• Aug 24th 2010, 02:51 PM
Revy
I need to find radius of convergence and interval of convergence of $\sum_{n=1}^{\inf}\frac{x^n}{\sqrt{n}}$

In d'Alembert feature I find radius like this $\lim_{n->\inf}\frac{x^{n+1}\sqrt{n}}{\sqrt{n+z}x^n}=|x|< 1$
$|x|<1$
$-1

If I try in Cauchy feature, I get same:
$\lim_{n->\inf}\sqrt{|\frac{x^n}{\sqrt{n}}|}<1$ first square $n$ root
$\lim_{n->\inf}|\frac{x}{\sqrt_{n+2}{n}}}|<1$ square $n+2$ root
$|x|<1$
$-1

Interval of convergence:
I take -1 and 1
$\sum_{n=1}^{\inf}\frac{x^2 * 1^n}{\sqrt{n}}$ and $\sum_{n=1}^{\inf}\frac{x^2 * (-1)^n}{\sqrt{n}}$

I do check if diverge/converge:
$\lim_{n->\inf}\frac{x^n}{\sqrt{n}}=\inf$ it diverge
$\lim_{n->\inf}|\frac{x^n}{\sqrt{n}}|!=0$ here != means 'not even'

Interval is $x\E(-1;1)$

Questions:
1. Did I do task correct?
2. $\lim_{n->\inf}|\frac{x^n}{\sqrt{n}}|!=0$ it diverge or converge?
3. How interval of convergence change because of divergence and convergence?
• Aug 24th 2010, 02:58 PM
Plato
You need to ask if $\displaystyle\sum\limits_n {\frac{{( - 1)^n }}
{{\sqrt n }}}$
converges.
• Aug 26th 2010, 02:51 AM
Revy
Quote:

Originally Posted by Plato
You need to ask if $\displaystyle\sum\limits_n {\frac{{( - 1)^n }}
{{\sqrt n }}}$
converges.

It doesn't have limit, but it still get two answers: 1 and -1. If it's like this, it is still said that it converges?

How does answer change because of that?
How would answer change if it converged with limit 1/2?
And how would answer change if it would converge with limit 2?
• Aug 26th 2010, 03:12 AM
Prove It
$a_{n + 1} = \frac{x^{n + 1}}{\sqrt{n + 1}}$ and $a_n = \frac{x^n}{\sqrt{n}}$.

By the ratio test, the series converges where

$\lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right| < 1$

$\lim_{n \to \infty}\left|\frac{\frac{x^{n + 1}}{\sqrt{n + 1}}}{\frac{x^n}{\sqrt{n}}}\right| < 1$

$\lim_{n \to \infty}\left|\frac{x\sqrt{n}}{\sqrt{n + 1}}\right| < 1$

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

$\lim_{n \to \infty}\sqrt{\frac{n + 1 - 1}{n + 1}}|x| < 1$

$\lim_{n \to \infty}\sqrt{1 - \frac{1}{n + 1}}|x| < 1$

$1|x| < 1$

$|x| < 1$.

So the radius of convergence is $1$ and the interval of convergence is $-1 < x < 1$.
• Aug 26th 2010, 04:22 AM
Revy
Quote:

Originally Posted by Prove It
So the radius of convergence is $1$ and the interval of convergence is $-1 < x < 1$.

Isn't interval (-1;1) only when it diverge with -1 and 1?
• Aug 26th 2010, 04:28 AM
Prove It
By the ratio test, the series converges where the limit $< 1$, diverges where the limit is $> 1$ and inconclsive where the limit $= 1$.

So all we can say at the moment is that the series is convergent when $|x| < 1$, because this is where the limit $< 1$.

You will need to use a different test to test the endpoints.
• Aug 26th 2010, 12:29 PM
HallsofIvy
Specifically, when x= 1, the series becomes $\sum \frac{1}{\sqrt{n}}= \sum \frac{1}{x^{1/2}}$ and 1/2< 1. What does that tell you?

When x= -1, the series becomes $\sum\frac{(-1)^n}{\sqrt{n}}$ which is an alternating series and $\frac{1}{\sqrt{n}}$ decreases to 0 as n goes to infinity. What does that tell you?
• Aug 27th 2010, 07:12 AM
Revy
Both of them converge.
Is convergence and divergence important finding interval of convergence? If so, how?
• Aug 27th 2010, 07:19 AM
Defunkt
Are you sure $\displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ converges?

And what do you mean by that question? Convergence and divergence of what?
• Aug 27th 2010, 08:03 AM
Prove It
Quote:

Originally Posted by Revy
Both of them converge.
Is convergence and divergence important finding interval of convergence? If so, how?

Well of course it is. You're testing if the function will converge at the endpoints. So obviously you need to see if the endpoints converge or diverge.

Specifically for the case of $x = 1$ when the series becomes $\sum{\frac{1}{\sqrt{n}}} = \sum{\frac{1}{n^{\frac{1}{2}}}}$, this is a p-series with $p < 1$. What do you know about p-series?

For the case of $x = -1$ when the series becomes $\sum{\frac{(-1)^n}{\sqrt{n}}}$, what are the conditions for which Leibnitz's Alternating Series Theorem implies convergence?
• Aug 27th 2010, 12:35 PM
Revy
Quote:

Originally Posted by Prove It
What do you know about p-series?

If $p<=1$, then series diverge.
Quote:

Originally Posted by Prove It
What are the conditions for which Leibnitz's Alternating Series Theorem implies convergence?

If sequence monotonically decrease to $0$, then series converge.
• Aug 27th 2010, 08:02 PM
Prove It
Correct, so putting it all together you know that the series converges for $-1 < x < 1$, diverges for $x < -1$ and $x > 1$, diverges at $x = 1$ and converges at $x = -1$.

So the interval of convergence is

$-1 \leq x < 1$.