# How to show that a series of functions does not converge uniformly?

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?

• Hint: What can you say about the general term at $x=1/n^2$? Jan 2, 2018 at 10:54
• it is $1/2$ there
– user370967
Jan 2, 2018 at 11:02
• Then the $sup$ is greater than $\frac{1}{2}$ what do you deduce ? Jan 2, 2018 at 11:07
• @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.... Jan 2, 2018 at 11:08
• 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
– user370967
Jan 2, 2018 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 $$y\mapsto y^{-1}+y$$.

– user370967
Jan 2, 2018 at 11:47
• yes, it is correct. Jan 2, 2018 at 11:47
• Thanks a lot for your effort!
– user370967
Jan 2, 2018 at 11:49