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.

  • $\begingroup$ It looks familiar. One of self-challenging problem that I found online was: Determining convergence of $\sum \frac{\sin(\ln n)}{n\ln n}$. $\endgroup$ – i707107 Feb 12 '14 at 5:15
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    $\begingroup$ $$\sum _{n=2}^{\infty } \frac{\sin (\log (n))}{n \log (n)}=\Im\left(\int_1^{\infty } (-1+\zeta (x-i)) \, dx\right)\approx 1.14964647671184932067946796534639573228838143131427$$ $\endgroup$ – Darius Feb 12 '14 at 10:18
  • $\begingroup$ @i707107 $\sum\frac{\sin(\ln(n))}{n}$ is bounded (read both answers), so en.wikipedia.org/wiki/Dirichlet%27s_test is passed. $\endgroup$ – Darius Feb 13 '14 at 8:22
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    $\begingroup$ @i707107: I've recently posted a proof in chat that $$\sum_{n=1}^\infty\frac{\sin(H_n)}{nH_n}$$ diverges and the proof should be almost identical for your series. $\endgroup$ – robjohn Apr 14 '14 at 8:05

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 $t \neq 0$.]

  • $\begingroup$ You mean en.wikipedia.org/wiki/Cauchy%27s_convergence_test summands ai is convergent if and only $\forall_{\varepsilon > 0} \exists_{n_0 \in N} \forall_{n\geqslant n_0} \forall_{k\in N} \left \vert \sum_{i=n}^{n+k} a_i \right \vert < \varepsilon$ $\endgroup$ – Darius Jan 9 '14 at 4:50
  • $\begingroup$ @user119734 No, not really. $\endgroup$ – Pedro Tamaroff Feb 10 '14 at 16:53
  • $\begingroup$ Ok, so summands $a_i$ is divergent if and only $\forall _{{n_{0}\in N}}\exists _{{\varepsilon >0}}\exists _{{n\geqslant n_{0}}}\exists _{{k\in N}}\left\vert \sum _{{i=n}}^{{n+k}}a_{i}\right\vert \ge\varepsilon$ $\endgroup$ – Darius Feb 10 '14 at 17:03
  • $\begingroup$ @user119734 The point is not that, rather the delicate observations Elkies makes about the sine, and how he proves the assertion made in the first paragraph. $\endgroup$ – Pedro Tamaroff Feb 10 '14 at 17:10
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    $\begingroup$ Do you mean $N_1/N_2\to\exp(-2\pi/3)$? $\endgroup$ – Matemáticos Chibchas Nov 3 '16 at 21:59

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 $f(n)=\Im\left(\frac{1}{n^{1-i}}\right)$.


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