Show that the series of functions $$\sum_{n=0}^\infty \frac{n^2x}{1+n^4x^2}$$ converges pointwise, but not uniformly, on the interval $(0,1]$

This is the first time I have to show that a series of functions does not convere uniformly.

I managed to show that the series converges pointwise, using the comparison test.

However, I'm unsure how to prove it does not converge uniformly.

Since I don't know what the pointwise limit is, I suspect we have to use Cauchy's criterium:

$$\forall \epsilon > 0: \exists n_0: \forall p,q \geq n_0: \sup_{x \in (0,1]}\left|\sum_{k=0}^p f_k(x) - \sum_{k=0}^q f_k(x)\right| < \epsilon$$

But we should of course prove the negation of this criterium.

So, I started with seeing what I could do:

Let $p > q$, then I tried to estimate $$\sup_{x \in (0,1]}\left|\sum_{k=q+1}^p f_k(x) \right|$$

but couldn't find anything useful. Any clues?

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    $\begingroup$ Hint: What can you say about the general term at $x=1/n^2$? $\endgroup$ – Jyrki Lahtonen Jan 2 '18 at 10:54
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    $\begingroup$ it is $1/2$ there $\endgroup$ – user370967 Jan 2 '18 at 11:02
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    $\begingroup$ Then the $sup$ is greater than $\frac{1}{2}$ what do you deduce ? $\endgroup$ – Atmos Jan 2 '18 at 11:07
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    $\begingroup$ @Math_QED And so...? Take $p>q$ for arbitrary $n$, consider the Cauchy slice as before, and observe the $\sup_x$ is at least the value at $x_p=1/p^2\in(0,1]$, which is (all terms are positive) as least $1/2$. Therefore.... $\endgroup$ – Clement C. Jan 2 '18 at 11:08
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    $\begingroup$ Well, let me try. Let $n \in \mathbb{N}$, let $p > q \geq n$. Then, for $x = 1/p²$, we have that $|\sum_{k=q+1}^{p} \frac{k^2x}{1+k^4x^2}| \geq 1/2$, hence: $\exists \epsilon > 0: \forall n \in \mathbb{N}: \exists p,q \geq n: \exists x \in (0,1]: |\sum_{k}^p f_k - \sum_{k}^q f_k| \geq 1/2$ hence the series does not converge uniformly by the cauchy criterion $\endgroup$ – user370967 Jan 2 '18 at 11:39

The uniform convergence for a series of functions can also be stated as:

The sequence of functions $(\sum_{k=0}^n f_k)_{n\in\Bbb N}$ converges uniformly in $A$ if and only if for each $\epsilon>0$ exists some $N\in\Bbb N$ such that $\sup_{x\in A}\left|\sum_{k=m}^\infty f_k(x)\right|<\epsilon$ for all $m\ge N$.

The logical negation of the above reads:

The sequence of functions $(\sum_{k=0}^n f_k)_{n\in\Bbb N}$ does not converges uniformly in $A$ if and only if exists $\epsilon>0$ such that for all $N\in\Bbb N$ there exists some $m\geqslant N$ such that $\sup_{x\in A}\left|\sum_{k=m}^\infty f_k(x)\right|\ge\epsilon$.

Then if we find some $x_m\in A$ such that $\left|\sum_{k=m}^\infty f_k(x_m)\right|\ge\epsilon$ for each $m\in\Bbb N$, for some fixed $\epsilon>0$, we are done.

A general procedure is try to find this $x_m$, to do this we can study what is $\sup_{x\in A}f_m(x)$. From a brief analysis you can find that $x_m:=1/m^2$ belongs to $A:=(0,1]$ and that $f_m(x_m)=1/2$.

As I tried to show in my comment the functions $f_k$ have the form $\frac1{y^{-1}+y}$ for some $y>0$, then it is enough to see what is the infimum of the function defined by $y\mapsto y^{-1}+y$.

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  • $\begingroup$ Thanks for your answer. Does my answer in the comments seem correct? $\endgroup$ – user370967 Jan 2 '18 at 11:47
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    $\begingroup$ yes, it is correct. $\endgroup$ – Masacroso Jan 2 '18 at 11:47
  • $\begingroup$ Thanks a lot for your effort! $\endgroup$ – user370967 Jan 2 '18 at 11:49

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