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

Question

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?

Answer 1

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 $\int_1^\infty f(x) \, dx$.

Answer 2

(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.

Answer 3

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 $c \in [0,\alpha]$

$$\left|a_n - \frac{\sin\ln(n+\alpha)}{n+\alpha}\right| = \left|\frac{\cos\ln(n+c) - \sin\ln(n+c)}{(n+c)^2}\right|\alpha \le \frac{\sqrt{2}\alpha}{n^2}$$ 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

Answer 4

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.