I asked this question yesterday, but didn't get an answer except the link that I had already referred to: Show if the series $f(x)=\sum\limits_{k=1}^\infty \frac{1}{k} \sin(\frac{x}{k})$ converges uniformly or not.

Above might be related, but is NOT what I really asked, and I don't understand how the answer in the question above can define x∈[0,1]

My question is: $$f(x)=\sum\limits_{n=1}^\infty \frac{1}{n} \sin\left(\frac{x}{n}\right)$$

Where is $f$ defined? Is it continuous? Differentiable? Twice-differentiable?

I'm basically self-teaching the math, so please don't give a one-sentence *hint... Please correct me with the full right answer so that I can study the solution :(

What I think is:

  1. Since $\sin(x/n) \in [-1,1]$ for any $n \geq 1 $, $f$ is defined for all $x \in \mathbb{R}$

  2. Since $\lim_{n \to \infty} \frac{1}{n} \sin(\frac{x}{n}) = 0$ for any $x \in \mathbb{R}$, it's continuous

  3. Since $|f_n(x)|≤ 1$ for all $n\geq1$,we can use the Weirstrass M-Test to conclude that $f(x)=\sum\limits_{n=1}^\infty \frac{1}{n} \sin(\frac{x}{n})$ converges uniformly for any $x\in \mathbb{R}$

    3-1. Hence, it's differentiable by Term-by-Term Differentiability Theorem

  4. $f''(x)=\sum\limits_{n=1}^\infty -\frac{1}{n^3} \sin(\frac{x}{n}) \\ \to |-\frac{1}{n^3} \sin(\frac{x}{n})| \leq \frac{1}{n^3} $

    4-1. Then again by Weirstrass-M Test and Term-by-Term Differentiability Theorem, it's twice-differentiable.

* Weierstrass M-Test: For each $n\in \mathbb{N}$, let $f_n$ be a function defined on a set $A\subset \mathbb{R}$, and let $M_n>0$ be a real number satisfying $|f_n(x)|\leq M_n $ for all $x\in A$. If $\sum\limits_{n=1}^\infty M_n$ converges, then $\sum\limits_{n=1}^\infty f_n$ converges uniformly on A.

* Term-by-Term Differentiability Theorem: Let $f_n$ be differentiable funcitons defined on an interval A, and assume $\sum\limits_{n=1}^\infty f'_n(x)$ converges unifomly to a limit $g(x)$ on A. If there exists a point $x_0 \in [a,b]$ where $\sum\limits_{n=1}^\infty f_n(x_0)$ converges, then the series $\sum\limits_{n=1}^\infty f_n(x)$ converges uniformly to a differentiable function $f(x)$ satisfying $f'(x)=g(x)$ on A. In other words, $f(x) = \sum\limits_{n=1}^\infty f_n(x)$ and $f'(x)=\sum\limits_{n=1}^\infty f'_n(x)$

Please correct me if I'm wrong.


$1$ is not enough to prove $f(x)$ is defined, i.e. the series converges, because you then only have $\Bigl|\dfrac 1n\sin \dfrac x n\Bigr|\le \dfrac1n$, and the latter is divergent.

But you can argue this way, using equivalence: $$\Bigl|\frac 1n\sin\frac x n\Bigr|\sim_\infty \frac1n\Bigl|\frac xn\Bigr|=\frac{|x|}{n^2}$$ which is a convergent Riemann series

$2$. To prove the sum of the series is continuous, you can prove it converges uniformly on every compact interval. Indeed , if $|x|\le M$ for some $M>0$, we have $$\Biggl|\,\sum_{k=1}^n \frac{1}{k} \sin\Bigl(\frac{x}{k}\Bigr)\Biggr|\le\sum_{k=1}^n \frac{1}{k}\biggl|\, \sin\Bigl(\frac{x}{k}\Bigr) \biggr|\le\sum_{k=1}^n \frac{1}{k}\frac{|x|}{k}\le \sum_{k=1}^n\frac{M}{k^2},$$ so it is normally convergent on the disk centred at origin, with radius $M$.

Proceed similarly for $3$ and $4$.

  • $\begingroup$ 1. Riemann hasn't been introduced yet. Is there any other (rather basic) argument? 2. There is no domain given for x. Can I just suppose x is in $[-M,M] \subset \mathbb{R}$? Can you write more details for 3 and 4? (As I said, I can't proceed further by myself......) $\endgroup$ – mathnub Jun 5 '18 at 12:12
  • $\begingroup$ @Winther Can you provide more explanation extending from Bernard's answer for Q1? I'd appreciate much if you could provide answers for Q3 and Q4 $\endgroup$ – mathnub Jun 5 '18 at 12:41
  • $\begingroup$ @mathnub He is using the limit comparison test here. We have that $\sum \frac{|x|}{n^2}$ is a convergent series and since $\lim_{n\to \infty} \frac{|\sin(x/n)|}{1/n} / (|x|/n^2) = 1$ (this is what is implies by the $\sim$ symbol) so $\sum \frac{|\sin(x/n)|}{n}$ converges by this test. If a series converges absolutely (with the absolute values) then it also converges without the absolute values. $\endgroup$ – Winther Jun 5 '18 at 12:55
  • $\begingroup$ @mathnub: What I call a Riemann series is also called a $p$-series. These come with the very basics of series with positive terms. $\endgroup$ – Bernard Jun 5 '18 at 13:20
  • $\begingroup$ @Winther So can I suppose $x \in [-M, M] \subseteq \mathbb{R}$ ? $\endgroup$ – mathnub Jun 5 '18 at 14:59

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.