Find the sum of the following series:
$\displaystyle 2+\frac{4}{2!}+\frac{8}{3!}+\frac{16}{4!}+...$
To two decimal places.
Remember that
$\displaystyle e^{x}=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{ x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$
so
$\displaystyle e^2=1+2+\frac{4}{2!}+\frac{8}{3!}+\frac{16}{4!}+.. .$
so if you subtract 1 from both sides you get
$\displaystyle e^2-1=2+\frac{4}{2!}+\frac{8}{3!}+\frac{16}{4!}+...$