# Convergence of $\sum_{k = 1}^{\infty} \cos (\log k)$ and $\sum_{k = 1}^{\infty} \sin (\log k)$

I was trying to determine if

$$\sum_{k = 1}^{\infty} \cos (\log k)$$

(with $$k = 1, 2, \ldots$$) converges. $$\log$$ is the natural logarithm.

I tried the ratio test,

$$\lim_{k \to \infty} \frac{\cos \left[ \log ( k + 1 ) \right]}{\cos \left[ \log ( k ) \right]} = 1$$

but it is not useful to determine the convergence.

According to Wolfram Alpha, it diverges. How to prove it?

And what about $$\displaystyle \sum_{k = 1}^{\infty} \sin (\log k)$$?

What I can empirically observe is that, regardless of how big $$\log k$$ can become, its cosine will always be a number $$x$$ such that $$-1 \leq x \leq 1$$. But I can't immediately see that this leads to an infinite sum.

Trivially, the term test is:

$$\lim_{k \to \infty} \cos (\log k)$$

and this limit is not defined, due to the cosine. Maybe this is sufficient to state that the series diverges.

• Yes, for an infinite series to converge, it is necessary that its terms tend to $0$.
– Gary
Feb 10, 2021 at 11:54

Suppose the series converges $$\sum_{k=1}^\infty \cos(\log k) = s \in \Bbb R$$ and let $$s_n = \sum_{k=1}^n \cos(\log k)$$.

This would imply $$\cos(\log(n)) = s_{n} - s_{n-1} \to s-s = 0$$ by arithmetic of limits, but $$\lim_{n\to\infty} \cos(\log n)$$ doesn't exist.

The argument that $$\sum \sin(\log k)$$ diverges should be similar.

$$\log \lfloor e^{2\pi k} \rfloor = \log( e^{2\pi k}+O(1))= \log e^{2\pi k} +\log(1+ O(e^{-2\pi k}))=2\pi k+O(e^{-2\pi k})$$

Thus $$\lim_{k\to \infty}\cos(\log \lfloor e^{2\pi k} \rfloor )=1$$

Similarly $$\lim_{k\to \infty}\sin(\log \lfloor e^{2\pi (k+1/4)} \rfloor )=1$$

For $$k$$ sufficiently big, $$\log(k+1)-\log k<\frac\pi2$$. On the other hand, $$\log k$$ is unbounded. Therefore, $$\log k$$ "walks" to $$\infty$$ in steps so small that it will hit every interval $$[2m\pi-\frac\pi4,2m\pi+\frac\pi4]$$ with $$m\gg0$$. That is, we will have $$\cos \log k\ge \frac{\sqrt 2}2$$ infinitely often.

• Thank you for these very useful observations. If it's possible, can you please provide a link about the assumption $\log (k + 1) - \log k < \pi / 2$? I didn't know this proof. Feb 10, 2021 at 16:46
• @BowPark $\log(k+1)-\log(k) = \log(\frac{k+1}{k}) \to \log(1) = 0$ (by continuity of $\log$) so this eventually gets small enough Feb 10, 2021 at 17:03

Suppose that $$a_n =\cos (\log n )\to 0$$ then of course $$a_{10^k} =\cos k \to 0.$$ But it is well known that the last is not true.

• Did you mean $a_{\lfloor e^{2\pi k} \rfloor}$ ? Feb 10, 2021 at 12:28
• log is rather $log_{10}$
– user235708
Feb 10, 2021 at 12:34
• Sorry, can you rephrase your answer? I can not understand it. In my question $\log$ is referred to the natural logarithm. Feb 10, 2021 at 15:40