Convergence/Divergence of infinite series $\sum_{n=1}^{\infty} \frac{(\sin n+2)^n}{n3^n}$

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

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.

Solution 1:

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.

Solution 2:

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:

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 $n\ge m^\vartheta$.
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.

Solution 3:

Solution 4:

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.

Solution 5:

(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