# Some questions in series of functions uniform convergence

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.

• oh I see that is if we fix the x that happens right ?
– user111750
Dec 8 '14 at 5:41
• 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.
– user111750
Dec 8 '14 at 5:43
• but see below my comment regarding that.
– user111750
Dec 8 '14 at 5:50
• 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 $x$. Dec 8 '14 at 6:04

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.

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.

• 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 ?
– user111750
Dec 8 '14 at 5:48
• 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. Dec 8 '14 at 5:52

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

• however, i am not sure a bout my arguments since I haven't dealt too much with those concepts Dec 8 '14 at 6:25