Convergence of $\sum(n^3\sin^2n)^{-1}$

2010-05-14T06:09:43Z

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.

Answer by Wadim Zudilin for Convergence of $\sum(n^3\sin^2n)^{-1}$

2010-05-15T08:40:38Z

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.

Convergence of $\sum(n^3\sin^2n)^{-1}$ - MathOverflow

Convergence of $\sum(n^3\sin^2n)^{-1}$

Answer by Wadim Zudilin for Convergence of $\sum(n^3\sin^2n)^{-1}$