# Is the series $\sum_n|\sin n|^n/n$ convergent?

Problem. Is the series $$\sum_{n=1}^\infty\frac{|\sin(n)|^n}n$$convergent?

(The problem was posed on 22.06.2017 by Ph D students of H.Steinhaus Center of Wroclaw Polytechnica. The promised prize for solution is "butelka miodu pitnego", see page 37 of Volume 1 of the Lviv Scottish Book. To get the prize, write to the e-mail: hsc@pwr.edu.pl).

• This touches on research-level mathematics, but given the specific nature of the problem it seems like it might do better on math.SE rather than on mathoverflow. Sep 28, 2017 at 18:45
• @StevenStadnicki Maybe, but this problem is not a standard university level exercise in Calculus. Moreover neither positive nor negative answers are not evident. Just try to resolve it yourself. It does not seem to be trivial. Sep 28, 2017 at 18:50
• The obvious random model suggests it is bounded. Sep 28, 2017 at 19:23
• @LvivScottishBook: $|\sin n|^n$ does not tend to $0$ (this follows from the argument given in the answer here: math.stackexchange.com/questions/1141045/…). Sep 28, 2017 at 20:01
• It almost assuredly would not have done better at Math.SE. It seems to have been a real test of professional skill and knowledge. Sep 30, 2017 at 16:32

Note that if $\pi$ were rational (with even numerator), then $\sin(n)$ would equal $1$ periodically, so the series would diverge. Similarly if $\pi$ were a sufficiently strong Liouville number. Thus, to establish convergence, one must use some quantitative measure of the irrationality of $\pi$.

It is known that the irrationality measure $\mu$ of $\pi$ is finite (indeed, the current best bound is $\mu \leq 7.60630853$). Thus, one has a lower bound $$| \pi - \frac{p}{q} | \gg \frac{1}{q^{\mu+\varepsilon}}$$ for all $p,q$ and any fixed $\varepsilon>0$. This implies that $$\mathrm{dist}( p/\pi, \mathbf{Z}) \gg \frac{1}{p^{\mu-1+\varepsilon}},$$ for all large $p$ (apply the previous bound with $q$ the nearest integer to $p/\pi$, multiply by $q/\pi$, and note that $q$ is comparable to $p$). In particular, if $I \subset {\bf R}/{\bf Z}$ is an arc of length $0 < \delta < 1$, the set of $n$ for which $n/\pi \hbox{ mod } 1 \in I$ is $\gg \delta^{-1/(\mu-1+\varepsilon)}$-separated. This implies, for any natural number $k$, that the number of $n$ in $[2^k,2^{k+1}]$ such that $|\sin(n)|$ lies in any given interval $J$ of length $2^{-k}$ (which forces $n/\pi \hbox{ mod } 1$ to lie in the union of at most two intervals of length at most $O(2^{-k/2})$) is at most $\ll 2^{k(1 - \frac{1}{2(\mu-1+\varepsilon)})}$, the key point being that this is a "power saving" over the trivial bound of $2^k$. Noting (from Taylor expansion) that $|\sin(n)|^n \ll \exp( - j)$ if $n \in [2^k,2^{k+1}]$ and $|\sin(n)| \in [1 - \frac{j+1}{2^k}, 1-\frac{j}{2^k}]$, we conclude on summing in $j$ that $$\sum_{2^k \leq n < 2^{k+1}} |\sin(n)|^n \ll 2^{k(1 - \frac{1}{2(\mu-1+\varepsilon)})}$$ and hence $$\sum_{2^k \leq n < 2^{k+1}} \frac{|\sin(n)|^n}{n} \ll 2^{- k\frac{1}{2(\mu-1+\varepsilon)}}.$$ The geometric series on the RHS is summable in $k$, so the series $\sum_{n=1}^\infty \frac{|\sin(n)|^n}{n}$ is convergent. (In fact the argument also shows the stronger claim that $\sum_{n=1}^\infty \frac{|\sin(n)|^n}{n^{1-\frac{1}{2(\mu-1+\varepsilon)}}}$ is convergent for any $\varepsilon>0$.)

EDIT: the apparent numerical divergence of the series may possibly be due to the reasonably good rational approximation $\pi \approx 22/7$, which is causing $|\sin(n)|$ to be close to $1$ for $n$ that are reasonably small odd multiples of $11$. UPDATE: I now agree with Will that it is the growth of $-2^{3/2}/\pi^{1/2} n^{1/2}$, rather than any rational approximant to $1/\pi$, which was responsible for the apparent numerical divergence at medium values of $n$, as is made clear by the updated numerics on another answer to this question.

