# Convergence of serie

1. Apr 4, 2005

### quasar987

I need help with the following problem.

Consider the serie of function

$$\sum_{n=1}^{\infty}\frac{1}{1+n^2x}$$

The serie is undefined for $x \in \{0\}\cup \{-1/n^2, \ n\in \mathbb{N}\}$. I want to find wheter it converges pointwise in (-1, 0) or not and if it does, does it converge uniformly?

The way I would start this problem is by saying: For a given number $m \in \mathbb{N}$, consider

$$x_0 \in \left(\frac{-1}{m^2} \ ,\frac{-1}{(m+1)^2}\right)$$

Consider

$$f_n(x) = \frac{1}{1+n^2x}$$

Then

$$|f_n(x_0)| = \frac{1}{|1+n^2x_0|} = \frac{1}{|1-n^2|x_0||}= \left\{ \begin{array}{rcl} \frac{1}{1-n^2|x_0|} & \mbox{for} & n<\sqrt{\frac{1}{|x_0|} \\ \frac{1}{n^2|x_0|-1} & \mbox{for} & n>\sqrt{\frac{1}{|x_0|} \end{array}\right$$

and

$$\sum_{n=1}^{\infty}|f_n(x_0)| = \sum_{n=1}^{\left[\sqrt{1/|x_0|}\right]}\frac{1}{1-n^2|x_0|} + \sum_{n=\left[\sqrt{1/|x_0|}\right]+1}^{\infty}\frac{1}{n^2|x_0|-1}$$

I'm guessing this serie converges, but I'm having trouble finding a convergent serie to bound it with. The other convergence tests have failed and the use of the integral convergence criterion is forbiden. I know that if there is a serie to bound it with, it would be of the form

$$\sum_{n=1}^{\infty}a_n = \sum_{n=1}^{\left[\sqrt{1/|x_0|}\right]}\frac{1}{1-n^2|x_0|} + \sum_{n=\left[\sqrt{1/|x_0|}\right]+1}^{\infty}b_n$$

with

$$\frac{1}{n^2|x_0|-1} \leq b_n$$

for n > N.

Edit:

And if there exists such an N that also satisfies

$$N\leq \left[\sqrt{1/|x_0|}\right]$$

then according to Weirstrass M-test, the convergence is uniform.

Last edited: Apr 4, 2005
2. Apr 4, 2005

### mathman

Because of all the singularities (0, -1/n2) in the interval, it can't converge uniformly.

3. Apr 4, 2005

### quasar987

That was not well said. What I meant to say is, does it converge pointwise and uniformly for the intervals in (-1,0) where the serie is defined. I.e. in the intervals

$$\left(\frac{-1}{m^2} \ ,\frac{-1}{(m+1)^2}\right), & m \in \mathbb{N}$$

4. Apr 5, 2005

### mathman

In the intervals of interest it converges pointwise, but not uniformly because of the blow ups at the end points of each interval.

5. Apr 5, 2005

### quasar987

On the basis of which theorem(s) are these statements made true? I would apreciate a quick answer because I need to hand out this question tomorrow!!

Thanks!

Last edited: Apr 5, 2005
6. Apr 5, 2005

### quasar987

By the way, I have found how to prove the pointwise convergence, I just don't know how to prove that it's not uniformly convergent on these intervals.

7. Apr 6, 2005

### mathman

I don't know what approach you are using to prove pointwise convergence. However, if you are using the old fashioned epsilon delta argument, you will see that there is a dependence on x when x is near a singular value.

8. Apr 6, 2005

### quasar987

I noticed that like 10 minutes before handing it out