Show that $\sum_{k=1}^{\infty}\frac{1}{k^2}\frac{1}{\sqrt{|\sin(kx)|}}$ converges a.e (Proof verification) Solution: 
Note that if we fix $x \neq 2\pi n$ for all integers $n$, then $\frac{1}{\sqrt{|\sin(kx)|}}$ is bounded by some constant $C$. Also note that the periodicity of sin$(x)$ means that such a constant exists that the expression is bounded for all $k$. Thus we get 
$$\sum_{k=1}^{\infty}\frac{1}{k^2}\frac{1}{\sqrt{|\sin(kx)|}} \leq C\sum_{k=1}^{\infty}\frac{1}{k^2}$$
and the right hand side clearly converges. Noting that the set of $x$ that are integer multiples of $\pi$ has measure zero, the result has been shown. 
 A: Put $f_k(x)=\frac{1}{k^2}\frac{1}{\sqrt{|\sin(kx)|}}$. By the change of variable $kt=u$, we get 
$$\int_0^{2\pi}\frac{dt}{\sqrt{|\sin(kt)|}}=\frac{1}{k}\int_0^{2\pi k}\frac{du}{\sqrt{|\sin(u)|}}=\int_0^{2\pi}\frac{du}{\sqrt{|\sin(u)|}}$$
and that the last integral is finite.
Hence the series $\displaystyle \int_0^{2\pi}|f_k(t)|dt=\int_0^{2\pi}f_k(t)dt=u_k$ is convergent. By a known result, this imply that the sum $\displaystyle \sum f_k(x)=f(x)$ is in $L^1([0,2\pi])$, hence finite a.e.
A: It is enough to prove that the exceptional set $E$ given by 
$$ E=\{x\in\mathbb{R}:\,\left|\sin(nx)\right|\leq n^{-2+\frac{1}{9}}\text{ holds for infinite values of } n\in\mathbb{N}^+\}$$ has measure zero. $\pi$ is a trascendental number with a finite irrationality measure: by invoking Salikov's (2008) result we have that
$$ \left|\pi-\frac{p}{q}\right|\leq \frac{1}{q^8} $$
holds just for a finite number of $\frac{p}{q}\in\mathbb{Q}^+$. By considering some $x\in\mathbb{R}\setminus \pi\mathbb{Z}$ and the convergents of its continued fraction, then by exploiting the Lipschitz-continuity of $f(z)=\sin z$, we have that $E$ is given by a neighbourhood of $\pi\mathbb{Z}$ with measure zero, hence
$$ \sum_{n\geq 1}\frac{1}{n^2\sqrt{|\sin(nx)|}} $$
is a.e. absolutely convergent by asymptotic comparison with $\sum_{n\geq 1}\frac{1}{n^{1+1/18}}$, which is convergent.
