Question

Prove that:

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

Answer 1

Given the series is $\frac{2}{1!}+\frac{4}{3!}+\frac{6}{5!}+.....+\infty$.

So the $n$ th term of the series is $t_n=\frac{2n}{(2n-1)!}$ can be written as,

$t_n=\frac{2n}{(2n-1)!}$

or, $t_n=\frac{2n-1+1}{(2n-1)!}$

$t_n=\frac{1}{(2n-2)!}$$+\frac{1}{(2n-1)!}$.

Now the required series is

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

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

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

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

$=e$.