Uniform Convergence: Find Domain of Intervals (a,b)

[kidsmoker]

Homework Statement

Consider

f(x)=\sum\frac{1}{n(1+nx^{2})}

from n=1 to n=infinity.

On what intervals of the form (a,b) does the series converge uniformly? On what intervals of the form (a,b) does the series fail to converge uniformly?

Homework Equations

Weierstrass M-test: If there exists numbers M_{r} for each f_{r}(x) such that f_{r}(x) \leq M_{r} and

\sum M_{r}

converges, then \sum f_{r} converges uniformly.

The Attempt at a Solution

Write

f_{r}(x)=\frac{1}{r(1+rx^{2})} .

If i just consider the case where x>1 then

\frac{1}{r(1+rx^{2})} \leq \frac{1}{r^{2}x^{2}} \leq \frac{1}{r^{2}}

so let

M_{r}=\frac{1}{r^{2}}

and the sum of this from r=1 to r=infinity converges. So

\sum f_{r}

converges uniformly for x>1? Or am I misunderstanding the M-test?

But then say I went with x>0.5 instead of x>1. I could then choose

M_{r}=\frac{2}{r^{2}}

to get uniform convergence for x>0.5? etc etc

So what's the required domain? I'm really confused :-(

 

[clamtrox]

But then say I went with x>0.5 instead of x>1. I could then choose

M_{r}=\frac{2}{r^{2}}

to get uniform convergence for x>0.5? etc etc

So what's the required domain? I'm really confused :-(

That's exactly correct if you just use M = 4/r^2 instead. How about uniform convergence for x>a where a is an arbitrary positive real number? You can generalise your answer easily. Then there's only one real number you're leaving out, and it turns out that the series does not converge there.