Prove that if $\{a_{k}\}$ is a sequence of real numbers such that
$$\sum_{k=1}^{\infty} \frac{|a_{k}|}{k} = \infty$$and
$$\sum_{n=1}^{\infty} \left( \sum_{k=2^{n-1}}^{2^n-1} k(a_k - a_{k+1})^2 \right)^{1/2} < \infty,$$then
$$\int_{0}^{\pi} \left| \sum_{k=1}^{\infty} a_k \sin(kx) \right| \,dx = \infty.$$
My idea to proveAlthough the condition$$\lim_{k \rightarrow \infty} a_k=0$$does not appear in the statement of the problem, by the well-known Cantor-Lebesgue theorem, this follows from the fact that the series$$\sum_{k=1}^{\infty} a_k \sin k x$$is convergent almost everywhere. We note that for the application of this theorem, it would be sufficient if the series (2) were convergent on a set of positive measure. To make reference easier, we list the remaining conditions:$$\sum_{k=1}^{\infty} \frac{\left|a_k\right|}{k}=\infty$$
$$\sum_{n=1}^{\infty}\left(\sum_{k=2^{n-1}}^{2^n-1} k\left|\Delta a_k\right|^2\right)^{1 / 2}<\infty$$where$$\Delta a_k:=a_k-a_{k+1} \quad(k=1,2, \ldots) .$$
From (4) it follows that the sequence $\left\{a_k\right\}$ has bounded variation, that is,$$\sum_{k=1}^{\infty}\left|\Delta a_k\right|<\infty$$
Really, by the Cauchy inequality,
\begin{aligned}\sum_{k=1}^{\infty}\left|\Delta a_k\right| & =\sum_{n=1}^{\infty} \sum_{k=2^{n-1}}^{2^n-1}\left|\Delta a_k\right| \\& \leq \sum_{n=1}^{\infty}\left(2^{n-1} \sum_{k=2^{n-1}}^{2^n-1}\left|\Delta a_k\right|^2\right)^{1 / 2} \\& \leq \sum_{n=1}^{\infty}\left(\sum_{k=2^{n-1}}^{2^n-1} k\left|\Delta a_k\right|^2\right)^{1 / 2}\end{aligned}
Consider the $n$ th partial sum of (2). By Abel's rearrangement, we obtain$$\sum_{k=1}^n a_k \sin k x = \sum_{k=1}^n \tilde{D}_k(x) \Delta a_k + a_{n+1} \tilde{D}_n(x)$$where $\tilde{D}_n(x)$ is the conjugate Dirichlet kernel:\begin{align*}\tilde{D}_n(x) &:= \sum_{k=1}^n \sin k x \\&= \frac{\cos \frac{x}{2} - \cos \left(n + \frac{1}{2}\right) x}{2 \sin \frac{x}{2}} \quad (n=1,2, \ldots) .\end{align*}
Introduce the notation$$\bar{D}_n(x):=-\frac{\cos \left(n+\frac{1}{2}\right) x}{2 \sin \frac{x}{2}} \quad(n=0,1, \ldots) .$$
Then$$\tilde{D}_n(x)=\bar{D}_n(x)-\bar{D}_0(x) \quad\left(n=0,1, \ldots ; \tilde{D}_0(x)=0\right),$$