Question

$\sqrt{e}-\frac{x}{5}=\frac{1}{1.2}+\frac{1.3}{\mathrm{1.2.3.4}}+\frac{\mathrm{1.3.5}}{\mathrm{1.2.3.4.5.6}}+...$

Find $x$

Answer 1

Let $T_n$ be nth term of the given series, then

$T_n=\frac{\mathrm{1.3.5....}(2n-1)}{\left(2n\right)!}$

$=\frac{\mathrm{1.2.3.4.5....}(2n-1)2n}{\left(2n\right)!\left(\mathrm{2.4.6....}2n\right)}$

$=\frac{\left(2n\right)!}{\left(2n\right)!2^{n}n!}$

$=\frac{1}{2^{n}n!}$

$=\frac{(1/2{)}^n}{n!}$

Hence sum of the series

$S_{\infty }=\sum _{n=1}^{\infty }{T}_n$

$=\sum _{n=1}^{\infty }\frac{(1/2{)}^n}{n!}$

$=[\frac{(1/2{)}^1}{1!}+\frac{(1/2{)}^2}{2!}+\frac{(1/2{)}^3}{3!}+...]$

$=e^{1/2}-1$

$=\sqrt{e}-1$

$\therefore x=5$