Question

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

Answer 1

Step-by-step explanation:

By definition of e, we have

e=∑n=0∞1n!=∑n=0∞1(2n+1)!+∑n=0∞1(2n)!e=∑n=0∞1n!=∑n=0∞1(2n+1)!+∑n=0∞1(2n)!

by separating even and odd terms (both series still converge).

Bringing them together again,

e=∑n=0∞(1(2n+1)!+1(2n)!)=∑n=0∞(2n)!+(2n+1)!(2n+1)!(2n)!=∑n=0∞1+(2n+1)(2n+1)!=∑