Power Series Convergence

**Roam** · May 3rd 2009, 07:49 PM

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

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

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

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

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

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

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

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

**TheEmptySet** · May 3rd 2009, 07:57 PM

Originally Posted by Roam

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

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

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

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

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

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

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

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

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

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

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