Question

Find the sum:

$\frac{2}{1!}+\frac{4}{3!}+\frac{6}{5!}+\dots \dots \infty$

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

Answer 1

The correct option is A $e$

$e^x=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+....$

$\frac{2}{1!}+\frac{4}{3!}+\frac{6}{5!}+\dots \dots \infty$
$=\sum _{i=1}^{\infty }\frac{2i}{(2i-1)!}={\sum }_{i=1}^{\infty }\left(\frac{2i-1+1}{(2i-1)!}\right)={\sum }_{i=1}^{\infty }[\frac{1}{(2i-2)!}+\frac{1}{(2i+1)!}]$

$=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+.....\infty$

$=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+......\infty$

$=e$