proof that$ \sum_{n=0}^{\infty} \frac{2^n}{n!} = \sum_{n=0}^{\infty} \frac{ e (2-1)^n}{n!} $

Question

I wanted to use series expansion of exp function. I have a doubt about what i found.

Knowing exp series has a radius of convergence infinite, we have that expansion serie center at 0 is $exp(x)=\sum_{n=0}^{\infty} \frac{x^n}{n!} $

So choosing x=2 it gives : $exp(2)=\sum_{n=0}^{\infty} \frac{2^n}{n!} $

Now i use maclaurin formula to determine exp exapnsion at c= 1. I obtained $ exp(x) =\sum_{n=0}^{\infty} \frac{ e (x-1)^n}{n!} $ So choosing x=2 it gives : $ exp(2) =\sum_{n=0}^{\infty} \frac{ e (2-1)^n}{n!} $

Finally, it means that

$ \sum_{n=0}^{\infty} \frac{2^n}{n!} = \sum_{n=0}^{\infty} \frac{ e}{n!} $

Answer 1

Simply note that

• $\sum_{0}^{\infty} \frac{2^n}{n!} = e^2$

and

• $\sum_{0}^{\infty} \frac{ e}{n!} =e\sum_{0}^{\infty} \frac{1}{n!} =e^2$

Answer 2

It may be helpful to prove the more general result that $\sum_{k\ge 0}\frac{x^k}{k!}\sum_{l\ge 0}\frac{y^l}{l!}=\sum_{n\ge 0}\frac{(x+y)^n}{n!}$. What is the $x^k y^l$ coefficient? On the left, $\frac{1}{k!l!}$; on the right, $\sum_{n=k+l}\frac{\binom{n}{k}}{n!}=\frac{\binom{k+l}{k}}{(k+l)!}=\frac{1}{k!l!}$.