1. ## Power Series Convergence

Hello!
I want to find the radius of convergence of the following power series:

$\displaystyle \sum^{\infty}_{n=1} \frac{1}{\sqrt{n}} x^n$

using the ratio test, $\displaystyle \frac{S_{n+1}}{S_{n}}$:

$\displaystyle \frac{1}{\sqrt{n+1}} x^{n+1} / \frac{1}{\sqrt{n}} x^n$

$\displaystyle = \sqrt{\frac{n}{n+1}} |x|$

Now I'm confused, is the limit as n->infinity = 1 or 0?

Because $\displaystyle lim_{n \rightarrow \infty} \sqrt{\frac{n}{n+1}}=0$
0.|x| = 0 !?

Or do we just ignore the $\displaystyle \frac{1}{\sqrt{n}}$part and say the interval of convergence is |x|<1 and the radius is 1 ?

2. Originally Posted by Roam
Hello!
I want to find the radius of convergence of the following power series:

$\displaystyle \sum^{\infty}_{n=1} \frac{1}{\sqrt{n}} x^n$

using the ratio test, $\displaystyle \frac{S_{n+1}}{S_{n}}$:

$\displaystyle \frac{1}{\sqrt{n+1}} x^{n+1} / \frac{1}{\sqrt{n}} x^n$

$\displaystyle = \sqrt{\frac{n}{n+1}} |x|$

Now I'm confused, is the limit as n->infinity = 1 or 0?

Because $\displaystyle lim_{n \rightarrow \infty} \sqrt{\frac{n}{n+1}}=0$
0.|x| = 0 !?

Or do we just ignore the $\displaystyle \frac{1}{\sqrt{n}}$part and say the interval of convergence is |x|<1 and the radius is 1 ?
$\displaystyle \lim_{n \to \infty}\sqrt{\frac{n}{n+1}}$

$\displaystyle \lim_{n \to \infty}\sqrt{\frac{1}{1+\frac{1}{n}}}=\sqrt{\frac{ 1}{1+0}}=1$