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I am just solving many question on series of uniform convergence in order to get more intuition and experience with this stuff so I wanted to know what do you guys think of my argument. Its easy for like sequence but series sometimes is hard to deal with I am kinda confused on series for uniform convergence if someone could clarify this issue that would be great.

1)$\sum 1/(nx)^2$ for $x \in (0,1]$ so this will diverge and the reason for that we just substitute x = 1/n so we get $\sum 1$ so this will won't even converge point-wise therefore we and will diverge so since $\lim \sum 1$ is infinity Hence won't converge uniformly.

2)$\sum x^2/n^2$ for $x \in [5,\infty]$ I think diverge if we choose for x > n we choose x = 1/n and therefore will diverge by same argument I had above.

3)$\sum 1/(1 + n^2x^2)$ for $x \in (0,1]$ This will diverge if we choose x = 1/n and use the same argument as above.

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  • $\begingroup$ oh I see that is if we fix the x that happens right ? $\endgroup$
    – user111750
    Dec 8, 2014 at 5:41
  • $\begingroup$ oh I see ye that makes sense can you maybe explain more for the uniform convergence of those series, because I am confused on this part I mean they are pretty easy to see for sequence, but I am getting confused for series. $\endgroup$
    – user111750
    Dec 8, 2014 at 5:43
  • $\begingroup$ but see below my comment regarding that. $\endgroup$
    – user111750
    Dec 8, 2014 at 5:50
  • $\begingroup$ For the second question, in my comment I assumed that $x$ was still in $(0,1)$. But it is in $[5,\infty)$ and for large $x$ the truncation error is large. If we truncate at $q$ the error is greater than $\frac{x^2}{(q+1)^2}$. So to get truncation error $\lt \epsilon$, we need to go very far out if $x$ is large, the "$N$" depends on $x$. $\endgroup$
    – André Nicolas
    Dec 8, 2014 at 6:04

3 Answers 3

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Regarding question 1 for instance. It certainly converges pointwise. Fix an x, you can factor it outside and you obtain 1/x^2 times a series which converges to pai/6. so we get pai/6x^2 but this pointwise limit cannot be the uniform limit since it is unbounded as we go closer to 0.

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They all converge pointwise. Remember that for pointwise convergence we study the expression for fixed $x$,

For example, fix $x$ in $(0,1)$. By Limit Comparison with $\sum_1^\infty \frac{1}{n^2}$ we conclude that our series converges.

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  • $\begingroup$ For the first one we fixed an x between (0,1) and if we use the M-Test there will be a problem because its not < 1/n^2 so we can't use that to conclude that first one will converge uniformly right ? $\endgroup$
    – user111750
    Dec 8, 2014 at 5:48
  • $\begingroup$ Right. In the first, if $x$ is very close to $0$, we have to go a lot further out to make the error small than if $x$ is more reasonable, like $\frac{1}{2}$. Try to make this argument more formal. $\endgroup$
    – André Nicolas
    Dec 8, 2014 at 5:52
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In question 2 , seems like you are on the right track. assume X>n and bound it bellow by n^2/n^2. So doesn't seem that It converges uniformly

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  • $\begingroup$ however, i am not sure a bout my arguments since I haven't dealt too much with those concepts $\endgroup$
    – Daniel
    Dec 8, 2014 at 6:25

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