Is $\sum_{n=0}^\infty \int_0^1 \frac{\cos\left(n\pi x\right)}{\log\log\frac{4}{x}}dx$ is divergent or not? I'm searching for a continuous function whose Fourier series diverges at x=0.
Now I've come up with an idea and don't know if it's right.
$$\sum_{n=0}^\infty \int_0^1 \frac{\cos\left(n\pi x\right)}{\log\log\frac{4}{x}}dx$$
Is the series above divergent or not? I expect it be divergent.
I calculated its sum of first N terms (N varies from 1 to 1000) with help of Mathematica, there seems no sign that it converges.  

$\sum_{n=0}^\infty \int_0^1 \frac{\cos\left(n\pi x\right)}{\log\log\frac{4}{x}}dx$=$\int_0^1 \frac{1}{\log\log\frac{4}{x}}dx$+$\sum_{n=1}^\infty \int_0^1 \frac{\cos\left(n\pi x\right)}{\log\log\frac{4}{x}}dx$
Using Riemann sum for the integral:
$\int_0^1 \frac{\cos\left(n\pi x\right)}{\log\log\frac{4}{x}}dx$=$\lim_{n -> \infty} {\frac{1}n}\sum_{k=0}^n$ $\frac{(-1)^k}{\log\log\frac{4}{\frac{k}{n}}}$
We get:
=$\int_0^1 \frac{1}{\log\log\frac{4}{x}}dx$+$\sum_{n=1}^\infty {\frac{1}n}\sum_{k=0}^n$ $\frac{(-1)^k}{\log\log\frac{4n}{k}}$= 
The sum can be decreased by using $log(log\frac{4n}{k})<\frac{4n}{k}$ where ${k}\le {n}.$ 
$>$ $\int_0^1 \frac{1}{\log\log\frac{4}{x}}dx$+$\sum_{n=1}^\infty {\frac{1}{4n^2}}\sum_{k=0}^n$ ${k}{(-1)^k}$= =$\int_0^1 \frac{1}{\log\log\frac{4}{x}}dx$+$\sum_{n=1}^\infty {\frac{1}{4n^2}}\frac{n}{2}$=
$\int_0^1 \frac{1}{\log\log\frac{4}{x}}dx$+$\frac{1}{8}\sum_{n=1}^\infty {\frac{1}{n}} \longmapsto \infty$ if ${n \longmapsto \infty}$
Finally 
$\sum_{n=0}^\infty \int_0^1 \frac{\cos\left(n\pi x\right)}{\log\log\frac{4}{x}}dx$ $\gt \infty$
