For the function $f (x)$ on the real line $\mathbb{R}$ defined below, which of the following statements about $f$ is true?Choose all the correct options: $$f (x) :=\sum_{n\ge 1}\frac{\text{sin}(x/n)}{n}$$

(a) f is continuous but not uniformly continuous on $\mathbb{R}$.

(b) f is uniformly continuous on $\mathbb{R}$.

(c) f is differentiable on $\mathbb{R}$.

(d) f is an increasing function on $\mathbb{R}$.

My attempt:

Option (c) is correct since $f(x)$ is $C^\infty$ function.

Option (d) is incorrect since sine function is periodic, f(x)is neither decreasing nor increasing function.

I applied $M-test$ but reached to a conclusion that $f(x)\le \frac{x}{n^2}$. I did this by finding $f'(x)=0$ and $f''(0)<0$. The other inequality that seems obvious is $f(x)\le 1/n$ and $\sum \frac{1}{n}$ is divergent sequence.

Am I doing something wrong here while doing $M-test$ ? Please help me to solve this question. Thankyou.

  • $\begingroup$ Does $n \in \Bbb{Z}$? $\endgroup$
    – ModCon
    Mar 31, 2020 at 8:08
  • $\begingroup$ Since $n\ge 1$, I believe $n \in \mathbb{N}$ $\endgroup$ Mar 31, 2020 at 8:10
  • $\begingroup$ Your argument about (d) seems incorrect/very weak. $\sin(x)$ being periodic doesn't say anything about the sum, which has $\frac{x}n$ as argument of the sine-function. Also how do you justify your argument about (c)? You are just claiming a very specific thing, without any kind of argument for that. $\endgroup$
    – Ingix
    Mar 31, 2020 at 9:14
  • $\begingroup$ @Ingix you're right about my argument (d). It seems incorrect to me now. $\endgroup$ Mar 31, 2020 at 10:54

1 Answer 1


User ModCon had a (now deleted) solution to some parts of the problem. There was a small mistake in it that made a few, but not all conclusions incorrect. I'll try to repeat his contributions, and add a few of my own.

We have $$\left|\frac{\sin(x/n)}n\right| \le \frac{|x/n|}n = \frac{|x|}{n^2}.$$

That means for each $x$ the series $\sum_{n=1}^{\infty}\frac{\sin(x/n)}{n}$ has a convergent majorant

$$\sum_{n=1}^{\infty}\frac{|x|}{n^2} = |x| \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}6|x|,$$

which means it converges pointwise. This also means that we can use the Weierstrass M-Test on an interval $[-a,a]$ for some $a>0$ and choose $M_n=\frac{a}{n^2}$. That means the series converges absolutely and uniformly on $[a,-a]$, which means $f(x)$ is continuous on that inverval. Since we can chose any real $a$, this means $f(x)$ is continuous on the whole $\mathbb R$.

If we differentiate the series term-wise, we get another function:


Using $|\cos(x/n)| \le 1$ we can directly apply the M-test on whole real line for $g(x)$, when setting $M_n=\frac1{n^2}$. This shows that g(x) is actually well defined and the defining series converges uniformly on $\mathbb R$ to $g(x)$.

So is now $f'(x)=g(x)$? By Theorem 1 on page 2 in this university script, the answer is yes. The term-wise derivate series (for $g(x)$) converges uniformly on the whole $\mathbb R$, the original series (for $f(x)$) converges on a point (we already know it converges anywhere), so we now know that

$$f'(x)=g(x),\; \forall x \in \mathbb R$$

This means $f(x)$ is differentible, which answers (c) in the affirmative.

We also know that

$$|g(x)| \le \sum_{n=1}^{\infty}\frac{|\cos(x/n)|}{n^2} \le \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}6,$$

so the derivate of $f(x)$ is bounded. That means $f(x)$ is uniformly continuous: Given any $\epsilon > 0$, we can choose $\delta=\frac{6\epsilon}{\pi^2}$ and if we assume $x_0 < x_1 < x_0+\delta$, we have $f(x_1)=f(x_0)+(x_1-x_0)f'(\xi)$ with $\xi \in [x_0,x_1]$ (Mean Value Theorem), so

$$|f(x_1)-f(x_0)| = |f(x_0)+(x_1-x_0)f'(\xi) - f(x_0)| = |x_1-x_0||f'(\xi)| < \delta\frac{\pi^2}6 = \epsilon.$$

This ansers (b) in the affirmative and (a) in the negative.

For (d), because we know $f(x)$ is differentiable, it is enough to find a point where $f'(x)=g(x)$ is negative, to prove that (d) is not true.

Consider $x=\pi$. Then the first element of the series is $\cos(\pi)=-1$. We know that all the other elements of the series can sum up to at most $\sum_{n=\color{red}{2}}^{\infty}\frac{1}{n^2} = \frac{\pi^2}6 - 1 < 1$, so $g(\pi) < 0$ and (d) is not true.

If you don't want to use the non-elementary exact sum of that series, you can can also just say

$$\sum_{n=2}^{\infty}\frac{1}{n^2} < \sum_{n=2}^{\infty}\frac{1}{n(n-1)} = \sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n}\right) = (1-\frac12)+(\frac12-\frac13)+(\frac13-\frac14)\ldots=1.$$

Finally, a plot of the function (hopefully good enough truncated) by Wolfram Alpha:

enter image description here

  • $\begingroup$ Thanks for such a detailed answer. I just wanted to ask: (we already know it converges anywhere)- does this statement mean we can make it converge anywhere OR it means we know it converges to some point? $\endgroup$ Mar 31, 2020 at 13:38
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    $\begingroup$ The question was much more difficult than I thought it would be. @ModCon and Ingix cheers to you. $\endgroup$ Mar 31, 2020 at 13:40
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    $\begingroup$ @s1mple For the theorm I mentioned, you basically have to do "all the work" for the termwise derivative, and for the function itself you only need to check convergence at 1 point. In preparation, we already proofed much more than we needed, so I made that remark. It means we proved it convergens for any $x$, which is much more than we need to apply that theorem. $\endgroup$
    – Ingix
    Mar 31, 2020 at 13:44
  • $\begingroup$ +1 here. The way you have dealt with uniform convergence is really nice. One can show the uniform convergence in any finite interval by Weierstrass M test, but for whole of $\mathbb {R} $ we need something different. Luckily the function turned out to have a bounded derivative which made the task easy. $\endgroup$
    – Paramanand Singh
    Apr 1, 2020 at 2:27

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