Thread: Find the sum of this series

1. Find the sum of this series

Find the sum of the following series:

$\displaystyle 2+\frac{4}{2!}+\frac{8}{3!}+\frac{16}{4!}+...$

To two decimal places.

2. Originally Posted by ZosoPage
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

$\displaystyle e^2-1=2+\frac{4}{2!}+\frac{8}{3!}+\frac{16}{4!}+...$

3. Originally Posted by ZosoPage
Find the sum of the following series:

$\displaystyle 2+\frac{4}{2!}+\frac{8}{3!}+\frac{16}{4!}+...$

To two decimal places.
$\displaystyle e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$

$\displaystyle e^x - 1 = x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$

let $\displaystyle x = 2$

4. Originally Posted by ZosoPage
Find the sum of the following series:

$\displaystyle 2+\frac{4}{2!}+\frac{8}{3!}+\frac{16}{4!}+...$

To two decimal places.
This series looks like

$\displaystyle \frac{2}{1!}+\frac{4}{2!}+\frac{8}{3!}+\frac{16}{4 !}+...$

$\displaystyle \frac{2^1}{1!}+\frac{2^2}{2!}+\frac{2^3}{3!}+\frac {2^4}{4!}+...$

$\displaystyle \sum_{n=1}^{\infty} \frac{2^n}{n!}$

I hope this helps!

5. So the sum is $\displaystyle e^2-1$?

6. Originally Posted by ZosoPage
So the sum is $\displaystyle e^2-1$?
Yes that is the exact sum