# Let $f(x)=\sum\limits_{n=1}^\infty \frac{1}{n} \sin(\frac{x}{n})$. Where is $f$ defined? Is it continuous? Differentiable? Twice-Differentiable?

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 Answer

$1$ 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 $4$.

• 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......) – mathnub Jun 5 '18 at 12:12
• @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 – mathnub Jun 5 '18 at 12:41
• @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. – Winther Jun 5 '18 at 12:55
• @mathnub: What I call a Riemann series is also called a $p$-series. These come with the very basics of series with positive terms. – Bernard Jun 5 '18 at 13:20
• @Winther So can I suppose $x \in [-M, M] \subseteq \mathbb{R}$ ? – mathnub Jun 5 '18 at 14:59