# Power series convergence

• April 29th 2011, 07:07 AM
Ulysses
Power series convergence
Hi there. Well, I was trying to determine the radius and interval of convergence for this power series:
$\displaystyle\sum_{0}^{\infty} \displaystyle\frac{x^n}{n-2}$

So this is what I did till now:
$\displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{n-1}{n-2}}\right |}=1$

Then $\left |{x}\right |<1$
The thing is this result is clearly wrong, I think the series diverges for any x, there is a discontinuity at n=2 for the sum. But I don't know what I'm doing wrong.

Bye there, thanks for your help!
• April 29th 2011, 07:30 AM
Ted
Quote:

$\displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{n-1}{n-2}}\right |}=1$

Where is x here?

And the sum should NOT start at 0
It should start at 3 or higher.
• April 29th 2011, 07:33 AM
Ulysses
It does start at zero. x is not needed to get the radius of convergence. It follows from the ratio test for the convergence of series. What I'm trying to find is the radius for the convergence of the power series, and its interval

If exists $\displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n+1}(x-x_0)^{n+1}}{a_{n}(x-x_0)^{n}}}\right |}<1$

Then
$R=\displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n}}{a_{n+1}}}\right |}<(x-x_0)$

The thing is I think it actually doesn't exists the limit I gave before. I mean, the series doesn't converges for any x. But I have to demonstrate it.
• April 29th 2011, 07:41 AM
topsquark
Quote:

Originally Posted by Ulysses
It does start at zero.

It can't start at n = 0. The n = 2 term is singular.

-Dan
• April 29th 2011, 07:46 AM
Plato
We can take care to say $n\ne 2$.
Example: $f(x) = \sum\limits_{n = 0}^1 {\frac{{x^n }}{{n - 2}}} + \sum\limits_{n = 3}^\infty {\frac{{x^n }}{{n - 2}}}$
Now $f(x)$ is defined for all $|x|<1$.
The root test works nicely here.
• April 29th 2011, 07:46 AM
Ulysses
Ok. I know it diverges at n=2, but in the exercise it starts at n=0. What I'm supposed to do?
• April 29th 2011, 07:47 AM
Ted
There is nothing called RATIO FOR THE CONVERGENCE OF THE POWER SERIES !
Do you mean radius of the convergence or interval of the convergence?

Your series takes the form $\sum \, a_n$

First you will compute $\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|$.

$\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x^n \cdot (n-2)}{n-1} \right| \rightarrow |x|$ as $n \rightarrow \infty$. (Note that we are dealing with x as a constant while evaluating the limit.).

For the convergence, |x| must be < 1 ,that is -1<x<1

so we reach the form a<x<b , then the radius equal (b-a)/2 = (1-(-1))/2=2/2=1

It is impossible for the power series to be divergent for all x.

at least, it must converge for 1 value for x.

and the sum should not start at n=0. (your $a_n$ is not continuous at n=2).
• April 29th 2011, 07:48 AM
Ulysses
Yes Ted, I'm sorry, my English sucks :P

Thank you all, I think that made things clear to me.
• April 29th 2011, 07:51 AM
topsquark
Quote:

Originally Posted by Ulysses
Ok. I know it diverges at n=2, but in the exercise it starts at n=0. What I'm supposed to do?

The convergence test will still work quite nicely if n starts at 3. (The limit is, after all, taken at infinity.) The summation is just not correctly stated is all.

-Dan

Edit: Or you can use Plato's summation. I didn't see that before.
• April 29th 2011, 10:46 AM
HallsofIvy
Of course, changing any finite number of terms of an infinite sequence change what it converges to but does not change whether or not it converges.