Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
It looks familiar. One of self-challenging problem that I found online was: Determining convergence of $\sum \frac{\sin(\ln n)}{n\ln n}$. –  i707107 Feb 12 at 5:15
    
$$\sum _{n=2}^{\infty } \frac{\sin (\log (n))}{n \log (n)}=\Im\left(\int_1^{\infty } (-1+\zeta (x-i)) \, dx\right)\approx 1.14964647671184932067946796534639573228838143131427$$ –  Darius Feb 12 at 10:18
    
@i707107 $\sum\frac{\sin(\ln(n))}{n}$ is bounded (read both answers), so en.wikipedia.org/wiki/Dirichlet%27s_test is passed. –  Darius Feb 13 at 8:22
    
@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. –  robjohn Apr 14 at 8:05

2 Answers 2

up vote 45 down vote accepted

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 $t \neq 0$.]

share|improve this answer
7  
+1 This is worth remembering. –  Pedro Tamaroff Jan 8 at 23:32
    
You mean en.wikipedia.org/wiki/Cauchy%27s_convergence_test summands ai is convergent if and only $\forall_{\varepsilon > 0} \exists_{n_0 \in N} \forall_{n\geqslant n_0} \forall_{k\in N} \left \vert \sum_{i=n}^{n+k} a_i \right \vert < \varepsilon$ –  Darius Jan 9 at 4:50
    
@user119734 No, not really. –  Pedro Tamaroff Feb 10 at 16:53
    
Ok, so summands $a_i$ is divergent if and only $\forall _{{n_{0}\in N}}\exists _{{\varepsilon >0}}\exists _{{n\geqslant n_{0}}}\exists _{{k\in N}}\left\vert \sum _{{i=n}}^{{n+k}}a_{i}\right\vert \ge\varepsilon$ –  Darius Feb 10 at 17:03
    
@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. –  Pedro Tamaroff Feb 10 at 17:10

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.