Does the series $ \sum\limits_{n=1}^{\infty} \frac{1}{n^{1 + |\sin(n)|}} $ converge or diverge?

Question

Does the following series converge or diverge? I would like to see a demonstration.

$$ \sum_{n=1}^{\infty} \frac{1}{n^{1 + |\sin(n)|}}. $$

I can see that: $$ \sum_{n=1}^{\infty} \frac{1}{n^{1 + |\sin(n)|}} \leqslant \sum_{n=1}^{\infty} \frac{1}{n^{1 + {\sin^{2}}(n)}} \leqslant \sum_{n=1}^{\infty} \frac{1}{n^{1 + {\sin^{2n}}(n)}}. $$

Answer 1

This is problem 11162, posed by Paolo Perfetti, in the June-July 2005 issue of the American Mathematical Monthly. The solution below, due to the Microsoft Research Problems Group, is found in the February 2007 issue of the same magazine.

For positive integer $n$, define $$A_n=[0,2^n)\cap\{k\in \mathbb{N}:|\!\sin k|<\textstyle{1\over n}\},\quad B_n=[2^{n-1},2^n)\cap A_n.$$ If $k\in B_n$, then $k^{-1-|\sin k|}>(2^n)^{-1-1/n }=2^{-n-1}$. If $n>1$, then $A_n$ is contained in the disjoint union of $A_{n-1}$ and $B_n$, so $|B_n|\geq |A_n|-|A_{n-1}|$. To estimate $|A_n|$, partition the unit circle into $7n$ arcs, each with angle $2\pi/(7n)$. Of the values $e^{ik}$ for $0\leq k<2^n$, at least $2^n/(7n)$ lie in the same arc by the Pigeonhole Principle. If $e^{ik_1}$ and $e^{ik_2}$ lie in the same arc, then $$|\sin(k_1-k_2)|\leq |e^{i(k_1-k_2)}-1|=|e^{ik_1}-e^{ik_2}|<{2\pi\over7n}<{1\over n}$$ and $|k_1-k_2|\in A_n$. Subtracting the smallest $k$ from the others (and itself), we find that $|A_n|\geq2^n/(7n)$. Now if $N\geq2$, then \begin{eqnarray*} \sum_{k=2}^{2^N-1}k^{-1-|\sin k|}&=&\sum_{n=2}^N\sum_{k=2^{n-1}}^{2^n-1}k^{-1-|\sin k|} \geq \sum_{n=2}^N\sum_{k\in B_n}k^{-1-|\sin k|}\geq\sum_{n=2}^N {|B_n|\over2^{n+1}}\\[9pt] &\geq& \sum_{n=2}^N{|A_n|-|A_{n-1}|\over 2^{n+1}}=\sum_{n=2}^N \left(\left( {|A_n|\over 2^{n+2}}-{|A_{n-1}|\over 2^{n+1}}\right)+{|A_n|\over 2^{n+2}}\right) \\[9pt] &=&{A_N\over 2^{N+2}}-{|A_1|\over8}+\sum_{n=2}^N{|A_n|\over 2^{n+2}}\\[9pt] &\geq& -{|A_1|\over8}+\sum_{n=2}^N {2^n/(7n)\over2^{n+2}}= -{|A_1|\over8}+\sum_{n=2}^N {1\over28n} \end{eqnarray*}

This grows without bound as $N\to\infty$.

Answer 2

There are some details that I haven't fully vetted but here's a long sketch which I believe should show divergence.

Define $c_n := \frac{n}{2\pi} \pmod 1$ and let $[a,b]$ be any interval in the torus $\mathbb R/\mathbb Z$. The discrepancy $D(N)$ of the sequence $c_n$ is defined to be the difference between $\# \{ n \le N : c_n \in [a,b]\}$ and the expected count $(b-a)N$.

Weyl's criterion tells us that $D(N) = o(N)$. More quantitatively, the Erdős-Turán inequality states that for any integer $K>0$,

$$D(N) \ll \frac{N}{K} + \sum_{k=1}^K \frac1k\left| \sum_{n=1}^N e^{kni}\right|.$$

While I imagine there are better ways to control the rightmost sum on average, it's a nice fact that we can control it pointwise using a result of Mahler that there exists an absolute constant $C>0$ such that

$$\left|\pi - \frac{p}{q}\right| \gg \frac{1}{q^C}.$$

(The current best value of $C$ is about $7.6$ due to Salikhov.) Note that $\left|\sum_{n=1}^N e^{kni}\right| \le 2(1-e^{ki})^{-1}$ by geometric series, and Mahler's theorem controls how close $k$ can be to a multiple of $2\pi$ and thereby how close $e^{ki}$ can be to $1$. This should give us $\left|\sum_{n=1}^N e^{kni}\right| \ll k^{C}$ uniformly in $N$, possibly with a different $C$.

By choosing $K$ to be a small power of $N$ (something like $N^{1/C}$) in Erdős-Turán, we get that $D(N) \le N^{1-c}$ for some absolute constant $c>0$.

