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$$ \sum_{n=1}^{\infty} \frac{(\sin(n)+2)^n}{n3^n}$$

Does it converge or diverge? Can we have a rigorous proof that is not probabilistic? For reference, this question is supposedly a mix of real analysis and calculus.

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    Whoa - Ary and I both edited and I think it did something odd. I haven't seen MSE do that before.2012-02-13
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    Thanks for the edits, BTW. I'm new to this board syntax.2012-02-13
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    @mixedmath: There was also a pending (incorrect) edit in the queue at the same time, which might have helped confuse the system.2012-02-13
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    Might've been me pressing random buttons. :)2012-02-13
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    I posted a totally incorrect argument below claiming that it converged. Gerry Myerson pointed out the mistake. I deleted it. Apologies to those who may have wasted their time puzzling over my goof.2012-02-14
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    Presumably the answer has to do with how well $\pi$ can be approximated by rationals.2012-02-14
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    For simplicity, let's consider the similar sum $\sum (\cos(n)+1)^n/n2^n$. For any $x$, define $y(x)$ to be the smallest positive number such that $x=k\pi \pm y$, where $k$ is an integer. Then $y$ is a sawtooth function oscillating between 0 and $\pi/2$. We can throw away all terms in the sum with $y(n)$ greater than some fixed value, and by the ratio test there is no effect on convergence. Therefore let's only consider terms with small $y(n)$, so that $\cos n\approx 1-(1/2)y^2$. Then the $n$th term is roughly $(1/n)\exp(-ny^2/4)$.2012-02-14
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    Continuing the train of thought of the preceding comment, it doesn't really matter whether we replace $y^2$ with some similar oscillating function that has nearly parabolic-looking minima of zero, so roughly speaking we're trying to prove convergence of something of the flavor $\sum (1/n)\exp[-n(1+\sin n)]$. The corresponding integral, $\int_1^\infty (1/x)\exp[-x(1+\sin x)]dx$, does converge, because the integrand has humps of height $1/x$ and width $\sim 1/x$, so we're basically summing $1/n^2$. Statistically, that suggests that the sum converges.2012-02-14
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    @GEdgar If it helps, Salikhov proved that there are only finitely many $p/q$ for which $|\pi-p/q| < 1/q^{7.6063}$. See http://mathworld.wolfram.com/IrrationalityMeasure.html , and thanks to George Lowther for pointing this out to me on a previous question http://math.stackexchange.com/questions/2270/convergence-of-sum-limits-n-1-infty-sinnk-n/2275#2275 . However (continued)2012-02-14
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    I don't find this so useful. What we need is not bounds on how close $p/q$ can be to $\pi$ in rare cases, but on how frequently it can be somewhat close. For example, whenever $|p/q - \pi/2| < 1/q^{1.5}$, with $p \equiv 1 \mod 4$, the $p$-th summand is bounded below by $\mathrm{constant}/p$. So, if this happened for a set of $q$ with positive density, the sum would diverge. Note that $|\pi/2 - p/q| < 1/q^{1.5}$ is a relatively weak approximation -- the convergents of $\pi/2$ achieve $1/q^2$.2012-02-14
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    One approach that I think might work would be to write the sum as something like $\sum_{j=0}^\infty A_j$, where $A_j=\sum_{k=j(j+1)/2}^{j(j+1)/2+j} B_k$, and $B_k$ are the terms of the original sum. I think it should be possible to show that $A_j$ falls off quickly enough without resorting to statistical arguments, because the $B_k$ are correlated for nearby values of $k$.2012-02-14
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    I don't want to type anything more complicated, because I haven't started learning LaTeX yet, but I'd like to note that where I saw this question asked first they used [Hata's theorem](http://mathworld.wolfram.com/IrrationalityMeasure.html) to do... something. I don't remember. Sorry. **EDIT** Ah, my bad, I see it was mentioned by Mr. Speyer. I'll refrain from commenting until I open the "view more comments" box from now on.2012-02-14
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    It would seem pretty goofy to me if the irrationality measure of $\pi$ were relevant. Then the convergence of divergence of the series might change if we replaced $\sin(n)$ with, say, $\sin((\pi/e) n)$, or $\sin((\pi/u) n)$, where $u$ was some other irrational number. So would we need a different proof for every possible $u$? The statistical argument for convergence seems persuasive to me as long as $u$ is irrational. I can imagine that if $u$ was something like Liouville's constant, the sum might be very difficult to evaluate accurately, but I'd still bet a six-pack that it would converge.2012-02-15
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    The probabilistic argument is that the series converges, but interestingly enough Mathematica tells me that it diverges. Not entirely relevant, but thought you might be interested.2012-02-15
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    I've seen Wolfram|Alpha tell me that a series diverges while in the very next line giving its correct finite value, so I wouldn't put too much stock in that :-)2012-02-15
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    Related: http://math.stackexchange.com/questions/109827/asymptotic-behavior-of-sum-j-1n-cosp-pi-u-j-for-large-n-and-p2012-02-16
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    Also related: $\sum_{n=1}^\infty \frac{|\sin(n t)|^n}{n}$ converges for almost every real $t$ (in the sense of Lebesgue measure), but diverges for $t$ in a dense $G_\delta$ subset of $\mathbb R$ (thus for "generic" $t$ in the sense of Baire category). Which case is $t=1$ in? Nobody knows (but my bet would be on Lebesgue)2012-02-16
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    @jorki Can we talk about this in the chat? I just graphed the function and got what you said. There are spikes all over the place that are bounded by $\dfrac{1}{x}$2012-02-16
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    @joriki I tagged you wrongly. See my comment.2012-02-16
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    @RobertIsrael: Really? Nobody knows whether $\sum |\sin n|^n/n$ converges?? Crazy, man, crazy!2012-02-16
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    @Robert: That's very interesting. Do you have a reference for that?2012-02-16
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    Wow, this question has turned into a graveyard of flawed answers -- four have been deleted so far.2012-02-16
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    Its also open if the series $1/n^3sin^2(n)$ converges2012-02-16
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    The fact that $\sum_n |\sin (nt)|^n/n$ converges for almost every $t$ is a consequence of the fact that it converges in $L^1[0,2\pi]$: note that for even $n$, $\int_0^{2\pi} \sin(nt)^{n}\ dt = \frac{\pi}{2^{n-1}} {n \choose n/2} \sim C/\sqrt{n}$. The set where a series of nonnegative continuous functions diverges is always a $G_\delta$, and this one contains $\pi p/(2q)$ for positive integers $p$ and $q$ with $p$ odd.2012-02-16
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    Hmm, maybe I should have taken $\sum_n |\sin(nt)|^{n^3}$ instead as my example. If I'm not mistaken, this one diverges if there are infinitely many pairs of integers $p,q$ with $q$ odd and $|\pi - 2pt/q| < 1/q^{2.5}$. So here Salikhov's result wouldn't be sufficient.2012-02-16
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    An example of a well-studied series of this kind whose convergence is currently open is $\sum_{n=1}^\infty 1/(n^3\:\sin^2 n)$: Alekseyev, "On convergence of the Flint Hills series," http://arxiv.org/abs/1104.5100v12012-02-20

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