How to prove $f'(x) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \cos(x/k)\overset{?}{=} \Omega(\log \log x)$?

Question

While investigating $f(x) = \sum_{k=1}^{\infty} (-1)^{k+1} \sin(x/k)$ such as on these questions on its unboundedness and its infinite number of real zeros, I referenced one of the first post on MSE about it that mostly did some numerical explorations. In this answer, was plotted the graph of $f'(x)$: $$ f'(x) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \cos\left(\frac{x}{k}\right) $$ At first glance, the graph of $f'(x)$ appears to be bounded, but comments hinted that there exists a sequence of points $x_n \to \infty$ where the $f'(x_n)$ grows as $\log \log x_n$. I was not able to reconstruct the full argument from the brief discussion, nor being able to prove that this rate is optimal, meaning $f'(x)=O(\log \log x)$. I would appreciate a self-contained answer explaining how to rigorously establish this result.

$$ \boxed{\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \cos\left(\frac{x}{k}\right) \overset{?}{=} \Omega(\log \log x)} $$

Answer 1

Let $g(x)=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \cos\left(\frac{x}{k}\right)$

For simplicity, we will state two lemmas (inspired by the Hardy-Littlewood second paper on Lambertian summation where a result similar to the OP one is proven) to be proven later and show how the result $$g(x)=\Omega (\log \log x)$$ is deduced from them. Note that the problem of the "true" order of $g$ is related to tricky exponential sums, so there is no claim on that.

Lemma $1$: $$g(x)=\sum_{k \le x} \frac{(-1)^{k+1}}{k} \cos\left(\frac{x}{k}\right)+O(1)$$

Lemma $2$: If $c \sqrt x \le n <m \le x$ then $$g_{n,m}(x)=\sum_{n < k \le m} \frac{(-1)^{k+1}}{k} \cos\left(\frac{x}{k}\right)=O_c(1)$$

Let $A_N=[1,3,...,2N+1]$ (the least common multiple of all odds up to $2N+1$) and $K_N=2\pi A_N$ and for each odd $1 \le j \le A_N$ let $x_j=jK_N$ so $K_N \le x_j \le K_N^2$. Hence $\frac{\sqrt x_j}{2\pi} \le A_N \le \frac{x_j}{2\pi}$ so by Lemma $1$ we have$$g(x_j)=\sum_{k \le x_j} \frac{(-1)^{k+1}}{k} \cos\left(\frac{x_j}{k}\right)+O(1)=\sum_{k \le A_N} \frac{(-1)^{k+1}}{k} \cos\left(\frac{x_j}{k}\right)+g_{A_N,x_j}(x_j)+O(1)$$ and by Lemma $2$ we have $$g_{A_N,x_j}(x_j)=O_{\frac{1}{2\pi}}(1)=O(1)$$

Hence $$g(x_j)=\sum_{k \le A_N} \frac{(-1)^{k+1}}{k} \cos\left(\frac{x_j}{k}\right)+O(1)$$ and we will prove that (with $B_N=\frac{A_N-1}{2}$) $$G_N=\frac{1}{B_N+1}\sum_{k=0}^{B_N}g(x_{2k+1})=\frac{1}{B_N+1}\sum_{k=0}^{B_N}\sum_{m=1}^{A_N}\frac{(-1)^{m+1}}{m} \cos\left(\frac{(2k+1)K_N}{m}\right)+O(1)$$ $$G_N>> \log N+O(1) $$

This means that for at least one $j_N$ we have $g(x_{j_N}) >> \log N$ and since clearly $N \sim \log K_N$ while $\log K_N \le \log x_j \le 2\log K_N$ the result follows.

