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$\begingroup$

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

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

5 Answers 5

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The values for which $\sin(n)$ is close to $1$ (say in an interval $[1-\varepsilon ; 1]$) are somewhat regular :

$1 - \varepsilon \le \sin(n)$ implies that there exists an integer $k(n)$ such that $n = 2k(n) \pi + \frac \pi 2 + a(n)$ where $|a(n)| \leq \arccos(1- \varepsilon)$. As $\varepsilon \to 0$, $\arccos(1- \varepsilon) \sim \sqrt{2 \varepsilon}$, thus we can safely say that for $\varepsilon$ small enough, $|n-2k(n) \pi - \frac{\pi}2| = |a(n)| \leq 2 \sqrt{ \varepsilon} $

If $m \gt n$ and $\sin(n)$ and $\sin(m)$ are both in $[1-\varepsilon ; 1]$, then we have the inequality $|(m-n) - 2(k(m)-k(n)) \pi| \leq |m-2k(m)\pi - \frac{\pi}2| + |n-2k(n)\pi - \frac{\pi}2| \leq 4 \sqrt { \varepsilon} $ where $(k(m)-k(n))$ is some integer $k$.

Since $\pi$ has a finite irrationality measure, we know that there is a finite real constant $\mu \gt 2$ such that for any integers $n,k$ large enough, $|n-k \pi| \ge k^{1- \mu} $.

By picking $\varepsilon$ small enough we can forget about the finite number of exceptions to the inequality, and we get $ 4\sqrt{\varepsilon} \ge (2k)^{1- \mu}$. Thus $(m-n) \ge 2k\pi - 4\sqrt{\varepsilon} \ge \pi(4\sqrt{\varepsilon})^{\frac1{1- \mu}} - 4\sqrt{\varepsilon} \ge A_\varepsilon = A\sqrt{\varepsilon}^{\frac1{1- \mu}} $ for some constant $A$.

Therefore, we have a guarantee on the lengh of the gaps between equally problematic terms, and we know how this length grows as $\varepsilon$ gets smaller (as we look for more problematic terms)

We can get a lower bound for the first problematic term using the irrationality measure as well : from $|n-2k(n) \pi - \frac{\pi}2| \leq 2\sqrt {\varepsilon}$, we get that for $\varepsilon$ small enough, $(4k+1)^{1- \mu} \le |2n - (4k+1) \pi| \le 4\sqrt \varepsilon$, and then $n \ge B_\varepsilon = B\sqrt\varepsilon^{\frac1{1- \mu}}$ for some constant $B$.

Therefore, there exists a constant $C$ such that forall $\varepsilon$ small enough, the $k$-th integer $n$ such that $1-\varepsilon \le \sin n$ is greater than $C_\varepsilon k = C\sqrt\varepsilon^{\frac1{1- \mu}}k$

Since $\varepsilon < 1$ and $\frac 1 {1- \mu} < 0$, this bound $C_ \varepsilon$ grows when $\varepsilon$ gets smaller. And furthermore, the speed of this growth is greater if we can pick a smaller (better) value for $\mu$ (though all that matters is that $\mu$ is finite)


Now let us give an upper bound on the contribution of the terms where $n$ is an integer such that $\sin (n) \in [1-2\varepsilon ; 1-\varepsilon]$

$$S_\varepsilon = \sum \frac{(2+\sin(n))^n}{n3^n} \le \sum_{k\ge 1} \frac{(1- \varepsilon/3)^{kC_{2\varepsilon}}}{kC_{2\varepsilon}} = \frac{- \log (1- (1- \varepsilon/3)^{C_{2\varepsilon}})}{C_{2\varepsilon}} \\ \le \frac{- \log (1- (1- C_{2\varepsilon} \varepsilon/3))}{C_{2\varepsilon}} = \frac{- \log (C_{2\varepsilon} \varepsilon/3))}{C_{2\varepsilon}} $$

$C_{2\varepsilon} = C \sqrt{2\varepsilon}^\frac 1 {1- \mu} = C' \varepsilon^\nu$ with $ \nu = \frac 1 {2(1- \mu)} \in ] -1/2 ; 0[$, so :

$$ S_\varepsilon \le - \frac{ \log (C'/3) + (1+ \nu) \log \varepsilon}{C'\varepsilon^\nu} $$


Finally, we have to check if the series $\sum S_{2^{-k}}$ converges or not :

$$ \sum S_{2^{-k}} \le \sum - \frac { \log (C'/3) - k(1+ \nu) \log 2}{C' 2^{-k\nu}} = \sum (A+Bk)(2^ \nu)^k $$

Since $2^ \nu < 1$, the series converges.

