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

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.

  • 6
    $\begingroup$ Presumably the answer has to do with how well $\pi$ can be approximated by rationals. $\endgroup$ – GEdgar Feb 14 '12 at 1:36
  • 11
    $\begingroup$ 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) $\endgroup$ – Robert Israel Feb 16 '12 at 2:37
  • 5
    $\begingroup$ @RobertIsrael: Really? Nobody knows whether $\sum |\sin n|^n/n$ converges?? Crazy, man, crazy! $\endgroup$ – Ben Crowell Feb 16 '12 at 3:16
  • 5
    $\begingroup$ @Robert: That's very interesting. Do you have a reference for that? $\endgroup$ – joriki Feb 16 '12 at 3:22
  • 4
    $\begingroup$ Wow, this question has turned into a graveyard of flawed answers -- four have been deleted so far. $\endgroup$ – joriki Feb 16 '12 at 14:18

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.

| cite | improve this answer | |
  • $\begingroup$ @ 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. $\endgroup$ – mercio Feb 16 '12 at 17:29
  • 1
    $\begingroup$ +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 $\mu\gt2$?) $\endgroup$ – joriki Feb 16 '12 at 18:02
  • 1
    $\begingroup$ @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. $\endgroup$ – mercio Feb 16 '12 at 18:41
  • 1
    $\begingroup$ 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 $\mu\leqslant20$?). $\endgroup$ – Did Feb 17 '12 at 12:54
  • 2
    $\begingroup$ mercio: If I may insist: hence one cannot write that Salikhov proved that $\pi$ has a finite irrationality measure (fourth paragraph). $\endgroup$ – Did Feb 17 '12 at 15:10

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.

| cite | improve this answer | |

see also here


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.

| cite | improve this answer | |

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.

| cite | improve this answer | |

(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

| cite | improve this answer | |
  • $\begingroup$ It seems that by "$IN$" you mean $\mathbb N$? You can produce that with \mathbb N. $\endgroup$ – joriki Feb 16 '12 at 11:18
  • 1
    $\begingroup$ Case $2$ is the hard part, so I'm afraid you haven't made any progress with this. $\endgroup$ – joriki Feb 16 '12 at 11:20
  • $\begingroup$ i just share , maybe that will help someone $\endgroup$ – Hassan Feb 16 '12 at 11:29
  • 2
    $\begingroup$ 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. $\endgroup$ – joriki Feb 16 '12 at 11:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.