TITLE: Series convergent but not absolutely? $\sum_{n=1}^{\infty} \frac{\cos(n^p \pi)}{n^p}$
QUESTION [10 upvotes]: For which real numbers $p>0$ does the series
$$\sum_{n=1}^{\infty} \frac{\cos(n^p \pi)}{n^p}$$
converge? 
Obviously it converges absolutely for $p>1$ but what about $0<p<1$? I have the feeling that something qualitative happens at $p=1/2$.
Help wanted, thanks.

REPLY [3 votes]: I. The series is absolutely convergent if and only if $p>1$.
This is trivial for $p>1$. If $p\le1$ then the half of the terms we have $|\cos(..)|\ge\frac1{\sqrt2}$ and $\sum\frac1{n^p}$ diverges.
II. We prove that the series diverges for $0<p\le\frac12$.
Estimate a partial sum of consecutive terms where
$\cos(n^p\pi)\ge\frac12$. Take a large integer $k$ and consider those indices $n$ that satisfy $2k-\frac13\le n^p\le 2k+\frac13$; so let
$$
S_k = \sum_{2k-\frac13\le n^p \le 2k+\frac13} \frac{\cos(n^p\pi)}{n^p}.
$$
The number of terms in $S_k$ is roughly
$$
(2k+\frac13)^{1/p}-(2k-\frac13)^{1/p} \approx
C_1\cdot k^{\frac1p-1}
$$
with some positive constant $C_1$; the magnitude order of the terms is $k^{-1}$.
Hence,
$$
S_k > C_2 k^{\frac1p-2}.
$$
The last exponent, $\frac1p-2$ is nonnegative. So $S_k$ has a positive lower bound, so $S_k\not\to0$. Therefore, the series diverges.
III. Now assume $\frac12<p<1$. Replace the series by the improper integral. It is easy to verify that
$$
\left|\frac{\cos(n^p\pi)}{n^p} - \int_{n-\frac12}^{n+\frac12}\frac{\cos(x^p\pi)}{x^p}\,dx \right| \le 
C_3 \max_{n\le x\le n+1} \left|\bigg(\frac{\cos(x^p\pi)}{x^p}\bigg)''\right|
< \frac{C_4}{n^{2-p}}. 
$$
Since $\sum\limits\frac1{n^{2-p}}$ is convergent, we can conclude that $\sum\limits_{n=1}^\infty\frac{\cos(n^p\pi)}{n^p}$ is convergent if and only if 
$\int\limits_{1}^\infty\frac{\cos(x^p\pi)}{x^p}dx$ is convergent.
Integrating by parts,
$$
\int_1^\infty \frac{\cos(x^p\pi)}{x^p}dx =
\frac1{p\pi}\int_1^\infty (\sin(x^p\pi))' \frac1{x^{2p-1}}dx =
$$
$$
\frac1{p\pi}\left[\frac{\sin(x^p\pi)}{x^{2p-1}}\right]_1^\infty
+\frac{2p-1}{p\pi}
\int_1^\infty \frac{\sin(x^p\pi)}{x^{2p}} dx.
$$
Due to $2p>1$ the last integral is absolutely convergent.