Question

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

Answer 1

Answer:

My attempt:

I googled the problem, and I found that ∑∞(n=1)2n(2n−1)!=e

I also found that ∑∞n=02n+2(2n+1)! is equal to e.

By definition of e, we have

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)!=∑n=0∞2(n+1)(2n+1)!

(Note that ∑∞n=02(n+1)(2n+1)!=∑∞n=12n(2n−1)! by a simple change of indices.)Start from the known series for e

e=∑j=0∞1j!

and group every even term with the next one :

e=∑j=0∞1j!=∑k=0∞1(2k)!+1(2k+1)!=∑k=0∞2k+1(2k+1)!+1(2k+1)!=∑k

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