\frac{2^6}{3!}+\frac{2^8}{4!}+...\infty )$
=$\frac{1}{2^2}(\frac{2^2}{1!}+\frac{(2^2{)}^2}{2!}+\frac{(2^2{)}^3}{3!}+\frac{(2^2{)}^4}{4!}+...)$
=$\frac{1}{2^2}(e^{2^2}-1)$
=$\frac{e^4-1}{4}$ ....(1)
Denominator of R.H.S.= $1+\frac{1}{2!}+\frac{2}{3!}+\frac{2^2}{4!}+...\infty$
=$\frac{1}{2^2}(2^2+\frac{2^2}{2!}+\frac{2^3}{3!}+\frac{2^4}{4!}+...\infty )$
=$\frac{1}{4}(1+(1+2+\frac{2^2}{2!}+\frac{2^3}{3!}+\frac{2^4}{4!}+...))$
$(write{2}^2=1+1+2)$
=$\frac{1}{4}(1+e^2)$ . ...(2)
from (1) and (2);
$\frac{1+\frac{2^2}{2!}+\frac{2^4}{3!}+\frac{2^6}{4!}+...}{1+\frac{1}{2!}+\frac{2}{3!}+\frac{2^2}{4!}+...}$ = $\frac{\frac{1}{4}(e^4-1)}{\frac{1}{4}(e^2+1)}$ = $\frac{(e^2+1)(e^2-1)}{(e^2+1)}$ =$e^2-1$
Proved.