# Uniform convergence

1. Nov 15, 2009

### kidsmoker

1. The problem statement, all variables and given/known data

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?

2. Relevant 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.

3. 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 :-(

2. Nov 18, 2009

### clamtrox

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.