($\log A_N=\sum_{p \hspace{1mm} odd, p^a \le 2N+1}\log p=\sum_{n \hspace{1mm} odd \le 2N+1}\Lambda(n)=N(1+o(1))$ by carefully applying PNT or by looking at the asymptotics of $\zeta'_{odd}/\zeta_{odd}$ where $\zeta_{odd}(s)=\sum_{n \hspace{1mm} odd}n^{-s}$ etc)

Now we note that $\cos x+\cos 3x+...\cos (2B_N+1)x=\frac{\sin (2B_N+2)x}{2\sin x}$ where the sum is $B_N+1$ if $x=2k\pi$ and $-B_N-1$ if $x=(2k+1)\pi$

So fixing an $m$ and switching sums in the above, we have $$\sum_{k=0}^{B_N}\cos\left(\frac{(2k+1)K_N}{m}\right)=\frac{\sin (A_N+1)K_N/m}{2\sin K_N/m}$$ while if $m|2A_N$ we either have $m$ odd so $A_N/m$ is integral and $K_N/m$ is an even multiple of $\pi$ and the sum is $B_N+1$ or $m$ is even and then $2A_N/m$ is an odd integer so the sum is $-B_N-1$.

In all other cases using $\frac{1}{|\sin \pi y|} =O(1/||y||)$ where $||y||$ is the distance from $y$ to the closest integer, each sum is $O(1/||2A_N/m||)$ and since $2A_N/m$ is now a non-integral fraction with denominator at most $m$, we have $O(1/||2A_N/m||)=O(m)$

Since each even $m$ is taken with negative sign in our general sum, all the terms with $m$ even dividing $2A_N$ become positive as do all the terms with odd $m$ dividing $A_N$. Also by our choice $m|A_N$ for $m=1,3,...2N+1$ so the first odd terms up to $2N+1$ sum to $\sum_{k=0}^N\frac{1}{2k+1}=\frac{\log N}{2}+O(1)$, while of course each term $m$ which doesn't divide $2A_N$ contributes an $O(m)/m=O(1)$ to the total sum hence a $O(\frac{1}{B_N+1})$ to $G_N$. Since there are at most $A_N$ such terms we get that the part of $G_N$ given by these $m$ is $O(1)$.

Putting all together we get $$G_N \ge \frac{\log N}{2}+O(1)$$ as asserted and the conclusion $g(x) =\Omega(\log \log x)$ is proven modulo the two lemmas.

Lemma $1$ folows using $$\frac{1}{k}\cos \frac{x}{k}-\frac{1}{k+1}\cos \frac{x}{k+1}=O(\frac{1+x}{k^2})$$ so summing on $k>x$ we get an $O((1+x)/x)=O(1)$

Lemma $2$ follows immediately from Kuzmin Landau but we will give the proof from the Hardy-Littlewood cited paper $$g_{n,m}(x)=\sum_{n < \le m} \frac{(-1)^{k+1}}{k} \cos\left(\frac{x}{k}\right)=\sum_{n < k \le m} \frac{1}{k} \cos\left(\frac{x}{k}\right)-\sum_{n/2 < k \le m/2} \frac{1}{2k} \cos\left(\frac{x/2}{k}\right)$$

The condition $c\sqrt x \le n \le m \le x$ is satisfied for $n/2, m/2, x/2$ with $c/\sqrt 2$ so it is enough to prove the result $$h_{n,m}(x)=\sum_{n < k \le m} \frac{1}{k} \cos\left(\frac{x}{k}\right)=O_c(1)$$ under $c\sqrt x \le n <m \le x$

But $$h_{n,m}(x)-\int_{n}^{m+1}\frac{1}{t}\cos \frac{x}{t}dt=\sum_{k=n}^m\int_k^{k+1}(\frac{1}{k}\cos \frac{x}{k}-\frac{1}{t}\cos \frac{x}{t})dt$$

The derivative in $t$ of $\frac{\cos (x/t)}{t}$ is $-\frac{\cos (x/t)}{t^2}-\frac{x\sin (x/t)}{t^3}$ so is $O(1/t^2)+O(x/t^3)$ hence by the MVT each integral is $O(1/k^2)+O(x/k^3)$ and summing $k$ from $n$ to $m$ gets us $O(1)+O(x/n^2)=O_c(1)$.

But the change of variable $x=ty$ so $|dt/t|=|dy/y|$ gives $$\int_{n}^{m+1}\frac{1}{t}\cos \frac{x}{t}dt=\int_{x/(m+1)}^{x/n}\frac{\cos y}{y}dy=\frac{m+1}{x}\int_{(m+1)/x}^c\cos y dy=O((m+1)/x)=O(1)$$ by the second mean value theorem for integrals and our choices (or integrate by parts etc) and Lemma $2$ is proven too.