For small $\epsilon >0$, let $A_\epsilon$ be the Bohr set $\{n \in \mathbb N: |\sin(n)| < \epsilon\}$. By the discrepancy bound, the counting function satisfies (ignoring a tiny error term from the non-linearity of sine):

$$A_\epsilon(N) := \{n \le N : n \in A_\epsilon\} = \frac{\epsilon}{\pi}N + O(N^{1-c}).$$

In particular, $A_\epsilon(N) \gg \epsilon N$ provided that $N > \epsilon^{-C}$. Using this with partial summation, we can estimate the contribution to the original sum from $A_\epsilon$:

$$\sum_{n \in A_\epsilon} \frac{1}{n^{1+|\sin(n)|}} \ge \sum_{n \in A_\epsilon} n^{-1-\epsilon} = \int_1^\infty (1+\epsilon) A_\epsilon(t) t^{-2-\epsilon}\, dt \ge \int_{\epsilon^{-C}}^\infty A_\epsilon(t) t^{-2-\epsilon}\, dt \gg \int_{\epsilon^{-C}}^\infty \epsilon t^{-1-\epsilon}\, dt,$$

which simplifies to $\epsilon^{C\epsilon}$.

Note that as $\epsilon \to 0$, $\epsilon^{C\epsilon} \to 1$. We're very close to getting divergence. Just apply the above calculation to the set $A_\epsilon \setminus A_{\epsilon/2}$. This has about the same counting function and still gives $$\sum_{n \in A_\epsilon \setminus A_{\epsilon/2}} \frac{1}{n^{1+|\sin(n)|}} \gg \epsilon^{C\epsilon}.$$

Now sum this over a dyadic sequence of $\epsilon_k = 2^{-k}$ so that the resulting sets are disjoint, and we get a divergent sum.

Answer 3

I proceed the same way as in my other answer here https://math.stackexchange.com/a/110019/17445

$|\sin n| \le \varepsilon$ implies that there exists an integer $k(n)$ such that $n = k(n)\pi + a(n)$ where $a(n) \in [-\arcsin(\varepsilon) ; \arcsin(\varepsilon)] \subset [-\pi \varepsilon/2 ; \pi \varepsilon/2]$, and if both $|\sin n|$ and $|\sin m|$ are less than $\varepsilon$, then $m-n = k'(m,n)\pi + b(m,n)$ where $k(m,n) = k(m)-k(n)$ is an integer and $b(m,n) = a(m)-a(n) \in [-\pi \varepsilon ; \pi \varepsilon]$.

Since $\pi$ has a finite irrationality measure, we know that there is a finite real constant $\mu >2$ such that for any integers $n,k$ large enough, $|n−k\pi| \ge k^{1-\mu}$.

Therefore, if we pick $\varepsilon$ small enough we force $k(n)$ and $n$ to be high enough so that the inequality is true, and we have $\pi \varepsilon/2 \ge |a(n)| = |n-k(n)\pi| \ge k(n)^{1-\mu} \ge (n/\pi + \epsilon /2)^{1- \mu}$. After taking the $1/(1-\mu)$th power of this, we get $n \ge ((\pi \varepsilon/2)^\frac 1 {1-\mu} - \varepsilon/2)\pi \ge A \varepsilon^\frac 1 {1-\mu} $ for some constant $A > 0$.

Similarly, $\pi \varepsilon \ge |b(m,n)| = |m-n-k(m,n)\pi| \ge k(m,n)^{1-\mu} \ge ((m-n)/\pi + \epsilon )^{1- \mu}$, giving $m-n \ge ((\pi \varepsilon)^\frac 1 {1-\mu} - \varepsilon)\pi \ge B \varepsilon^\frac 1 {1-\mu} $ for some constant $B > 0$.

All of this to deduce that there are positive constants $C$ and $\eta$ such that if $0 < \varepsilon < \eta$ and $n(\varepsilon,k)$ is the $k$-th integer $n$ satisfying $|\sin(n)| \le \varepsilon$, then $n(\varepsilon,k) \ge kC \varepsilon^\frac 1 {1-\mu}$.

---

Now we can try and bound the sum :

$\sum_{|\sin n| \ge \eta} n^{-1-|\sin n|} \le \sum_{n \in \Bbb N} n^{-1- \eta} = \zeta(1+ \eta) < \infty$.

Given $\varepsilon \le \eta$, put $S_\varepsilon = \sum_{|\sin n| \in [\varepsilon/2 ; \varepsilon]} n^{-1-|\sin n|}$ :
$$S_\varepsilon \le \sum_{k \in \Bbb N} n(\varepsilon,k)^{-1- \epsilon/2} \le \sum_{k \in \Bbb N} (kC \varepsilon^\frac 1 {1-\mu})^{-1- \varepsilon/2} = \zeta(1+\varepsilon/2)(C \varepsilon^\frac 1 {1-\mu})^{-1- \varepsilon/2} \le \zeta(1+\varepsilon/2)(C \varepsilon^\frac 1 {1-\mu})^{-1} \le (D / \varepsilon) (C \varepsilon^\frac 1 {1-\mu})^{-1} = K \varepsilon^\alpha$$ where $D,K$ are suitable positive constants, and $\alpha = \frac{-1}{1- \mu}-1 < 0$ (because $\mu > 2$).

Finally, picking $\varepsilon = \eta 2^{-k}$ and summing all these inequalities, we obtain $\sum n^{-1-|\sin n|} \le \zeta(1+\eta) + K(\sum_{k \ge 0} (\eta 2^{-k})^\alpha) = \zeta(1+\eta) + K\eta^\alpha \sum_{k \ge 0} (2^{-\alpha})^k$.

However since $2^{-\alpha}>1$ this sum diverges so this method fails.