1. ## Uniform Convergence

I'm having trouble with this question:
Is the $\displaystyle$\displaystyle\sum\limits_{n=0}^\infty \frac{n^2 x}{1+{n^4}{x^2}}
uniformly convergent over the interval (0,1].
Now I know that it isn't uniformly convergent, and have started my proof with a contradiction.
So assuming for contradiction that it is uniformly convergent, we can say that $\displaystyle \forall \epsilon >0 \exists N \in \mathbb{N} s.t. \forall n,m>N |f_n(x) - f_m(x)|<\epsilon$ but I don't know where to go from there. Any help?

2. This series is pointwise convergent. We have, if $\displaystyle x$ is positive that $\displaystyle \left|\frac{n^2x}{1+n^4x}\right|\leq \left|\frac{n^2x}{n^4x}\right|=\frac 1{n^2}$ so for all $\displaystyle n$, $\displaystyle \displaystyle\sup_{0<x\leq 1}\left|\frac{n^2x}{1+n^4x}\right|\leq\frac 1{n^2}$ and the series is normally convergent.

3. Sorry, I can't see how this answers the question?

4. If a series is normally convergent on the set $\displaystyle A$, then it is uniformly convergent on $\displaystyle A$.

5. I thought that was only for closed intervals? So are you saying this is uniformly convergent over (0,1]?

6. Originally Posted by worc3247
Sorry, I can't see how this answers the question?

You can use Weierstass M-test.

7. Please accept my apologies for this, I missed out a squared on the denominator of the series and have only just realised. Really sorry, have corrected the mistake now.

8. As was mentioned the series is pointwise convergent to 0. Investigate what happens at the point $\displaystyle x=\frac{1}{n^2}\in (0,1]$ for all n!

9. Find max f_n(x), by taking the derivative, you'll get x=+/- 1/n^4, x in (0,1] so x=1/n^4

Placing that x=1/n^4 back to f_n(x). You will get f_n(1/n^4)=1/n^2

sum 1/n^2 is converges.

{n^2x}/{1+n^4x^2} < 1/n^2 for all x in (0,1], now the rest comes from Weierstrass test.