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Does $\sum_n e^{n(\sin n -1)}$ converge ?

I'm pretty sure this series diverges because it seems $n(\sin n -1)$ is "close" to $0$ infinitely often. Nevertheless I lack skills in diophantine approximation to prove that...

Plot of $- n(\sin n -1)$ for $1\leq n\leq 500$.

Dents in the blue background are small values. enter image description here Can someone point me in the right direction ?

Context: some textbook question asks for the radius of $\sum_{n} e^{n\sin n}z^n$. It's easy to prove that the answer is $\frac 1e$, and I'd like to know what happens at $z=\frac 1e$.

Gabriel Romon
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  • $a_n\not\to 0$ as $n\to\infty$ It doesn't have to be far away from $0$ all the time, just infinitely many times. In fact all you really need is that the exponent is positive infinitely often. – Adam Hughes Sep 26 '16 at 19:34
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    ^But the exponent is never positive, as $\sin n < 1$ for all integers $n$. – JimmyK4542 Sep 26 '16 at 19:37
  • See also http://math.stackexchange.com/questions/828563/ which hinges on a very similar conceit. – Steven Stadnicki Sep 26 '16 at 21:18

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$\frac{\pi}{2}$ is irrational, hence has an infinite simple continued fraction expansion. Denoting the convergents by $\frac{p_k}{q_k}$, the relation $p_{k-1}q_k - p_kq_{k-1} = (-1)^k$ - holding for all simple continued fractions - shows that the denominators of successive convergents are coprime, in particular there are infinitely many convergents of $\frac{\pi}{2}$ with an odd denominator. Further, for all convergents we have

$$\biggl\lvert \frac{\pi}{2} - \frac{p_k}{q_k}\biggr\rvert < \frac{1}{q_k q_{k+1}} \leqslant \frac{1}{{q_k}^2}.$$

If $\frac{p}{q}$ is a convergent of $\frac{\pi}{2}$ with a denominator $\equiv 3 \pmod{4}$, then with $a = 3p,\, b = 3q$ we obtain a (non-reduced) fraction with denominator $\equiv 1 \pmod{4}$ satisfying

$$\biggl\lvert \frac{\pi}{2} - \frac{a}{b}\biggr\rvert < \frac{9}{b^2}.$$

Thus, regardless of whether infinitely many convergents of $\frac{\pi}{2}$ have denominator $\equiv 1 \pmod{4}$ or not, there are infinitely many pairs $(n,m)$ of positive integers with

$$\biggl\lvert \frac{\pi}{2} - \frac{n}{4m+1}\biggr\rvert < \frac{9}{(4m+1)^2},\quad \text{or}\quad \biggl\lvert (4m+1)\frac{\pi}{2} - n\biggr\rvert < \frac{9}{4m+1}.\tag{1}$$

For $m \geqslant 1$, $(1)$ implies $n < 2\cdot(4m+1)$, and thus

$$\biggl\lvert (4m+1)\frac{\pi}{2} - n\biggr\rvert < \frac{18}{n}.\tag{2}$$

Writing $\xi_m := (4m+1)\frac{\pi}{2}$, the identity

$$1 - \sin x = 1 - \cos \bigl(\xi_m-x\bigr) = 2\sin^2\bigl(\tfrac{1}{2}(\xi_m - x)\bigr)$$

shows that for $(n,m)$ satisfying $(2)$ we have

$$1 - \sin n \leqslant \frac{162}{n^2}$$

and consequently

$$\limsup_{n \to \infty} n(\sin n - 1) = 0,$$

showing that the series is divergent, as expected.

Daniel Fischer
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