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Is the series

$$ \sum_{n=1}^\infty \frac{|\sin n|^n}n\tag{1}$$

convergent?

If one want to use Abel's test, is

$$ \sum_{n=1}^\infty |\sin n|^n\tag{2}$$

convergent?

Thank you very much

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The second series is divergent because the irrational number $\frac{\pi}{2}$ has an infinite number of convergents $\frac{p_n}{q_n}$ with an odd denominator, hence $\left|\frac{p_n}{q_n}-\frac{\pi}{2}\right|\leq\frac{1}{q_n^2}$ gives: $$ \left|\sin(p_n)\right| = \left|\sin\left(\frac{\pi}{2}q_n+\frac{\theta}{q_n}\right)\right|=\left|\cos\frac{\theta}{q_n}\right|,\quad |\theta|\leq 1,$$ $$ \left|\sin(p_n)\right|\geq 1-\frac{1}{q_n^2}$$ hence $\left|\sin n\right|^n$ is bigger that $\left(1-\frac{1}{n^2}\right)^n$ infinitely often and the partial sums of $\sum_{n\in\mathbb{N}}\left|\sin n\right|^n$ cannot be bounded. However, they cannot grow too fast: numerical experiments give

$$\sum_{n=1}^{N}\left|\sin n\right|^n=O\left(N^{1/2}\right)\tag{1}$$

(consistent with the Weyl bounds for the partial sums of $\sum_{n\in\mathbb{N}}e^{in^2}$) that is enough to state that $$\sum_{n=1}^{+\infty}\frac{\left|\sin n\right|^n}{n}$$ converges by partial summation. This is also consistent with the model for which $\left|\sin n\right|$ acts like $\sin X$ where $X$ is a random variable, uniformly distributed over $[0,\pi]$: $$E\left[\sum_{n=1}^{+\infty}\frac{(\sin X)^n}{n}\right]=\frac{1}{\pi}\int_{0}^{\pi}-\log(1-\sin\theta)\,d\theta=\frac{4K}{\pi}+\log 2<+\infty.$$ A possible strategy in order to prove $(1)$ is to prove first, through Weyl's equidistribution theorem, that, provided that $M$ is big enough:

$$\sum_{n=N+1}^{N+M}\left|\sin n\right|^r\approx \frac{M}{\pi}\int_{0}^{\pi}\sin^r\theta\,d\theta,\tag{2}$$

then exploit the fact that:

$$ \int_{0}^{\pi}\sin^r \theta\,d\theta = O\left(\frac{1}{\sqrt{r}}\right).\tag{3}$$

By putting together $(2)$ and $(3)$ we get:

$$\sum_{i=1}^{M^2}\left|\sin n\right|^n\leq\sum_{j=0}^{M-1}\sum_{n=Mj+1}^{M(j+1)}\left|\sin n\right|^{Mj+1}\ll M\cdot\sum_{j=0}^{M-1}\frac{1}{\sqrt{Mj+1}}\ll M.\tag{4}$$

Moreover, since the supremum of the derivative of $\sin^r\theta$ over $[0,\pi]$ behaves like $\sqrt{\frac{r}{e}}$, keeping track of all the constants we get: $$\sum_{n=1}^{+\infty}\frac{\left|\sin n\right|^n}{n}\leq\frac{2}{\sqrt{\pi}}\left(1+\frac{1}{\sqrt{e}}\right)\zeta(3/2) = 4.73565\ldots$$ that is not too much far from the truth (given by partial summation and explicit numerical computations till $n=10^8$, with an error term $\frac{C}{10^4}$):

$$\sum_{n=1}^{+\infty}\frac{\left|\sin n\right|^n}{n}\leq \color{red}{2.151}.\tag{5}$$

Post-mortem addendum: it is interesting to study tailor-made numerical methods for the computation of such very slowly-converging series.

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  • $\begingroup$ Do you have an idea on how to prove the estimate with $O(N^{1/2+\epsilon})$? $\endgroup$ – Gabriel Romon Jun 16 '14 at 13:02
  • $\begingroup$ @G.T.R: I extended my answer including my thoughts about proving the crucial estimate. $\endgroup$ – Jack D'Aurizio Jun 16 '14 at 16:50
  • $\begingroup$ @Jack D'Aurizio: Writing in MMA Sum[Abs[Sin[n]]^n/n, {n, 1, 10^8}] I do not get an answer. How did you get to 2.151? Thank you! $\endgroup$ – TeM Jul 14 '17 at 20:10
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    $\begingroup$ @TeM: just follow the last link. $\endgroup$ – Jack D'Aurizio Jul 14 '17 at 20:11
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    $\begingroup$ It's not clear to me that this constitutes a rigorous proof; how is Weyl's equidistribution theorem sufficient here? $\endgroup$ – user41281 Nov 21 '17 at 1:36
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Note: Another great answer to this problem is given by Terry Tao in this MO post.

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Disclaimer:Not my solution. One elegant answer was posted by Robert Isreal on here . Apparently a holistic answer is not known.

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