Doubt on the convergence of this series (looks like a very hard and interesting issue). I would like to know if the series $\displaystyle 2\sum_{k=1}^{\infty} |k|^{-r} \cos(kx)$ or $\displaystyle \sum_{k\neq 0} |k|^{-r} e^{ikx}$ (actually they are the same) is convergent, where $0<r<1$, $x\in (-\pi,\pi)$ (one dimension).
Let us ignore the point $x=0$. Since if $x=0$, this series is divergent immediately so we do not need to consider the original seires anymore.
The answer is yes but I do not know why. Could you tell me which page of the book should I find on this knowledge?
 A: The following things are true:
Let $f(x)$~$\displaystyle \sum_{k=1}^{\infty} k^{-r} \cos(kx), 0<r<1, x \in [-\pi,\pi]$
Then as $\displaystyle \sum_{k=1}^{\infty} k^{-r} \cos(kx)$ converges uniformly on any compact set in $[-\pi, \pi]$ that doesn't contain $0$ (or more generally doesn't contain $2k\pi$ if we extend to the real axis by periodicity), $f$ is continuos except at $0$ (or more generally $2k\pi$). 
Since $f(x)=(\Gamma(1-r) \sin\frac{\pi r}{2})x^{r-1}+O(1), 0<r<1$ when $0 <x \le \pi$ and $f$ is even, $f \in L^1[-\pi,\pi]$ and $\displaystyle \sum_{k=1}^{\infty} k^{-r} \cos(kx)$ is its Fourier series
The first claim follows from the usual Dirichlet kernel and summation by parts since $\frac{1}{\sin \frac{x}{2}}$ is uniformly bounded on compact sets in $[-\pi,\pi]$ that do not contain $0$
The second statement is trickier but it follows from the expansion of $\frac{1}{(1-z)^r}$ on $|z|=r<1$ using the usual Binomial formula , estimates for said Binomial coefficients and then taking the limit $r \to 1$ and separating real and imaginary parts (which gives a similar formula for the corresponding $\sin$ series)
