For what real values of $a$ does $\sum\limits_{k=1}^\infty k^{-(1+|\sin k|^a)}$ converge?

Question

I was thinking about the $p$-series, and came up with this question:

For what real values of $a$ does $S=\sum\limits_{k=1}^\infty \dfrac{1}{k^{1+|\sin k|^a}}$ converge?

My attempt:

I know that if $a\le 0 $, then $S\le\sum\limits_{k=1}^\infty \dfrac{1}{k^{2}}=\pi^2/6$, so $S$ converges.

If $a>0$, the standard tests do not work:

• The $n$th term divergence test is inconclusive, because the terms approach $0$.
• The comparison test with a $p$-series test does not apply, because the exponent, $1+|\sin k|^a$, is not a constant. ($1+|\sin k|^a$ is always greater than $1$ since $k\in\mathbb{N}$, but this does not imply that the series converges. For example, $\sum\limits_{k=2}^\infty \dfrac{1}{k^{1+\frac{1}{\ln k}}}=\sum\limits_{k=2}^\infty \dfrac{1}{ek}$ diverges)
• The integral test does not apply, because $\dfrac{1}{x^{1+|\sin x|^a}}$ is not a decreasing function.
• The ratio test does not apply, because the limit of the ratio does not exist.
• The root test is inconclusive, because the limit of the root is $1$.

Numerical investigation:

If $a=1$, numerical investigation suggests that, for large $n$, $\sum\limits_{k=1}^n \dfrac{1}{k^{1+|\sin k|^a}}\approx c_1\ln (\ln n)+c_2$ where $c_1$ and $c_2$ are constants. If this approximation is valid, then $S$ would diverge when $a\ge 1$.

If $0<a<1$, numerical investigation does not elucidate what happens.

Answer 1

Not an answer, but definitely too long for a comment and interesting so thought it was worth it.

The gist of it is that the problem is quite deep actually, it has to do with rational approximations of $\pi$. Using Dirichlet's approximation theorem, see

https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem

We can produce an infinite ammount of natural numbers $k$ such that:

$$ |sin(k)| = |sin(k-q\pi)| \leq |k-q\pi| \leq \frac{1}{q} < \frac{4}{k}$$

Hence infinitely often one has that:

$$ \frac{1}{k^{1+|sin(k)|^a}} \geq \frac{1}{k^{1+Ck^{-a}}}$$

If the sequence of the RHS was summed over $\mathbb{N}$ then it diverges, but we are summing over a subsequence, so without knowing the form of that subsequence, we cannot possibly expect to deduce that the RHS diverges to get a comparison test to work. The proof that I have in mind of Diriclet theorem is by Minkowski's theorem, so we get no bounds on what the subsequence should look like. What I mean by that is that the proof is an existence proof not a constructive and exhaustive proof that would at least give some insight into what the "bad" integers $k$ look like.

The fact that numerical tests suggest that the series diverges for $a=1$ is also quite interesting, because it can probably work the other way arround, you get some results about the series and you should be able to translate it to approximations of $\pi$.

There are many similar cases about series involving $sin(k)$ one that came to mind was:

https://mathoverflow.net/questions/282259/is-the-series-sum-n-sin-nn-n-convergent/282290#282290

But this does not work in your case for any $0<a<1$ the point beeing that you have:

$$ |\pi-\frac{p}{q}| \leq \frac{1}{q^{\mu + \varepsilon}}$$

For all $p,q$ sufficiently large , $\varepsilon$ fixed and the best we can say about $\mu$ (from what I know atleast) is that $\mu \leq 7.103 \cdots$ in view of

https://arxiv.org/abs/1912.06345

But in order to use this to conclude you would need that $\mu <1$ as the above implies that for $k$ sufficiently large:

$$ \frac{1}{k^{1+|sin(k)|^{a}}} \leq \frac{1}{k^{1+Ck^{a(1-\mu-\varepsilon)}}}$$

For some absolute constant $c$ , and notice that if $a(1-\mu-\varepsilon) <0$ the series on the RHS diverges so you get nothing, it converges if $a(1-\mu-\varepsilon) >0$ but as the best we know about $\mu$ is that $\mu \leq 7.102 \cdots$ we are quite far from that.

Answer 2

$S$ converges for $a<1$, and diverges for $a\ge 1$.

The proof is in this question, and this answer.