Does $\sum _{n=1}^{\infty } \frac{\sin(\text{ln}(n))}{n}$ converge? Does $\sum _{n=1}^{\infty } \dfrac{\sin(\text{ln}(n))}{n}$ converge?
My hypothesis is that it doesn't , but I don't know how to prove it. $ζ(1+i)$ does not converge but it doesn't solve problem here. 
 A: The sum cannot converge because there is a constant $C>0$ such that
for each $N$ there are $N_1,N_2 > N$ such that the terms
$\frac1n \sin(\ln n)$ with $N_1 \leq n \leq N_2$
are of constant sign and their sum exceeds $C$ in absolute value.
For any integer $k$ we know that
$\sin x$ is of constant sign for $k\pi < x < (k+1)\pi$,
and of absolute value at least $\sin(\pi/6) = 1/2$
for $(k+\frac16)\pi < x < (k+\frac56)\pi$.
Thus the terms with $(k + \frac16) \pi < \ln n < (k + \frac56)\pi$
sum to at least $\frac12 \sum_{n=N_1}^{N_2} \frac1n$,
where
$$
N_1 = \Bigl\lceil \exp\Bigl((k+\frac16)\pi\Bigr) \Bigr\rceil, \quad
N_2 = \Bigl\lfloor \exp\Bigl((k+\frac56)\pi\Bigr) \Bigr\rfloor.
$$
Then $N_1 / N_2 \rightarrow \exp(2\pi/3)$ as $k \rightarrow \infty$,
so $\sum_{n=N_1}^{N_2} \frac1n \rightarrow 2\pi/3$
(and even without the asymptotic formula for the harmonic sums
we have the crude lower bound
$\sum_{n=N_1}^{N_2} \frac1n
> \sum_{n=N_1}^{N_2} \frac1{N_2} > (N_2-N_1)/N_2$
which approaches a positive limit).
This gives the desired estimate and completes the proof.
[The argument readily generalizes to prove the non-convergence of the sum
$\sum_{n=1}^\infty \frac1n \sin(\ln(tn))$ associated to
${\rm Im}(\zeta(1+it))$ for any $t \neq 0$.]
A: 
Recovered mainly from a now deleted post.

$f(n) = \frac{\sin (\ln(n))}{n}$ 
Let's consider integrating by parts: 
$$\int_{n}^{n+1}v'(x,n)f(x)dx=v(x,n) f(x)\left|_{n}^{n+1}\right.-\int_n^{n+1}v(x,n)f'(x)dx$$
$$v'(x,n)=1\ ,\ v(x,n)=x+c(n)\ , \ v(n+1,n)=-v(n,n)\rightarrow\ c(n)=-n-0.5$$
So:
$$\int_{n}^{n+1}f(x)dx=\frac{f(n+1)+f(n)}{2}-\int_n^{n+1}(x-n-0.5)f'(x)dx$$
$$\int_{1}^{+\infty}f(x)dx=-\frac{f(1)}{2}+\sum_{n=1}^{+\infty}f(n)-\sum_{n=1}^{+\infty}\int_n^{n+1}(x-n-0.5)f'(x)dx$$
$$\left|\int_n^{n+1}(x-n-0.5)f'(x)dx\right|\leq\int_n^{n+1}\left|(x-n-0.5)\right|\left|f'(x)\right|dx$$
$$\int_n^{n+1}\left|(x-n-0.5)\right|\left|f'(x)\right|dx\leq\sup\{|f'(x)|:n \leq x \leq n+1\}\cdot\int_n^{n+1}\left|(x-n-0.5)\right|dx$$
$$\int_n^{n+1}\left|(x-n-0.5)\right|\left|f'(x)\right|dx\leq\frac{\sup\{|f'(x)|:n \leq x \leq n+1\}}{4}$$
$$f'(x)=\frac{\sqrt{2}\sin\left(\ln(x)-\frac{\pi}{4}\right)}{x^2}\rightarrow\ \sup\{|f'(x)|:n \leq x \leq n+1\} \leq\frac{\sqrt{2}}{n^2}$$
So finally: 
$$\left|\int_n^{n+1}(x-n-0.5)f'(x)dx\right|\leq\int_n^{n+1}\left|(x-n-0.5)\right|\left|f'(x)\right|dx \leq \frac{\sqrt{2}}{4n^2}$$
Which means that the sum of the series: $\sum_{n=1}^{+\infty}\int_n^{n+1}(x-n-0.5)f'(x)dx$ is absolutely convergent, because $\sum_{n=1}^{+\infty}\frac{\sqrt{2}}{4n^2}$ is convergent (see integral test for convergence,The Basel problem).
......but: 
$\int_1^{+\infty}f(x)dx$ is divergent,because: $\int f(x)dx = -\cos (\ln (x))+C$
And finally , the sum of the series:
$$\sum_{n=1}^{+\infty}f(n) = \int_{1}^{+\infty}f(x)dx+\frac{f(1)}{2}+\sum_{n=1}^{+\infty}\int_n^{n+1}(x-n-0.5)f'(x)dx$$
is divergent. 
Counting bounds:
According to Noam D. Elkies and Wolfram Alpha:
$$\sum_{n=1}^{+\infty}\int_n^{n+1}(x-n-0.5)f'(x)dx=\Im\left(\zeta (1-i)) -\frac{1}{(1-i)-1}\right)$$
Which is strongly related with zeta function regularization and the fact that $f(n)=\Im\left(\frac{1}{n^{1-i}}\right)$.
