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quasar987
Apr4-05, 01:47 PM
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.

mathman
Apr4-05, 05:14 PM
Because of all the singularities (0, -1/n2) in the interval, it can't converge uniformly.

quasar987
Apr4-05, 06:18 PM
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?

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}

mathman
Apr5-05, 05:28 PM
In the intervals of interest it converges pointwise, but not uniformly because of the blow ups at the end points of each interval.

quasar987
Apr5-05, 06:26 PM
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!

quasar987
Apr5-05, 08:25 PM
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.

mathman
Apr6-05, 04:57 PM
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.

quasar987
Apr6-05, 06:09 PM
I noticed that like 10 minutes before handing it out :biggrin: