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I saw a while ago in a book by Clifford Pickover, that whether $\displaystyle \sum_{n=1}^\infty\frac1{n^3\sin^2 n}$ converges is open.

I would think that the question of its convergence is really about the density in $\mathbb N$ of the sequence of numerators of the standard convergent approximations to $\pi$ (which, in itself, seems like an interesting question). Naively, the point is that if $n$ is "close" to a whole multiple of $\pi$, then $1/(n^3\sin^2n)$ is "close" to $\frac1{\pi^2 n}$.

[Numerically there is some evidence that only some of these values of $n$ affect the overall behavior of the series. For example, letting $S(k)=\sum_{n=1}^{k}\frac1{n^3\sin^2n}$, one sees that $S(k)$ does not change much in the interval, say, $[50,354]$, with $S(354)<5$. However, $S(355)$ is close to $30$, and note that $355$ is very close to $113\pi$. On the other hand, $S(k)$ does not change much from that point until $k=100000$, where I stopped looking.]

I imagine there is a large body of work within which the question of the convergence of this series would fall naturally, and I would be interested in knowing something about it. Sadly, I'm terribly ignorant in these matters. Even knowing where to look for some information on approximations of $\pi$ by rationals, or an ad hoc approach just tailored to this specific series would be interesting as well.

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The sum you have written obviously converges. – Xandi Tuni May 14 '10 at 6:32
No, it doesn't obvously converge :-) – Robin Chapman May 14 '10 at 6:33
Good lord! Sorry, I was too quick with that comment. – Xandi Tuni May 14 '10 at 6:41
If we replace $n^3$ with $n^2$, what happens? (This isn't a rhetorical question; I'm honestly wondering if anyone knows the answer.) More generally, consider sums $F(a,b) = \sum_{n=1}^\infty 1/(n^a |sin n|^b)$. Clearly $F(a,b)$ increases with $a$ and decreases with $b$, but when is $F(a,b)$ known to be either finite or infinite? – Michael Lugo May 14 '10 at 13:17
The convergence of this general form is related to the irrationality measure of $\pi$, that is the infimum of exponents $k$ such that $|\pi-a/b|<1/b^k$ has only finitely many integer solutions. (For $|\sin n|$ to be small, $n$ must be close to an integer multiple $m\pi$ of $\pi$ and then $|\sin n|\sim m|\pi-n/m|$.) Results are known (see for instance and these will yield explicit values of $a$, $b$ for which the series converges, but the proofs are delicate and don't yield the best expected result. – Robin Chapman May 14 '10 at 15:26
up vote 97 down vote accepted

As Robin Chapman mentions in his comment, the difficulty of investigating the convergence of $$ \sum_{n=1}^\infty\frac1{n^3\sin^2n} $$ is due to lack of knowledge about the behavior of $|n\sin n|$ as $n\to\infty$, while the latter is related to rational approximations to $\pi$ as follows.

Neglecting the terms of the sum for which $n|\sin n|\ge n^\varepsilon$ ($\varepsilon>0$ is arbitrary), as they all contribute only to the `convergent part' of the sum, the question is equivalent to the one for the series $$ \sum_{n:n|\sin n|< n^\varepsilon}\frac1{n^3\sin^2n}. \qquad(1) $$ For any such $n$, let $q=q(n)$ minimizes the distance $|\pi q-n|\le\pi/2$. Then $$ \sin|\pi q-n|=|\sin n|< \frac1{n^{1-\varepsilon}}, $$ so that $|\pi q-n|\le C_1/n^{1-\varepsilon}$ for some absolute constant $C_1$ (here we use that $\sin x\sim x$ as $x\to0$). Therefore, $$ \biggl|\pi-\frac nq\biggr|<\frac{C_1}{qn^{-\varepsilon}}, $$ equivalently $$ \biggl|\pi-\frac nq\biggr|<\frac{C_2}{n^{2-\varepsilon}} \quad\text{or}\quad \biggl|\pi-\frac nq\biggr|<\frac{C_2'}{q^{2-\varepsilon}} $$ (because $n/q\approx\pi$) for all $n$ participating in the sum (1). It is now clear that the convergence of the sum (1) depends on how often we have $$ \biggl|\pi-\frac nq\biggr|<\frac{C_2'}{q^{2-\varepsilon}} $$ and how small is the quantity in these cases. (Note that it follows from Dirichlet's theorem that an even stronger inequality, $$ \biggl|\pi-\frac nq\biggr|<\frac1{q^2}, $$ happens for infinitely many pairs $n$ and $q$.) The series (1) converges if and only if $$ \sum_{n:|\pi-n/q|< C_2n^{-2+\varepsilon}}\frac1{n^5|\pi-n/q|^2} $$ converges. We can replace the summation by summing over $q$ (again, for each term $\pi q\approx n$) and then sum the result over all $q$, because the terms corresponding to $|\pi-n/q|< C_2n^{-2+\varepsilon}$ do not influence on the convergence: $$ \sum_{q=1}^\infty\frac1{q^5|\pi-n/q|^2} =\sum_{q=1}^\infty\frac1{q^3(\pi q-n)^2} \qquad(2) $$ where $n=n(q)$ is now chosen to minimize $|\pi-n/q|$.

