0
$\begingroup$

$$\sum\limits_{n=1}^{\infty} n^2 \sin(\frac{x}{n^4})$$ I am asked to study the pointwise and uniform convergence of the series.

For the pointwise requirement, it is convenient to study the asymptotically equivalent: $$\sum\limits_{n=1}^{\infty} n^2 \frac{x}{n^4}=\sum\limits_{n=1}^{\infty} \frac{x}{n^2}$$ Since it is an harmonic series of order two, the original series converges for all $x\in R$.

I wonder what is the appropriate way to study uniform convergence. I think that studying uniform convergence of an asymptotically equivalent series is not correct but I'd like to have some comments about it [for example, is it acceptable to use convergence radius of the asymptotically equivalent power series to discuss uniform convergence of a series?]

Using Weierstrass test and the inequality $\sin t < t$, I noticed that when considering set like $S=[a,b]$: $$\sup_{x \in S} |f_n(x)| \leq sup \frac{x}{n^2}=\frac{c}{n^2}$$ where $c=\max (|a|,|b|)$. Finally $\sum\limits_{n=1}^{\infty}\frac{c}{n^2}$ converges, so that the original series is uniformly convergent in $[a,b]$. Same argument doesn't apply when $S$ is not a limited set. Is this study of uniform convergence sufficiently exahustive? Can other uniform convergence sets exist? For example, I think that union of a finite number of disjoint intervals is still a solution. [Any compact set, actually].

Moreover, a general question: could a set of isolated points be the set of uniform convergence for a particular series?

$\endgroup$
  • $\begingroup$ You don’t need to examine this taxonomy of sets. Any bounded set is fine (you proved it already with the sup bound of the general term). Convergence is not uniform on an unbounded set. $\endgroup$ – RRL Apr 7 at 21:12
1
$\begingroup$

Your series is uniformly convergent on any bounded subset. On all of $\mathbb{R}$, however, it is not uniformly convergent. One way to see this is: if it were uniformly convergent, the general term would converge to zero uniformly. But this is clearly not the case, since given any $n$, you can choose $x=n^4$ and, at that point, the general term is $n^2\sin 1$ which is greater than, say, $\sin 1$.

$\endgroup$
1
$\begingroup$

Convergence on an unbounded set -- $(0,\infty)$, for example -- is not uniform.

Note that

$$\sup_{x \in (0,\infty)} \left|\sum_{k=n+1}^\infty n^2 \sin \frac{x}{n^4}\right| \geqslant \sup_{x \in (0,\infty)} \sum_{k=n+1}^{2n} k^2 \sin \frac{x}{k^4}\geqslant \sup_{x \in (0,\infty)}n \cdot n^2 \sin \frac{x}{n^4} \geqslant n \cdot n^2 \sin \frac{n^4}{n^4}=n^3 \sin (1)$$

Since the RHS does not converge to $0$ as $n \to \infty$, the convergence is not uniform.

$\endgroup$
  • $\begingroup$ The only difference between RRL's approach and my own is that he/she is using the "Cauchy condition" with $m=2n$ and I am doing the same with $m=n+1$, since this is clearly enough in this example. $\endgroup$ – GReyes Apr 7 at 22:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.