Krdytkyu

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

What I was thinking is, since $sumlimits_{n=1}^{infty} frac{1}{n}$ 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 $(2kpi, 2kpi + pi)$ to be greater than a fixed $epsilon$. But when it comes to the detail, it troubles me. Any hint?

(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 $$sin(log n)geqslantfrac12$$ for every $n$ such that $a_kleqslant nleqslant b_k$ for some $k$, with $$a_k=lceil e^{2kpi+pi/6}rceilqquad b_k=lfloor e^{2kpi+5pi/6}rfloor$$
Thus, the slice of the series from $a_k$ to $b_k$ can be lower bounded as follows: $$sum_{n=a_k}^{b_k}frac1nsin(log n)geqslantfrac12sum_{n=a_k}^{b_k}frac1ngeqslantfrac12(b_k-a_k+1)frac1{b_k}$$ Since $a_ktoinfty$ and the RHS above converges to $$frac12(1-e^{-2pi/3})ne0$$ the series of interest diverges.

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

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

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

$$left|a_n - frac{sinln(n+t)}{n+t}right|
= left|frac{sinln(n+c) + sinln(n+c)}{(n+c)^2}right|alpha le frac{sqrt{2}t}{n^2}$$
As a consequence,
$$|b_n| le frac{sqrt{2}}{n^2}int_0^1 alpha dalpha = frac{1}{sqrt{2}n^2}$$
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 $A = sum b_n < infty$

$$A= lim_{Ntoinfty}left[sum_{n=1}^N a_n - sinln(N+1)right]
$$
Since the limit of $ sinln(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 $nin Bbb N$ 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 $0leq jleq y_n-1$ we have $$pi (2n+frac {1}{2})=log x_nleq log y_nleq$$ $$leq log (j+y_n)<log y_n+log 2=$$ $$=d_n+log x_n+log 2=$$ $$=pi (2n+frac {1}{2})+d_n+log 2.$$

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

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

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

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

Now $y_nto infty$ as $nto infty$ 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.