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

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    $\begingroup$ Dear Giacomo, You may want to add a Diophantine approximation tag to this question, since (it seems to me) its answer will depend on the Diophantine approximation properties of $\pi$. (You may like to consider this question and its answer: math.stackexchange.com/questions/48035/… to understand what I mean, although I was unable to adapt that answer to your question.) Regards, $\endgroup$ – Matt E Jan 4 '13 at 5:17
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    $\begingroup$ It is highly likely that no one knows if the series converges or diverges. The main reason being our lack of knowledge of behavior of $n^{1+ \vert \sin(n) \vert}$ as $n \to \infty$, which in turn is due to lack of sufficient understanding of rational approximations to $\pi$. $\endgroup$ – user17762 Jan 4 '13 at 5:43
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    $\begingroup$ I think this is very interesting. I'm inclined to believe that it diverges. I might guess this because of the (not completely true) heuristic that $n \mod \pi$ is equidistributed, or approximately equidistributed, so that the Bohr sets for any given $\epsilon$ are infinite. But this is not at all rigorous, and nor do I think can be made rigorous (without knowing that $\pi$ is, say, a normal number or something). $\endgroup$ – davidlowryduda Jan 4 '13 at 6:18
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    $\begingroup$ I do not know if it helps, but it looks relevant, in that perhaps at least the techniques could help: math.stackexchange.com/a/77954 (The answer links to another website which links to a pdf.) $\endgroup$ – Jonas Meyer Jan 4 '13 at 7:02
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    $\begingroup$ @all: I have removed some comments discussing a now-deleted answer. $\endgroup$ – Willie Wong Jan 4 '13 at 16:22

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 $N\to\infty$.

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  • $\begingroup$ Ah, this was the ingredient I was missing. You don't need uniformity on the Bohr set, just a lower bound. This strategy should work for any non-zero $\sin(\alpha n)$, regardless of irrationality measure. $\endgroup$ – Erick Wong Jan 4 '13 at 21:42
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    $\begingroup$ Math is beautiful indeed. $\endgroup$ – Haskell Curry Jan 5 '13 at 8:40

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.

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  • $\begingroup$ Do you think this method could be used to answer the question math.stackexchange.com/questions/324901/… ? $\endgroup$ – Greg Martin Mar 8 '13 at 19:26
  • $\begingroup$ @GregMartin Hi Greg, yes it does look like this method will handle the other question. The discrepancy bound should give equidistribution on intervals as small as $N^{-c}$ and for the exponent $\lvert\sin n\rvert^a$ I think $(\log N)^{-A}$ is small enough. $\endgroup$ – Erick Wong Mar 9 '13 at 5:51
  • $\begingroup$ I don‘t seem to be able to use Mahler to obtain $\left|\sum_{n=1}^N e^{kni}\right| \le 2(1-e^{ki})^{-1}<<k^C$. Can you briefly elaborate? $\endgroup$ – Diger Dec 23 '18 at 16:59
  • $\begingroup$ @Diger $e^{ki}$ can only be very close to $1$ if $k$ is very close to $2\pi m$ for some integer $m$, hence if $k/2m$ is very close to $\pi$. $\endgroup$ – Erick Wong Dec 24 '18 at 1:32
  • $\begingroup$ I meant how do you precisely estimate. E.g. I would get $$\left|\frac{2}{1-{\rm e}^{- i(2\pi m - k)}}\right| < m^{C-1} \sim \left( \frac{k}{2\pi} \right)^{C-1} \, .$$ $\endgroup$ – Diger Dec 24 '18 at 12:00

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.

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    $\begingroup$ Well, this is interesting and perhaps slightly awkward... $\endgroup$ – Erick Wong Jan 4 '13 at 10:13
  • $\begingroup$ @ErickWong : I think one of us must have made some silly sign error somewhere. Probably me because it's hard to see what is increasing/decreasing when exponentiating things to exponents like $1/(1- \mu)$ $\endgroup$ – mercio Jan 4 '13 at 10:15
  • $\begingroup$ +1 in any case as your answer provides valuable insight on the structure of the approximating sets! $\endgroup$ – Erick Wong Jan 4 '13 at 10:20
  • $\begingroup$ Possible error? When $\mu > 2$, $\alpha < 0$. $\endgroup$ – Erick Wong Jan 4 '13 at 10:27
  • $\begingroup$ You're right, I was too hasty when I made that check. $\endgroup$ – mercio Jan 4 '13 at 12:13

Thanks for pointing out my ignorance. I can NOT assert that the integration is equivalent to the series sum so roughly, since the integration is not decreasing. However, they seems to be equivalent in light of the graph. $$$$NEED FURTHER PROOF.$$$$ The origin answer:$$$$ It is equal to prove that $$\int_{\pi}^\infty\dfrac{dx}{x^{1+|\sin x|}}$$ is diverge.$$$$ $$\int_{\pi}^\infty\dfrac{dx}{x^{1+|\sin x|}}\geq\sum_{n=1}^\infty\dfrac{1}{(n+1)\pi}\int_{n\pi}^{(n+1)\pi}\dfrac{dx}{x^{|\sin x|}}$$ Since $$\dfrac{d}{dx}(\frac{1}{x^{\sin x}})=-(\frac{\sin x}{x}+\cos x\ln x)(\frac{1}{x^{\sin x}})$$ We make 2 tangents of the graph $y=\dfrac{1}{x^{|\sin x|}}$ on each $(n\pi,1)$, then calculate the area of the triangles surrounded by the tangents and x axis. $$$$We have $$\int_{n\pi}^{(n+1)\pi}\dfrac{dx}{x^{|\sin x|}}\geq\dfrac{1}{\ln ((n+1)\pi)}$$ Since $$\sum_{n=1}^\infty \frac{1}{(n+1)\pi\ln ((n+1)\pi)}\geq\int_{2\pi}^\infty\dfrac{dx}{x\ln x}=\int_{2\pi}^\infty d\ln\ln x$$ is diverge $$$$We have construct a diverge sequence $$\sum_{n=1}^\infty \frac{1}{(n+1)\pi\ln ((n+1)\pi)}$$ Since $$\int_{\pi}^\infty\dfrac{1}{x^{1+|\sin x|}}\geq\sum_{n=1}^\infty \frac{dx}{(n+1)\pi\ln ((n+1)\pi)}$$ The origin integration is diverge.

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    $\begingroup$ I may be missing something, but I don't see why the convergence of the series equivalent to the convergence of the integral? The function is not decreasing, so the obvious estimates don't tie the two problems together. $\endgroup$ – Jyrki Lahtonen Dec 25 '19 at 15:45

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