Let $Q(x.\,m)=\sum_{k=m}^{2m-1}\frac{\cos{kx}}{2m-k}+\sum_{k=2m+1}^{3m}\frac{\cos{kx}}{2m-k}$. How do I prove that the function $f(x)=\sum_{n=1}^\infty\frac{Q(x,\,2^{n^2})}{n^2}$ is continuous and $2π$-periodic, but its Fourier series diverges at the point $x=0$?
The task is actually far from simple, I found an analogue of it in a book "R.E. Edwards. Fourier series: A modern introduction" on page $161$. But only there instead of $n^2$ there is $n^3$, because of this I also had doubts that there was a typo in this task, as it turned out - no.
My attempt:
Upon closer examination, it turned out that this example differs from the one presented in the book, although the principle of reasoning is the same here. And there are no typos in the problem - everything is solved well. Plus, I will add a justification of continuity, since the uniform boundedness of sums of the form $\sum_{k=1}^{N} \frac{\sin k x}{k}$ is an interesting and unobvious fact in itself.
First, rewrite $Q(x, m)=2 \sin (2 m x) \sum_{k=1}^{m} \frac{\sin k x}{k}$ and show that $Q(x, m)$ is uniformly bounded. Then the series $\sum_{n=1}^{\infty} \frac{Q\left(x, 2^{n^{2}}\right)}{n^{2}}$ will converge uniformly according to the Weierstrass criterion and $f(x)$ will be continuous. So, it is necessary to show that the sum $\sum_{k=1}^{N} \frac{\sin k x}{k}$ is uniformly bounded by a constant independent of either $x$, or $N$. It is enough to consider only the interval $0<x<\pi .$
We denote $m=\min \{N,[\pi / x]\}$ and divide the sum into two: $\sum_{k=1}^{m} \frac{\sin k x}{k}+\sum_{k=m+1}^{N} \frac{\sin k x}{k}$ (the second sum can be empty). Then $\left|\sum_{k=1}^{m} \frac{\sin k x}{k}\right| \leq \sum_{k=1}^{m} \frac{k x}{k}=m x \leq \pi$, because $m \leq \frac{\pi}{x} .$ To estimate the second sum, we use the Abel transform and obtain that $$\sum_{k=m+1}^{N} \frac{\sin k x}{k}=\frac{S_{N}}{N}-\frac{S_{m}}{m+1}-\sum_{k=m+1}^{N-1} S_{k}\left(\frac{1}{k+1}-\frac{1}{k}\right)$$, where $S_{n}=\sum_{k=1}^{n} \sin k x$ and $\left|S_{n}\right| \leq \frac{1}{\left|\sin \frac{x}{2}\right|}$, it follows that $$\left|\sum_{k=m+1}^{N} \frac{\sin k x}{k}\right| \leq \frac{3}{(m+1)\left|\sin \frac{x}{2}\right|} \leq \frac{3}{(m+1) \frac{x}{\pi}} \leq \frac{3 \pi}{m x} \leq 3 \pi$$, since $m x \geq 1 .$
Thus, the uniform boundedness of $Q(x, m)$ is proved. In particular, this makes it possible to perform term-by-term integration of the sum of the series when calculating the Fourier coefficients. Let $S_{k}$ - - be the partial sum of the Fourier series of the function $f(x)$ for $x=0 .$ Note that the Fourier series contains only cosines, therefore (without taking into account the normalizations) $$\left|S_{3 \cdot 2^{n^{2}}}-S_{2 \cdot 2^{n^{2}}}\right|=\left|\sum_{k=2 \cdot 2^{n^{2}}+1}^{3 \cdot 2^{n^{2}}} c_{k}\right| .$$ If $k$ varies from $2 \cdot 2^{n^{2}}+1$ to $3 \cdot 2^{n^{2}}$, then the first sum in the definition of $Q\left(x, 2^{n^{2}}\right)$, multiplied scalarly by $\cos k x$ gives $0$, and from the second sum for each fixed $n$ there remains only one term with the corresponding number (it is important the fact that for different $n$ in $Q\left(x, 2^{n^{2}}\right)$ there are no common harmonics). Accurate calculations give $\left|S_{3 \cdot 2^{n^{2}}}-S_{2 \cdot 2^{n^{2}}}\right|=\frac{1}{n^{2}} \cdot \sum_{k=1}^{2^{n^{2}}} \frac{1}{k} \rightarrow \ln 2$, since $\sum_{k=1}^{m} \frac{1}{k} \sim \ln m .$ Thus, for sufficiently large $n$, $$\left|S_{3 \cdot 2^{n^{2}}}-S_{2 \cdot 2^{n^{2}}}\right| \geq \frac{\ln 2}{2}$$ and according to the Cauchy criterion, the Fourier series diverges at zero.