Summarizing, the original series converges if and only if the series in (2) converges.

It is already an interesting question of what can be said about the convergence of (2) if we replace $\pi$ by other constant $\alpha$, for example by a "generic irrationality". The series $$ \sum_{q=1}^\infty\frac1{q^3(\alpha q-n)^2} $$ for a real quadratic irrationality $\alpha$ converges because the best approximations are $C_3/q^2\le|\alpha-n/q|\le C_4/q^2$, and they are achieved on the convergents $n/q$ with $q$ increasing geometrically. A more delicate question seems to be for $\alpha=e$, because one third of its convergents satisfies $$ C_3\frac{\log\log q}{q^2\log q}<\biggl|e-\frac pq\biggr|< C_4\frac{\log\log q}{q^2\log q} $$ (see, e.g., [C.S.Davis, Bull. Austral. Math. Soc. 20 (1979) 407--410]). The number $e$, quadratic irrationalities, and even algebraic numbers are `generic' in the sense that their irrationality exponent is known to be 2. What about $\pi$?

The irrationality exponent $\mu=\mu(\alpha)$ of a real irrational number $\alpha$ is defined as the infimum of exponents $\gamma$ such that the inequality $|\alpha-n/q|\le|q|^{-\gamma}$ has only finitely many solutions in $(n,q)\in\Bbb Z^2$ with $q\ne0$. (So, Dirichlet's theorem implies that $\mu(\alpha)\ge2$. At the same time from metric number theory we know that it is 2 for almost all real irrationals.) Assume that $\mu(\pi)>5/2$, then there are infinitely many solutions to the inequality $$ \biggl|\pi-\frac nq\biggr|<\frac{C_5}{q^{5/2}}, $$ hence infinitely many terms in (2) are bounded below by $1/C_5$, so that the series diverges (and (1) does as well). Although the general belief is that $\mu(\pi)=2$, the best known result of V.Salikhov (see this answer by Gerry and my comment) only asserts that $\mu(\pi)<7.6064\dots$,.

I hope that this explains the problem of determining the behavior of the series in question.

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Thanks, Robin and Wadim! – Andrés E. Caicedo May 15 '10 at 15:50
So am I right in reading this as saying that the series is expected to converge? – Lev Borisov Jan 30 '14 at 3:44
Lev, yes, this is the expectation. – Wadim Zudilin Feb 5 '14 at 11:12
why don't you write $|\pi q - n(q)|$ in (2) ? it would be clearer – user1952009 Jul 12 at 6:41

Wadim Zudilin's answer is further extended in (Max A. Alekseyev, On convergence of the Flint Hills series).

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Nice reference. Thanks! – Andrés E. Caicedo Jan 30 '14 at 4:07

There is an even bigger reduction that can be done:

Theorem: The Flint Hills series converges if and only if the series $$ \sum_{n = 1}^\infty \frac{1}{q_n^3 (q_n\pi - p_n)^2} \qquad{(1)} $$ converges, where $(p_n/q_n)_1^\infty$ is the sequence of convergents of $\pi$.

Proof: Let $$ S = \sum_{q = 1}^\infty \frac{1}{q^3 (q\pi - p)^2}, \qquad{(2)} $$ where $p\in\mathbb N$ is chosen to minimize $|q\pi - p|$. As Wadim Zudilin argued, the Flint Hills series converges if and only if $S$ converges. Now consider the unimodular lattice $\Lambda = \{(q,q\pi - p) : p,q\in\mathbb Z\}$. We can rewrite $S$ as $$ S = \sum_{\substack{(q,r)\in\Lambda^* \\ q > 0 \\ -1/2 < r < 1/2}} \frac{1}{q^3 |r|^2}\cdot $$ Here $\Lambda^* = \Lambda\setminus\{\mathbf 0\}$. Next, using the identity $$ \frac{1}{q^3 r^2} = \int_{s > q} \int_{t > r} \frac{\partial}{\partial s}\frac{\partial}{\partial t}\frac{1}{s^3 t^2} \;dt\;ds = \int_{s > 1} \int_{t > 0} \frac{\partial}{\partial s}\frac{\partial}{\partial t}\frac{1}{s^3 t^2} [s > q][t > r] \;dt\;ds $$ we get $$ S = \int_{s > 1} \int_{t > 0} \frac{\partial}{\partial s}\frac{\partial}{\partial t}\frac{1}{s^3 t^2} \sum_{\substack{(q,r)\in\Lambda^* \\ q > 0 \\ -1/2 < r < 1/2}} [s > q][t > |r|] \;dt\;ds\\ = \int_{s > 1} \int_{t > 0} \frac{\partial}{\partial s}\frac{\partial}{\partial t}\frac{1}{s^3 t^2} \#\big\{(q,r)\in\Lambda^* : 0 < q < s,\; |r| < \min(t,1/2)\big\} \;dt\;ds. \qquad{(3)} $$ We can bound the integrand in two different ways, depending on whether or not $$ N_{s,t} := \#\big\{(q,r)\in\Lambda^* : 0 < q < s,\; |r| < \min(t,1/2)\big\} \leq \max(0,3st - 1/2). \qquad{(4)} $$ If (4) holds, then it can be used to bound the entire integral; I leave it to the reader to verify that the resulting integral converges. So let us consider the cases where (4) fails.

