How to show a series of functions is not uniformly convergent

gregmoon
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Ok, I chose to ask about ways to determine if a series of functions is NOT uniformly convergent because I think that would best answer the overall difficulties I have with uniform convergence. I have a good idea of what uniform convergence is, I can give the definition, and if the problem is uniformly convergent there is a good chance I can show it. But if the problem is not then I don't think that "None of my tests work" is a rigorous mathematical argument to deny uniform convergence and none of the tests I have give explicit conditions for non-uniformity.

Homework Statement


Here is just one example. Show whether \sum \sin( \frac{x}{n^2}) converges uniformly for the interval "All x".

Homework Equations


The first time around I showed that this series was convergent by comparing it to \sum\frac{x}{n^2} and was thus able to show that it was absolutely convergent for all x.
\sum \left|\sin( \frac{x}{n^2})\right|\leq\sum\frac{x}{n^2}

The Attempt at a Solution


Now that I want to prove uniform convergence I'm wanting to do the same thing except I need to make the comparing series not dependent on x, make it a series of constants (which is essentialy the Weierstrass test, correct?). Unfortunately the way I see it, no matter what x I choose to make \sum\frac{x}{n^2} a constant, there will always be a greater x that will ruin the M-Test inequality. It is the connection from here to non-uniform convergence which I can't seem to make.

I appreciate any helpful comments. I also have a few other confusions which seem to result from the interval to be checked. Particularly when it is an open interval where the series of functions is undefined at either endpoint.
 
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Oh I want to ask the same thing. How can I prove something is *not* uniformly convergent?

Too bad there's no answer here. :confused:
 
Limn→∞sin(x/n2) = 0 for any fixed x.

Given any n can you find x s.t. |sin(x/n2) - 0| > 1/2? I'm letting epsilon = 1/2.
 
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