Answer to Question #143646 in Calculus for 1805040444b

Question

Prove this identity between two infinite sums (with x ∈ R and n! stands for factorial):
(∞Σn=0x^n/n!)^2=∞Σn=0(2x)^n/n!

Answer 1

$\left(\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}\right)^2=\left(\sum\limits_{n=0}^{\infty}\frac{x^n}{n!}\right)\left(\sum\limits_{m=0}^{\infty}\frac{x^m}{m!}\right)=\sum\limits_{k=0}^{\infty}\sum\limits_{\stackrel{m,n}{m+n=k}}\frac{x^n}{n!}\frac{x^m}{m!}=$

$=\sum\limits_{k=0}^{\infty}\sum\limits_{n=0}^k\frac{x^n}{n!}\frac{x^{k-n}}{(k-n)!}=\sum\limits_{k=0}^{\infty}\left(\sum\limits_{n=0}^k\frac{1}{n!}\frac{1}{(k-n)!}\right)x^k=$

$=\sum\limits_{k=0}^{\infty}\left(\sum\limits_{n=0}^k1^n\binom{k}{n}1^{k-n}\right)\frac{x^k}{k!}=\sum\limits_{k=0}^{\infty}(1+1)^k\frac{x^k}{k!}=\sum\limits_{k=0}^{\infty}\frac{(2x)^k}{k!}$