# Thread: Convergence of a series

1. ## Convergence of a series

Hi Am working through heaps of series problems in which i'm expected to show the series converges or diverges by applying both root and ratio test,

For the series;
$\displaystyle \sum_{n=1}^{Inf} \frac{2^n}{n!}$

I went ahead and applied to ratio test getting;

limit as n --> inf of 2/(n+1) = 0 hence 0 < 1 means convergence,

but i then applied the root test and ended up with;
lim as n--> inf of 2/((n!)^(1/n)) which i figured to be = 2/1 hence > 1 and divergent,

Clearly i ahve stuffed one of these up and i'm a bit sketchy with limits so could be there but can someone tell me which i've stuffed up and how?

2. Originally Posted by monster
Hi Am working through heaps of series problems in which i'm expected to show the series converges or diverges by applying both root and ratio test,

For the series;
$\displaystyle \sum_{n=1}^{Inf} \frac{2^n}{n!}$

I went ahead and applied to ratio test getting;

limit as n --> inf of 2/(n+1) = 0 hence 0 < 1 means convergence,
ok.

but i then applied the root test and ended up with;
lim as n--> inf of 2/((n!)^(1/n)) which i figured to be = 2/1 hence > 1 and divergent,
No, you are wrong about the limit of $\displaystyle \sqrt[n]{n!}$ as $\displaystyle n\to\infty$.

One way to see this is to know the so called "Sterling Approximation" of $\displaystyle n!$ to be $\displaystyle n!\sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$. From this it follows that $\displaystyle \sqrt[n]{n!}\to\infty$ for $\displaystyle n\to\infty$.

3. With little of patience You can find that is...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$