Does $\sum_{n=1}^\infty \frac{\cos(\ln(n))}{n}$ converge? [duplicate]

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EDIT: the question is answered here Divergence of $\sum\limits_{n=1}^{\infty} \frac{\cos(\log(n))}{n}$

Using integrals, I managed to prove that $$\displaystyle \forall m, \sin(\ln(m+1))\leq \sum_{n=1}^m \frac{\cos(\ln(n))}{n}\leq 1+\sin(\ln(m))$$ and I noticed that $\sin(\ln(m+1)) - \sin(\ln(m)) \to 0$

I tend to believe that the series diverges, but I can't prove it with the previous inequality.

marked as duplicate by Steven Stadnicki, Jack D'Aurizio, Jonas Meyer, Frunobulax, user147263 Sep 18 '14 at 15:31

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• I would say $\displaystyle \lim_{m \to \infty}\sin(\ln(m+1)) - \sin(\ln(m))$ not exist. – georg Sep 18 '14 at 14:12
• @georg, Why not? We know that $\ln(m+1)-\ln(m)\to 0$, so shouldn't their sines get very close together, too? – G Tony Jacobs Sep 18 '14 at 14:13
• @barakmanos there are lot of things Wolfram can't do. – User976796 Sep 18 '14 at 14:27
• Nahhhh... That website can make you a cup of Fibonacci coffee using the last $2$ cups of coffee that you left on your desk, and serve it to you in the right temperature. – barak manos Sep 18 '14 at 14:33
• related: math.stackexchange.com/q/28599 – Jonas Meyer Sep 18 '14 at 14:47

5 Answers

This is not an answer. Just my observation.

$$\sum_{n=1}^\infty \frac{\cos(\ln(n))}{n}=\sum_{n=1}^\infty \frac{1}{2}\left(\frac{\exp(i\ln(n))}{n}+\frac{\exp(-i\ln(n))}{n}\right)$$ $$=\sum_{n=1}^\infty \left(\frac{1}{n^{1-i}}+\frac{1}{n^{1-i}}\right)=\frac{1}{2}(\zeta(1-i)+\zeta(1+i))$$

I also believe that it is convergent.

• The series is $(R,1)$ summable and this is indeed the limit in $(R,1)$ sense. – Sungjin Kim Sep 18 '14 at 14:59

Assuming what you've already proven is correct,

$\sin(\ln(m+1)) \ge 2/3$ for infinitely many $m$, and $1+\sin(\ln(m)) \le 1/3$ also for infinitely many $m$.

(because those correspond to some small intervals for $\ln(m)$ to fall into, all of the form $[a;a+L]$. Because $\ln(m+1) - \ln(m) < 2/(m+1)$, as soon as $1/a < 2L$, you are guarantueed to get in there by picking the first $n$ greater than $e^a$)

Your series takes values greater than $2/3$ or less than $1/3$ infinitely many times, hence it can't converge.

• Can you sketch a proof of your estimations ? – User976796 Sep 18 '14 at 14:43
• @i707107 I also think so. However the oscillating behaviour of the integral should be more and more close to what happens to the series as $n$ grows large. I'll work on a complete answer. – mercio Sep 18 '14 at 14:46

For each $n \in \mathbb{Z}_{+}$, let

$$a_n = \frac{\cos\log n}{n} - \int_n^{n+1}\frac{\cos\log x}{x} dx = \int_0^1 \left(\frac{\cos\log n}{n} - \frac{\cos\log(n+t)}{n+t}\right) dt$$ Since for each $t \in (0,1]$, we can apply MVT to find a $\chi \in (0,t)$ such that

$$\left|\frac{\cos\log n}{n} - \frac{\cos\log(n+t)}{n+t}\right| = \left|\frac{\sin\log(n+\chi) + \cos\log(n+\chi)}{(n+\chi)^2}\right|t \le \frac{\sqrt{2}t}{n^2}$$

We find $$|a_n| \le \frac{\sqrt{2}}{n^2}\int_0^1 t dt = \frac{1}{\sqrt{2}n^2}$$

As a result, $\sum\limits_{n=1}^\infty a_n$ is an absolutely converging series. Let $\beta$ be the corresponding sum, we have

$$\lim_{N\to\infty}\left[\sum_{n=1}^N \frac{\cos\log n}{n} - \sin\log(N+1)\right] = \lim_{N\to\infty}\sum_{n=1}^N a_n = \beta$$ From this, we see $\displaystyle\;\sum_{n=1}^N \frac{\cos\log n}{n}$ doesn't converge but "oscillate" approximately between $-1+\beta$ and $1+\beta$.

As a side note, I know $\beta = \Re\zeta(1+i)$ but I don't have a proof for that.

It doesn't converge.

Take an integer $k > 0$.

If $2kπ - 1/2 <= x <= 2kπ + 1/2$ then $cos (x) ≥ 0.87$.

If $e^{2kπ - 1/2} <= n <= e^{2kπ + 1/2}$ then $cos (ln ((n)) ≥ 0.87$.

There will be more than $1.04 e^{2kπ}$ such numbers, each is divided by $n < 1.64 e^{2kπ}$, so the sum of $cos (ln (n)) / n$ over these numbers is more than 1.04 x 0.87 / 1.64 > 0.55.

So we will again and again have a sum over a subsequence that is > 0.55, therefore no convergence.

By writing $\cos(z)$ as $\Re(e^{iz})$ we have: $$\sum_{n=1}^{N}\frac{\cos\log n}{n}=\Re\sum_{n=1}^{N}n^{i-1}$$ but the last series is not converging (it behaves like $-1(N^i-1)$) in virtue of the Euler-MacLaurin summation formula. Ok, this is the same approach of sos440 in the other question.