Convergence of $\sum(n^3\sin^2n)^{-1}$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T07:36:22Z http://mathoverflow.net/feeds/question/24579 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/24579/convergence-of-sumn3-sin2n-1 Convergence of $\sum(n^3\sin^2n)^{-1}$ Andres Caicedo 2010-05-14T06:09:43Z 2012-08-11T23:51:45Z <p>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. </p> <p>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 $\frac1{\pi^2 n}$.</p> <p>[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 $S(354)&lt;5$. 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.]</p> <p>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 <em>ad hoc</em> approach just tailored to this specific series would be interesting as well.</p> http://mathoverflow.net/questions/24579/convergence-of-sumn3-sin2n-1/24712#24712 Answer by Wadim Zudilin for Convergence of $\sum(n^3\sin^2n)^{-1}$ Wadim Zudilin 2010-05-15T08:40:38Z 2010-05-15T08:40:38Z <p>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.</p> <p>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 $$\sum_{n:n|\sin n|&lt; n^\varepsilon}\frac1{n^3\sin^2n}. \qquad(1)$$ For any such $n$, let $q=q(n)$ minimizes the distance $|\pi q-n|\le\pi/2$. Then $$\sin|\pi q-n|=|\sin n|&lt; \frac1{n^{1-\varepsilon}},$$ 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, $$\biggl|\pi-\frac nq\biggr|&lt;\frac{C_1}{qn^{-\varepsilon}},$$ equivalently $$\biggl|\pi-\frac nq\biggr|&lt;\frac{C_2}{n^{2-\varepsilon}} \quad\text{or}\quad \biggl|\pi-\frac nq\biggr|&lt;\frac{C_2'}{q^{2-\varepsilon}}$$ (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 $$\biggl|\pi-\frac nq\biggr|&lt;\frac{C_2'}{q^{2-\varepsilon}}$$ and how small is the quantity in these cases. (Note that it follows from Dirichlet's theorem that an even stronger inequality, $$\biggl|\pi-\frac nq\biggr|&lt;\frac1{q^2},$$ happens for infinitely many pairs $n$ and $q$.) The series (1) converges if and only if $$\sum_{n:|\pi-n/q|&lt; C_2n^{-2+\varepsilon}}\frac1{n^5|\pi-n/q|^2}$$ 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 $|\pi-n/q|&lt; C_2n^{-2+\varepsilon}$ 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 $|\pi-n/q|$.</p> <p>Summarizing, <em>the original series converges if and only if the series in</em> (2) <em>converges.</em></p> <p>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 $$C_3\frac{\log\log q}{q^2\log q}&lt;\biggl|e-\frac pq\biggr|&lt; C_4\frac{\log\log q}{q^2\log q}$$ (see, e.g., [C.S.Davis, <em>Bull. Austral. Math. Soc.</em> 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 $\pi$?</p> <p>The <em>irrationality exponent</em> $\mu=\mu(\alpha)$ 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 $$\biggl|\pi-\frac nq\biggr|&lt;\frac{C_5}{q^{5/2}},$$ 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 <a href="http://mathoverflow.net/questions/23547/does-pi-contain-1000-consecutive-zeroes-in-base-10/23551#23551" rel="nofollow">this answer</a> by Gerry and my comment) only asserts that $\mu(\pi)&lt;7.6064\dots$,. </p> <p>I hope that this explains the problem of determining the behavior of the series in question.</p>