Show that the sum:
$\displaystyle S(x)=\sum_{n=1}^{\infty}\biggl({|\tan(n^3x)|\over{n^3}}\biggl)^{\!1/2}$
converges almost everywhere on $\mathbb{R}$.
My thoughts:
We define $\displaystyle f_N(x)=\sum_{n=1}^{N}\biggl({|\tan(n^3x)|\over{n^3}}\biggl)^{\!\!1/2}$.
Now, because then $f_N$ are all continuous except in a countable amount of discontinuities points, specifically, $\{{{\pi}\over{2n^3}}+{{\pi k}\over{n^3}}\}_{n\in \mathbb{Z}}$, then $f_N$ are integrable on $\mathbb{R}$.
Moreover, $f_N$ are non-negative, and increasing and their limit is $S(x)$.
Then, by the monotone convergence theorem, we get:
$$\int_{\mathbb{R}}S(x)dx=\lim_{N\rightarrow \infty}\int_{\mathbb{R}}f_N(x)=\lim_{N\rightarrow \infty}\int_{-\infty}^{\infty}\sum_{n=1}^{N}\biggl({|\tan(n^3x)|\over{n^3}}\biggl)^{1\over2}.$$
So, $\int_{\mathbb{R}}S(x)dx=\sum_{n=1}^{\infty}{n^{-{3\over2}}}\cdot \int_{-\infty}^{\infty} \sqrt{|\tan{(n^3x)}|}dx$.
Now, is it sufficient to show that the indefinite integral $\int_{-\infty}^{\infty} \sqrt{|\tan{(n^3x)}|}dx$ converges, right?
If so, how can I do that?
Please help!