• Jesus, you make it sound so easy.
– user78249
Sep 29, 2017 at 3:16
• Will you be claiming the bottle of mead? (with an assist from google translate) Sep 29, 2017 at 5:05
• Awesome! I wonder if also these very low frequency fluctuations clearly visible on Steve Huntsman's plot (especially in between $10^5$ and $10^6$) can be explained? Sep 29, 2017 at 5:33
• @TerryTao Great Solution! I have written to the Ph D students of Steinhaus Center about your solution of their problem. So, be ready to get the prize "butela miodu pitnego" (= a bottle of drinking honey :) Sep 29, 2017 at 7:32
• I don't think this 22/7 approximation is the culprit. The most suspicious region on მამუკა ჯიბლაძე's plot is for n>3000, elsewhere it looks highly convex. But $( (1/\pi) - (7/22) ) * 3000 = .3842\dots$ is quite large. Instead I think this is just the phenomenon of $1/\sqrt{n}$ going to zero pretty slowly and thus the graph continually increasing. Sep 29, 2017 at 8:35

On the OP request, here is the plot of first 10000 partial sums. Following Terry Tao's suggestion, here is the plot of ($n$th partial sum) $+2^{\frac32}/\sqrt{\pi n}$ for $n$ up to one million: The thick line in the beginning actually consists of high frequency oscillations - in the range up to 2000 it looks like this: (I hope there are no rounding artifacts, I calculated everything with 100 decimal digits precision)

Next, following suggestion by j.c. in a comment below, I tried to plot the (discrete) Fourier transform of the first 10000 points; the result is this: More precisely, height at the point with abscissa $n$ is the absolute value of the scalar product of the vector of first 10000 partial sums minus its average with the vector $\left(e^{\frac{2\pi i k}n}\right)_{1\le k\le 10000}$.

You see that $22$ and $355$, as well as $11$ ($=\frac{22}2$) and $177.5=\frac{355}2$ are all clearly visible.

If I will have more time I will try to do the same with more data, to detect $52163$ mentioned by Terry Tao in a previous comment. I am not sure about the arbitrary phase shift that I introduced, though - I could start with $k=0$ instead of $k=1$, or any other $k$.

• If you look at the random model, following Mateusz's calculations, the sum should grow like constant minus $2^{3/2} / (\pi n)^{1/2}$, which from 1000 to 10000 is a growth of .034, only slightly less then what you see. Sep 29, 2017 at 8:27
• Perhaps it is worth plotting the sum of the partial sums and $2^{3/2}/(\pi n)^{1/2}$, which should bring any deviations from the random model into clearer view, in particular the oscillations that are presumably coming from the rational approximants to pi. Sep 29, 2017 at 16:48
• The low frequency oscillations may be related to the approximation $\pi \approx 52163/16604$, which makes $|\sin(n)|$ approximately periodic with period $52163$. Sep 29, 2017 at 18:59
• Three years ago on Math.SE Jack D'Aurizio gave the estimate of $\leq 2.151$. See Is $\sum\limits_{n=1}^\infty \frac{|\sin n|^n}n$ convergent？.
– jeq
Sep 30, 2017 at 1:08
• Here's yet another plotting suggestion: presumably we will see peaks corresponding to rational approximants of $\pi$ in the Fourier transform of the data (with the $2^{3/2}/(\pi n)^{1/2}$ adjustment).
– j.c.
Sep 30, 2017 at 16:32 Semilog plot building on მამუკა ჯიბლაძე's picture, this time to $10^7$

Let $D_N$ be the discrepancy: $$D_N=\sup \left| \frac{ A(J:P)}{N} - |J|\right|$$ where $P=\{k/\pi \ \mathrm{mod} \ 1\}_{k=1,2,\ldots, n}$, $J$ is an interval in $[0,1]$.

If the irrationality measure $\mu$ of $\pi$ is finite, then we have $$D_N\ll N^{-\frac1{\mu-1} + \epsilon}.$$

From this result and Terry Tao's answer, the number of $n\in [2^k, 2^{k+1}]$ for which $|\sin n |$ falls in an interval of length $2^{-k}$, is $$\ll 2^{\frac k2} + 2^{k\left(1-\frac1{\mu-1} + \epsilon\right)}$$

Thus, if $r>\max\left\{\frac12, 1-\frac1{\mu-1} \right\}$, then the series $$\sum_{n=1}^{\infty} \frac{|\sin n|^n}{n^r}$$ is convergent.

It is conjectured that $\mu=2$. If we prove that $2\leq \mu <3$, then we can also prove that $$\sum_{n=1}^{\infty} \frac{|\sin n|^n}{\sqrt n}$$ diverges. I am not aware of any unconditional proof of the divergence of this series.