# Prove this series does not uniformly converge

$$\sum_{n=1}^{\infty}n^2 \sin \frac{x}{n^4}$$

It is easy to show that it absolutely converges. But what about uniform convergence?

With M-test: $$|| f_n|| = \sup (| n^2 \sin \frac{x}{n^4}|) \leq \sup (| \frac{x}{n^2} |)$$

It seems to me that in order to proceed I have to pick a restricted set $[a,b]$.

In this case, let $c= | max(|a|,|b|)$: in [a,b] the supremum is $\leq \frac{c}{n^2}$.

The series $$\sum_{n=1}^{\infty}\frac{c}{n^2}$$ converges, so in any restricted interval I have uniform convergence.

But how can I prove that in any other case there is not uniform convergence?

I tried with the necessary condition "$f_n$ must uniformly converge". This means: $$\lim \sup |f_n - f| = 0$$

In my case it is $\leq \lim \sup |x/n^2|$...

but if x is unrestricted I don't know how I can deal with lim sup.. Thanks a lot!

## EDIT

Suddenly I realized: what if I study this series passing to the asymptotically equivalent $$\sum_{n=1}^{\infty}n^2 \frac{x}{n^4}$$? In this case I could use the theorem for power series! Is it correct?

## EDIT 2

OMG, I've just noticed that $$\sum_{n=1}^{\infty}n^2 \frac{x}{n^4}$$ is NOT A POWER series..

However, let's suppose that the given series is $\sum_{n=1}^{\infty}n^2 \sin \frac{x^n}{n^4}$. Then I'd pass (#) to $$\sum_{n=1}^{\infty}\frac{x^n}{n^2}$$. I'd compute the convergence radius (1) and then say "oh, the given series uniformly converges just in any compact subset in (-1,1)". Is the (#) step correct?

• "It is easy to show that it absolutely converges" Much better to say it converges absolutely for each fixed $x.$ – zhw. Jun 22 '16 at 0:09
• $\sum n^2 \sin(x^n/n^4)$ is still not a power series. – zhw. Jun 22 '16 at 0:11

## 2 Answers

If it converges uniformly, then

$$\sup_{x\in \mathbb R} \left|n^2 \sin(x/n^4)\right|\to 0$$

as $n\to \infty$. This is clearly false.

• "clearly"? sorry but to me it isn't so clear :( could you explain? – Surfer on the fall Jun 14 '16 at 10:25
• For example, if $x = n^4(\pi/2)$, then $n^2 \sin (x/n^4) =n^2$. @Surferonthefall – user99914 Jun 14 '16 at 10:27
• I already thought about it, but I wasn't sure.. so the fact could be said as follows: if I take the sequence b_k made of $\sup {f_n: n >= k}$, even if k is huge the sequence doesn't approach zero.. so that sup doesn't approach zero, so it does not uniformly converge. Is it correct? I think I didn't understand "indeed .. is unbounded".. hmm, i think that the whole point here is to show that fact.. isn't it? – Surfer on the fall Jun 14 '16 at 10:34
• Note that your $b_k$ is infinite for all $k$. I want to say that $\| f_n\| = n^2$ for all $n$, where $\|f_n\| = \sup_{x\in \mathbb R} |f_n(x)|$. @Surferonthefall – user99914 Jun 14 '16 at 10:37
• Uhm ok but that would only mean I cannot use M-test.. isn't it? If I want to prove it cannot converge I have to show that some necessary condition (such as the uniform convergence of $f_n$) isn't respected.. the uniform convergence isn't respected if the limit of $b_k$ isn't 0.. – Surfer on the fall Jun 14 '16 at 10:44

This is meant to elucidate the A given By Arctic Char because of the long chat with the OP that followed it.

$f_n\to f$ uniformly on a domain $D$ iff $$\lim_{n\to \infty}\|f-f_n\|=0,$$ where $$\|f-f_n\|=\sup_{x\in D}|f(x)-f_n(x)|.$$ Suppose $f_n\to f$ uniformly on $D$ then also $\lim_{n\to \infty}\|f_{n+1}-f\|=0.$ Then we have $$\|f_{n+1}-f_n\|\leq \|f-f_n\|+\|f_{n+1}-f\|$$ (because $\;|f_{n+1}(x)-f_n(x)|\leq |f_{n+1}(x)-f(x)|+|f(x)-f_n(x)|\;),$ so we must have $$\lim_{n\to \infty}\|f_{n+1}-f_n\|=0.$$

When $f_n(x)=\sum_{j=1}^n j^2 \sin (x/j^4)$ and the domain $D=\mathbb R,$ we have $$\|f_{n+1}-f_n\|\geq |f_{n+1}(\pi (n+1)^4/2)-f_n(\pi(n+1)^4/2)|=(n+1)^2.$$ So $\|f_{n+1}-f_n\|$ does not converge to $0,$ so $f_n$ does not converge uniformly.