# How to prove the series $\sum\limits_{n=1}^\infty\frac1n\sin(\ln n)$ diverges

How to prove the following series is divergent: $$\sum_{n=1}^{\infty} \frac{1}{n} \sin(\ln n)\ ?$$

What I was thinking is, since $\sum\limits_{n=1}^{\infty} \frac{1}{n}$ diverges, $\sin$ is periodical and $(\ln (n+p)-\ln n)$ converges to $0$ when $n$ goes to infinity for every $p$, maybe I can look for a sum of partial terms within some $(2k\pi, 2k\pi + \pi)$ to be greater than a fixed $\epsilon$. But when it comes to the detail, it troubles me. Any hint?

• Maybe this question helps? – Sil Aug 26 '18 at 17:44
• @Sil Thank you! – Edward Wang Aug 26 '18 at 17:55
• You could also use the argument that $\sum_{n=1}^{+\infty}\frac{\sin(\ln(n))}{n}\sim \int \frac{\sin(\ln(t))}{t}\,\mathrm{d}t=-\cos(\ln(x))+C$. But since $-\cos(\ln(x))+C$ oscillates between $C-1$ and $C+1$, the series must be divergent. – Jam Aug 26 '18 at 18:09
• @Jam How do you justify the approximation by the integral? – Clement C. Aug 26 '18 at 18:24
• @ClementC. Truth be told, I'm still figuring that out. But I'm pretty sure there's a justification there somewhere – Jam Aug 26 '18 at 18:35

(2018-11-25) "Amusing" revenge downvote, three months later.

Since it seems that everyone feels compelled to rush to the comparison with an integral, let us present a simpler, low-tech, proof (which, additionally, is the OP's idea).

Start with the elementary fact that $$\sin(\log n)\geqslant\frac12$$ for every $$n$$ such that $$a_k\leqslant n\leqslant b_k$$ for some $$k$$, with $$a_k=\lceil e^{2k\pi+\pi/6}\rceil\qquad b_k=\lfloor e^{2k\pi+5\pi/6}\rfloor$$ Thus, the slice of the series from $$a_k$$ to $$b_k$$ can be lower bounded as follows: $$\sum_{n=a_k}^{b_k}\frac1n\sin(\log n)\geqslant\frac12\sum_{n=a_k}^{b_k}\frac1n\geqslant\frac12(b_k-a_k+1)\frac1{b_k}$$ Since $$a_k\to\infty$$ and the RHS above converges to $$\frac12(1-e^{-2\pi/3})\ne0$$ the series of interest diverges.

• Yippee... Yet another silent downvote. – Did Aug 26 '18 at 20:08
• I gave you a +1 to compensate for it. I rarely down-vote, but when I do, I explain it. – DanielWainfleet Aug 26 '18 at 20:29
• I don't understand. It seems the proof consists in saying that you can find a sub-series which diverges, and therefore the whole series diverges. But this series has negative terms. – Jack M Aug 26 '18 at 20:31
• @JackM This is showing that the sequence of partial sums is not Cauchy (and hence diverges). – Clement C. Aug 26 '18 at 20:42
• @JackM If a series $\sum x_n$ converges then the rests converge to zero in the sense that, for every $\epsilon>0$ there exists some $N$ such that, for every $n\geqslant m\geqslant N$, $$\left|\sum_{k=m}^nx_k\right|<\epsilon$$ – Did Aug 26 '18 at 21:15

As proved here, you can always apply the integral test to $\sum_{n \geqslant 1} f(n)$ even if $f$ is not monotone as long as

$$\tag{*}\int_1^\infty|f'(x)| \, dx < +\infty$$

In this case, $f'(x) = -\sin (\ln x)/x^2 + \cos(\ln x)/x^2$. Since $|f'(x)| \leqslant 2/x^2$, the condition (*) is met.

It then follows that the series diverges along with the integral $\int_1^\infty f(x) \, dx$.

• "Using some technology" as Prof. William Weiss once said in another context.................+1 – DanielWainfleet Aug 26 '18 at 20:27

Let $$a_n = \frac{\sin \ln n}{n}$$ For $n > 0$, let's take

$$b_n = a_n - \int_n^{n+1}\frac{\sin\ln x}{x} dx = \int_0^1 \left(a_n - \frac{\sin \ln(n+t)}{n+t}\right) dt$$ Since for each $\alpha \in ]0,1]$, we can apply MVT to find a $c \in [0,\alpha]$

$$\left|a_n - \frac{\sin\ln(n+t)}{n+t}\right| = \left|\frac{\sin\ln(n+c) + \sin\ln(n+c)}{(n+c)^2}\right|\alpha \le \frac{\sqrt{2}t}{n^2}$$ As a consequence, $$|b_n| \le \frac{\sqrt{2}}{n^2}\int_0^1 \alpha d\alpha = \frac{1}{\sqrt{2}n^2}$$ By the $p-$test, $\sum \frac{1}{\sqrt{2}n^2}$ is a convergent series. So by the comparison test, $\sum b_n$ is an absolutely converging series. Let $A = \sum b_n < \infty$

$$A= \lim_{N\to\infty}\left[\sum_{n=1}^N a_n - \sin\ln(N+1)\right]$$ Since the limit of $\sin\ln(N+1)$ oscillates, then the limit of $\sum_{n=1}^N a_n$ has to oscillate to maintain a finite $A$. Hence $\sum_{n=1}^{\infty} a_n$ diverges.

NOTE: Answer was inspired by @achille hui

For $n\in \Bbb N$ let $\log x_n=\pi(2n+\frac {1}{2}).$ Let $y_n$ be the least integer greater than or equal to $x_n.$

For brevity let $d_n=\log y_n-\log x_n.$

For $0\leq j\leq y_n-1$ we have $$\pi (2n+\frac {1}{2})=\log x_n\leq \log y_n\leq$$ $$\leq \log (j+y_n)<\log y_n+\log 2=$$ $$=d_n+\log x_n+\log 2=$$ $$=\pi (2n+\frac {1}{2})+d_n+\log 2.$$

We also have $$0\leq d_n=\log (1+\frac {y_n-x_n}{x_n})<$$ $$<\log (1+\frac {1}{x_n})<\frac {1}{x_n}\leq \frac {1}{x_1}=e^{-5\pi/2}.$$

So for $0\leq j<y_n-1$ we have $$\sin \log (j+y_n)=\cos (-E_{n,j})$$ where $0\leq E_{n,j}<d_n+\log 2 \leq e^{-5\pi/2}+\log 2<\pi /4.$

(Note: $E_{n,j}$ is an ad hoc abbreviation based on the preceding paragraphs.)

So for $0\leq j<2y_j-1$ we have $\sin \log (j+y_n)>\cos (\pi/4)>1/2.$

Now $y_n\to \infty$ as $n\to \infty$ and we have $$\sum_{v=y_n}^{2y_n-1}\frac {\sin \log v}{v}=\sum_{j=0}^{y_n-1}\frac {\sin \log (j+y_n)}{j+y_n}>\sum_{j=0}^{y_n-1}\frac {1/2}{2y_n}=\frac {1}{4}.$$

So the Cauchy Criterion is not met.

• I meant to give a hint but the details of the hint seemed to be difficult to say. As I re-wrote it, it became simpler and simpler, but it also turned into a complete solution – DanielWainfleet Aug 26 '18 at 20:20
• "it also turned into a complete solution" ...which happens to be completely similar (to stay polite) to another answer posted 1 hour earlier. – Did Aug 26 '18 at 21:17
• Which I didn't read in detail. Of course it's similar. – DanielWainfleet Aug 26 '18 at 21:36