Series convergent but not absolutely? $\sum_{n=1}^{\infty} \frac{\cos(n^p \pi)}{n^p}$ 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.
 A: 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.
A: We restrict our attention to the case $0 < p < 1$, since the answer is well-known for $p \geq 1$. Using the Riemann-Stieltjes integral, we can rewrite the partial sum as
$$ \sum_{k=1}^{n} \frac{\cos (\pi k^{p})}{k^{p}} = \int_{1^{-}}^{n} \frac{\cos \pi x^{p}}{x^{p}} d [x] = \int_{1^{-}}^{n^{p}} \frac{\cos \pi x}{x} d [x^{1/p}]. \tag{1} $$
Using the periodic Bernoulli polynomials $B_{m}(x)$, we have
$$ \sum_{k=1}^{n} \frac{\cos (\pi k^{p})}{k^{p}} = - \int_{1^{-}}^{n^{p}} \frac{\cos \pi x}{x} d B_{1}(x^{1/p}) + \int_{1}^{n^{p}} \frac{\cos \pi x}{x} \, d(x^{1/p}) =: I_{n} + J_{n}. $$
Step 1. Note first that
$$ J_{n} = \frac{1}{p} \int_{1}^{n^{p}} \frac{\cos \pi x}{x^{2-(1/p)}} \, dx. $$
This converges as $n \to \infty$ if and only if $2-(1/p) > 0$, or equivalently, $p > 1/2$. Indeed,


*

*For $1/2 < p < 1$, the convergence follows by usual integration-by-parts trick.

*For $0 < p \leq 1/2$, let
$$ a_{k} = \lceil (2\pi k - (\pi/4))^{1/p} \rceil \quad \text{and} \quad b_{k} = \lfloor (2\pi k + (\pi/4))^{1/p} \rfloor. $$
Then we observe that $a_{k}^{p} - (2\pi k - (\pi/4)) \to 0$ and $b_{k}^{p} - (2\pi k + (\pi/4)) \to 0$. Thus for large $k$,
$$ J_{b_{k}} - J_{a_{k}} = \frac{1}{p} \int_{a_{k}^{p}}^{b_{k}^{p}} \frac{\cos \pi x}{x^{2-(1/p)}} \, dx \gtrsim \int_{-\pi/4}^{\pi/4} (x+2\pi k)^{(1/p)-2} \, dx \gtrsim 1,$$
which cannot happen if $J_{n}$ converges. (That is, $J_{n}$ is not Cauchy and hence not convergent.)


Step 2. Now we look into the term $I_{n}$. Integration by parts shows that
$$ I_{n} = - \left[ \frac{\cos \pi x}{x} B_{1}(x) \right]_{1^{-}}^{n^{p}} - \int_{1}^{n^{p}} \left( \frac{\pi \sin \pi x}{x} + \frac{\cos \pi x}{x^{2}} \right) B_{1}(x^{1/p}) \, dx. $$
Here, the only term whose convergence is unclear is 
$$ \int_{1}^{n^{p}} \frac{\pi \sin \pi x}{x} B_{1}(x^{1/p}) \, dx = p \pi \int_{1}^{n} \frac{\sin (\pi x^{p})}{x} B_{1}(x) \, dx. $$
Introducing the function 
$$ C(x) = -\int_{x}^{\infty} \frac{B_{1}(t)}{t} \, dt = \mathcal{O}\left( \frac{1}{x} \right), $$
it follows that
$$ p \pi \int_{1}^{n} \frac{\sin (\pi x^{p})}{x} B_{1}(x) \, dx = p \pi \left[ \sin (\pi x^{p}) C(x) \right]_{1}^{n} - p^{2}\pi^{2} \int_{1}^{n} x^{p-1} \cos(\pi x^{p}) C(x) \, dx, $$
which converges absolutely as $n \to \infty$. Putting together, $I_{n}$ converges for any $0 < p < 1$ and therefore the sum (1) converges if and only if $p > 1/2$.
Remark. We have the same answer if cosine is replaced by sine.
Remaining questions.


*

*Asymptote for the infinite sum as $p \downarrow 1/2$.

*Same question for the sum
$$ \sum_{k=1}^{\infty} \frac{\cos (a k^{p})}{k^{q}}, $$
where $p$ and $q$ are now allowed to have different values.