Fix $s > 1$ and $t > 0$, and let $D_{s,t} = (-s,s)\times(-t,t)$. If $D_{s,t}$ contains two linearly independent elements of $\Lambda$, then $D$ contains a fundamental domain for $\Lambda$, say $F$; we have $$ 1 + 2N_{s,t} = \#(\Lambda\cap D_{s,t}) = \sum_{\mathbf x\in\Lambda\cap D_{s,t}} m(\mathbf x + F) = m\left(\bigcup_{\mathbf x\in\Lambda\cap D_{s,t}}(\mathbf x + F)\right) \leq m(2D_{s,t}) = 4st, $$ which implies that (4) holds. Similarly, if $D_{s,t}\cap\Lambda = \{\mathbf 0\}$, then (4) holds.

So if we assume that (4) fails for some pair $(s,t)$, then we have $D_{s,t}\cap\Lambda = D_{s,t}\cap \mathbb Z\mathbf x$ for some $\mathbf x = (q,r)\in D_{s,t}\cap\Lambda^*$. It follows that $$ \max(1,3st - 1/2) \leq N_{s,t} = \left\lfloor \min\left(\frac sq,\frac t{|r|}\right)\right\rfloor \leq \min\left(\frac sq,\frac t{|r|}\right) $$ and thus $$ \frac{st}{q|r|} \geq \min\left(\frac sq,\frac t{|r|}\right)^2 \geq \max(1,3st - 1/2)^2 \geq \max(1,3st - 1/2) \geq 2st, $$ so $q|r| = q|q\pi - p| \leq 1/2$. A well-known theorem now implies that $p/q$ is a convergent of $\pi$, i.e. $(q,p) = (q_n,p_n)$ for some $n\in\mathbb N$. So if we let $$ \Lambda_c = \{(k q_n, k(q_n\pi - p_n)) : n,k\in\mathbb N\} $$ then $$ N_{s,t} = \#(\Lambda_c\cap D_{s,t}). $$ In other words, the only points which are contributing to the integrand of (3) are points which come from $\Lambda_c$. Reversing the argument of (3) now gives $$ S \leq C + \sum_{\substack{(q,r)\in\Lambda_c \\ q > 0 \\ -1/2 < r < 1/2}} \frac{1}{q^3 |r|^2}, $$ where $C < \infty$ is a constant describing an upper bound on the contribution to the integral (3) of pairs $(s,t)$ satisfying (4). Thus, $$ S \leq C + \sum_{n = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(k q_n)^3 (k(q_n \pi - p_n))^2} = C+\zeta(5)\sum_{n = 1}^\infty \frac{1}{q_n^3 (q_n \pi - p_n)^2}\cdot $$ It follows that (1) converges if and only if (2) converges.

If this proof was too technical to follow, I'll try to summarize the main ideas: First of all, any rational number $p/q$ which is not a convergent of $\pi$ must satisfy $q|q\pi - p| > 1/2$ (this is a well-known fact). By itself this fact isn't enough to guarantee that the terms coming from non-convergents won't make the series (2) diverge, since you end up comparing it with the harmonic series, which (just barely) diverges. But that's just the crudest possible bound: most rationals $p/q$ will satisfy $q|q\pi - p| \gg 1$. Since (2) involves a summation over all $q$, there will be a lot of "averaging", and so the "spikes" which occur when $q|q\pi - p|$ is small will be washed out in the long run. In order to formalize this you need to talk about lattices and fundamental domains - basically, the idea is that the number of intersection points of a lattice with a convex centrally symmetric region is about the same as the area of the region except for certain exceptional cases; these exceptional cases turn out to correspond to the convergents of $\pi$.

Corollary: If the exponent of irrationality of $\pi$ is strictly less than $5/2$, then the Flint Hills series converges. Proof: If $\mu(\pi) < 5/2$, then there exists $\varepsilon > 0$ such that for all but finitely many $n$, we have $$ |q_n \pi - p_n| \geq q_n^{-3/2 + \varepsilon}. $$ This gives the following upper bound for (1): $$ \sum_{n = 1}^\infty \frac{1}{q_n^3 (q_n^{-3/2 + \varepsilon})^2} = \sum_{n = 1}^\infty \frac{1}{q_n^{2\varepsilon}}\cdot $$ But since the sequence $(q_n)_1^\infty$ must grow at least exponentially fast, this series converges.

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Thank you, David. – Andrés E. Caicedo Jun 23 '14 at 14:29

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