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    @ bgins : I don't think so. In order to recognize a Lebesgue-type thing, from what I remember I would have to sort the terms according to the value of ((2+sin(n))/3)^n and not simply sin(n), and it would be harder to evaluate the corresponding contribution. Maybe someone more knowledgeable can answer about this.2012-02-16
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    +1, very nice. I started down the same $\sin n \in [1-2\varepsilon ; 1-\varepsilon]$ route but didn't immediately see how to derive the required bounds that you derive in the first part -- I didn't pursue it any further when I saw Robert's comment about $\sum_{n=1}^\infty \frac{|\sin(n t)|^n}{n}$. Doesn't this proof also apply to that series for $t=1$? It seems the only differences are the factor of $1/3$ and the absolute value, but neither should matter. (By the way I think it should be $\mu\ge2$ instead of $\mu\gt2$?)2012-02-16
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    @joriki : I think the proof applies to $\sum \frac{|\sin(nt)|^n}n$ for any $t$ where the irrationality measure of $\frac t \pi$ is finite. mathworld says Salikhov proved his result in 2008, I don't know if the irrationality measure of $\pi$ was proven to be finite before then or not. About $\mu$, originally I wanted to be on the safe side and thought of it as a number strictly greater than the irrationality measure (which is an infimum, maybe not a minimum...). Maybe we can have $\mu = 2$ in some cases but it doesn't matter ultimately.2012-02-16
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    Very nice work!2012-02-16
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    I'm sorry--I just got past introductory proof based mathematics and into abstract algebra, so I'm having a bit of difficulty following this answer. We've not dealt with modulo multiplication groups yet, so I'm not sure what just happened at the beginning of this solution.2012-02-16
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    +1, obviously. Here are a problem and a question. Problem: when bounding $S_\varepsilon$, you use the inequality $-\log(1-u)\leqslant2u$ when $u\to1$. This inequality does not hold. (But the summation $\sum\limits_k S_{1/2^k}$ still converges, I believe.) Which brings the question: what bound on $\mu$ is really needed for your proof to work? Tracing back your steps, I got the impression that everything works fine as soon as $\mu\gt1$. Is that true?2012-02-17
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    @ Didier Piau : yes there was a problem, it is not as nice as I thought. You're thinking about $\mu$ in the wrong way, it's not "as long as $\mu$ is at least something", it is "at long as $\mu$ is at most something". In this case, as long as $\mu$ is finite it seems. The greater $\mu$ is, the worse everything is (the closer $2^\nu$ is to $1$ for example).2012-02-17
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    @ badreferences : I detailed a bit the algebra in the first part. Anyway we're not talking about multiplicative groups but additive groups, as in $(\mathbb{R},+)/(2\pi\mathbb{Z},+)$.2012-02-17
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    mercio: Right. Then the appeal to Salikhov might be debatable, since the finiteness of $\mu$ was known before Salikhov's explicit upper bound (am I right if I say this goes back at least (and maybe exactly) to Mahler 1953 with an estimate like $\mu\leqslant20$?).2012-02-17
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    @ Didier : according to http://mathworld.wolfram.com/Pi.html, you're right.2012-02-17
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    mercio: If I may insist: hence one cannot write that *Salikhov proved that $\pi$ has a finite irrationality measure* (fourth paragraph).2012-02-17
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    @mercio: Thanks for the detail. Cheers.2012-02-19
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    Sweet! So if we generalize the problem by replacing $\sin n$ with $\sin(\pi n/u)$, then I suppose that if $u$ is a Liouville number, we don't know whether the series converges.2012-02-20
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I knew this looked familiar. User Unoqualunque located the key reference. Here is a more recent reference that provides a fairly general approach:

Enrico Laeng, Vittorino Pata. A convergence–divergence test for series of nonnegative terms, Expositiones Mathematicae 29 (4), (2011) 420–424. MR2861768 (2012m:40002).

The authors discuss a test that does not require monotonicity of the sequence, and instead focuses on how "clustered" we find similar terms within the sequence.

They highlight that their test applies to show that $$ \sum_{n=1}^\infty\frac1{n^{2+\cos n}} $$ diverges, while $$ \sum_{n=1}^\infty\frac1n\left(\frac{2+\cos n}3\right)^n $$ converges. They say:

Case (i) has been recently addressed, in (Revue de la filière Mathématique (RMS) 119 (2008–2009), 3–8), where the authors give a proof that was (in their own words) at the frontier between analysis and number theory. Case (ii) apparently originated in a curious way: it was proposed in a calculus exam by mistake, and remained open for a long time thereafter. A solution was devised only ten years later (SIAM Problems and Solutions (2009)), once again by means of quite sophisticated tools.

