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.


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 $\frac12 \sum_{n=N_1}^{N_2} \frac1n$,
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 $t \neq 0$.]

Source : Link , Question Author : Darius , Answer Author : Bombyx mori

Leave a Comment