Question

prove that 2/1!+4/3!+6/5!+... ... ... .=e​

Answer 1

$\red{TO \: PROVE} :$

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

Solution:

$\\ \sf\sum _{k = 1}^{ \infin} \frac{2k}{(2k - 1)!}$

$\\ \sf= \lim_{ n\to \infty } \sum_{k = 1} ^{n} \frac{2k}{(2k - 1)!}$

$\\ = \sf 2 \lim _{n \to \infty } \sum _{k = 0} ^{ \infty } \frac{k + 1}{(2k + 1)!}$

$\\ = \sf2 \lim _{n \to \infty }(e - \frac{2 _{1}F_{2}(2 ; n + 2 ; n + \frac{5}{2} ; \frac{1}{4} ) }{(2n + 3)! }- \frac{2(n + 1) _{1}F_{2}(1; n + 1 ;n + \frac{3}{2} ; \frac{1}{4} ) }{(2n + 1)! })$

$= \sf 2(e - \frac{e}{2} )$

$= e$

$\bf \square \: proved$