As is to be expected, the test is very general but a bit cumbersome to state:

Let $(c_n)_{n\ge1}$ be a sequence of nonnegative terms such that $\sum_n c_n<+\infty$. Let $(a_n)_{n\ge1}$ be a series of nonnegative terms. Then:

  1. The series $\sum_n a_n$ converges if $(na_n)_{n\ge1}$ is a bounded sequence, and there exist $\rho,\vartheta\ge0$ and $\varepsilon\in(0,1]$ such that $$ |\{p \in\mathbb N\mid 1\le p\le m\mbox{ and }a_{n+p} > c_n \}| \le \rho m^{ 1 −\varepsilon} $$ for every $m$ sufficiently large, and every $n\ge m^\vartheta$.

  2. The series $\sum_n a_n$ diverges if there exist $\omega> 0$ and $\lambda\ge0$ such that the inequality $$ \max_{1\le p\le m} a_{km+p}\ge \frac{\omega}{(km+m)^{1+\lambda/m}} $$ holds for infinitely many $m$ and every $k$.

To apply the test to the series above, one needs to know something about rational approximations to $\pi$ (naturally). Actually, the authors show that to apply the test to show the divergence of the first series only requires that $\pi$ is irrational, and to show the convergence of the second series only needs that $\pi$ is not a Liouville number. The paper is reasonably self-contained.

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see also here

http://www.siam.org/journals/categories/99-005.php

A Calculus Exam Misprint (Solved)

Summary: A misprint from a calculus exam yields a problem that possibly cannot be answered by currently known methods. Specifically, the exam question asked whether the series $\sum_{n=1}^{\infty} \frac{(2 + \sin n)^n}{3^n \, n}$ converges.

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I propose the following heuristic argument that the series converges:

The natural numbers $n$ are uniformly distributed ${\rm mod}\ 2\pi$. Therefore the expected value of the $n$-th term of the series is $$a_n:={1\over n}\int_{-\pi}^\pi\left({2+\cos\phi\over 3}\right)^n\ d\phi\ .$$ Now a look at the graphs shows that $${2+\cos\phi\over 3}\leq e^{-\phi^2/9}\qquad(-\pi\leq\phi\leq\pi)\ .$$ Therefore $$a_n\leq{1\over n}\int_{-\pi}^\pi e^{-n\phi^2/9}\ d\phi<{1\over n} \int_{-\infty}^\infty e^{-n\phi^2/9}\ d\phi={\sqrt{3\pi}\over n^{3/2}}\ ,$$ which leads to convergence.

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(incomplete proof)

Consider this sequence:

$$v_k = \sum_{p=p_{k,min}}^{p_{k,max}} u_p$$

where $p_{k,min}=[2k\pi]+1$ and $p_{k,max}=[2(k+1)\pi]$ and $u_p = \frac{(\sin(p)+2)^p}{p3^p}$

1/ we have $\sum_{n=1}^{\infty} u_n = \sum_{n=1}^{\infty} v_n$

notice that $\mathbb N = \cup_{k \in \mathbb{N}} I_k$ where $I_k=[p_{k,min},p_{k,max}]$ and both $v_k>0$ and $u_n>0$

2/ $v_k$ can be bounded with a convergente term

Fact 1: $I_k$ can contain exactly 6 or 7 natural numbers

Fact 2: each interval of the solution of $sin(x)\geq 0.9$ have a lenght less than 2asin(0.9)-pi<1 so it can't contain 2 natural numbers.

we have two cases:

Case 1: for every p in $I_k$ $sin(p)<0.9$ $u_p < \frac{(2,9/3)^p}{p} $

so $v_k<7\frac{(2,9/3)^p_{k,min}}{p_{k,min}} $

Case 2: there is one p in $I_k$ such that $sin(p)\geq 0.9$ p+3 is also in $I_k$ and $sin(p+3)<0.5$

... this part need more thinking, i ll be back if i find something, or hope someone can use this

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    It seems that by "$IN$" you mean $\mathbb N$? You can produce that with `\mathbb N`.2012-02-16
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    Case $2$ is the hard part, so I'm afraid you haven't made any progress with this.2012-02-16
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    i just share , maybe that will help someone2012-02-16
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    Sharing is good, of course, but it's also good to read what others have already written and take it into account. If you read the comments under the question, you'll find that several people have already put thought into how to deal with the case where $\sin n$ gets close to $1$. Also, if you're already aware that you only dealt with the easy part, but you think it might be valuable to share nonetheless, it would make more sense to title it something like "an approach" or "an idea", not "incomplete proof" -- that sounds like you've made progress and the missing part is less than half the work.2012-02-16