Convergence of the series $\sum_{n=1}^\infty \frac{(\sin n)^n}{n}$.

Question

Please determine whether the series $\displaystyle\sum_{n=1}^\infty \frac{(\sin n)^n}{n}$ converges.

(Note: In Mathematica, the result tends to converge. Moreover, this is a problem mis-copied from Advanced Calculus Exam, so we don't know the difficulty of the problem [Maybe it can be solved in college mathematics]. )

Answer 1

This series converges absolutely, as do a variety of similar series. This follows from a mild generalization of a paper by Ravi B. Boppana which was about a similar problem. I'll offer a proof of the generalization I mentioned, but just a few months ago Ravi uploaded a YouTube video detailing his proof. The video is very informative and not too long, so I definitely recommend it. Below, I prove that for all real numbers $\alpha\in (0,1]$, the following series is absolutely convergent, where the title question is the special case $\alpha=1$. $$\sum_{n=1}^\infty \frac{((1-\alpha)+\alpha\sin(n))^n}{n}$$

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For convenience, let's say $s(x) = (1-\alpha) + \alpha\sin(x)$, hence we need to sum $s(n)^n/n$. To do this, we partition the sum into two components. The first component are the "tame" numbers, which by definition are those $n$ which obey $|s(n)|^n\leq \frac{1}{n}$, and otherwise $n$ is said to be "wild". It's easy to show that the summation over the tame numbers converges absolutely. $$\sum_{n=1}^\infty \frac{s(n)^n}{n} = \sum_{n\in\operatorname{tame}} \frac{s(n)^n}{n} + \sum_{n\in\operatorname{wild}} \frac{s(n)^n}{n}$$ $$\sum_{n\in\operatorname{tame}} \left|\frac{s(n)^n}{n}\right| \leq \sum_{n\in\operatorname{tame}} \frac{1}{n^2} \leq \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} < \infty $$

To prove absolute convergence of the "wild" series, we need to show that the wild numbers are very rare. If we can do this, then we can show that the sum of reciprocal wild numbers will converge. Since we can bound $|s(x)|\leq 1$, then this will prove absolute convergence of the previously mentioned summation over the wild numbers. To do this, we begin by showing the wild numbers are very close to an odd multiple of $\frac{\pi}{2}$. $$\begin{align} \text{$n$ is wild} &\iff |s(n)|^n >\frac{1}{n} \\ &\iff |s(n)| > n^{-1/n} \\ &\implies (1-\alpha)+\alpha|\sin(n)| > 1-\frac{\ln(n)}{n} \\ &\iff |\sin(n)| > 1 - \frac{\ln(n)}{\alpha n} \\ &\implies |\sin(n)| > 1-\frac{1}{\sqrt{n}} \end{align}$$

The last implication above holds so long as $n$ is sufficiently large, since we'll have $\frac{\ln(n)}{\alpha}<\sqrt{n}$ for all sufficiently large $n$. We can ignore any terms for which this inequality fails, since there's only finitely many exceptions and thus they can't affect convergence. Anyway, now we let $k$ be the unique integer satisfying $k\pi\leq n < (k+1)\pi$, which thus satisfies $|n-(2k+1)\frac{\pi}{2}|<\frac{\pi}{2}$. This leads to the following. $$|\sin(n)| = \left|\cos\left(n-(2k+1)\frac{\pi}{2}\right)\right| < 1-\left(n-(2k+1)\frac{\pi}{2}\right)^2$$ $$\begin{align} \text{$n$ is wild} &\implies 1-\left(n-(2k+1)\frac{\pi}{2}\right)^2 > 1-\frac{1}{\sqrt{n}} \\ &\iff \left(n-(2k+1)\frac{\pi}{2}\right)^2 < \frac{1}{\sqrt{n}} \\ &\iff \left|n-(2k+1)\frac{\pi}{2}\right| < n^{-1/4} \end{align}$$

This shows that the wild numbers must be quite close to an odd multiple of $\frac{\pi}{2}$. To show that the odd numbers are rare, we just need to show that the gaps between consecutive wild numbers are large. So, let's suppose that $N>n$ is another wild number, thus we have an integer $K$ such that $N\approx (2K+1)\pi/2$ in the same sense as above. Letting $g=N-n$ be the gap between these wild numbers, clearly $g\approx (K-k)\pi$, hence $g$ is an integer which is very near a multiple of $\pi$. More precisely, the following holds. $$\begin{align} |g-(K-k)\pi| &< n^{-1/4} + N^{-1/4} < 2n^{-1/4} \\ \implies \left|\pi-\frac{g}{K-k}\right| &< \frac{2}{(K-k)n^{1/4}} \end{align}$$

This shows that $\frac{g}{K-k}$ is a rational approximation of $\pi$. This relates to the irrationality measure of $\pi$, and Ravi cites a result showing that the irrationality measure of $\pi$ is at most $20$. More specifically, Ravi cites a result showing that $|\pi-p/q|>\frac{1}{q^{20}}$ for all integers $p,q$ where $q>1$. Since all of $n,k,N,K$ are integers while $N-n\approx (K-k)\pi$, this easily implies that $K-k>1$ so long as $n$ is sufficiently large. This lets us apply Ravi's result to prove the following. $$\begin{align} \left|\pi-\frac{g}{K-k}\right| &> \frac{1}{(K-k)^{20}} \\ \implies \frac{1}{(K-k)^{20}} &< \frac{2}{(K-k)n^{1/4}} \\ \implies (K-k) &> \left(\frac{n^{1/4}}{2}\right)^{1/19} \\ \implies (K-k) &> \frac{n^{1/76}}{2} \end{align}$$

This shows that the gap between consecutive wild numbers grows at least as fast as the $76$'th root of $n$. From here, it's much easier to show that the wild numbers are sufficiently rare. Let $W_1,W_2,\cdots$ denote the sequence of wild numbers in increasing order. Clearly we generally have $n\leq W_n$, and as above we have $W_{n+1}-W_n > \frac{1}{2}(W_n)^{1/76} \geq \frac{1}{2}n^{1/76}$. This gives a lower bound on the wild sequence like so. $$\begin{align} W_n &= W_1+\sum_{k=1}^{n-1} (W_{k+1}-W_k) \\ &\geq W_1 + \sum_{k=1}^{n-1} \frac{1}{2}n^{1/76} \\ &\approx \mathcal{O}(n^{77/76}) \end{align}$$

This proves that the wild numbers grow super-linearly, comparable to a power of $n$. This is enough to prove convergence of the sum of reciprocal wild numbers, which ultimately proves the absolute convergence of our original series. $$\begin{align} \sum_{n\in\operatorname{wild}}\left|\frac{s(n)^n}{n}\right| &\leq \sum_{n\in\operatorname{wild}} \frac{1}{n} \\ &= \sum_{k=1}^\infty \frac{1}{W_k} \\ &\leq \sum_{k=1}^\infty \mathcal{O}(n^{-77/76}) \\ &< \infty \end{align}$$

Answer 2

EDIT: This proof contains a mistake and I don't know if it can be saved (see the comments) - I'm leaving it up here in case it gives you an idea or if it can be corrected...

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I've managed to prove the (absolute!) convergence of the series based on an idea in the comment by user @xavierm02, but my proof uses a deep (and relatively modern) result in Diophantine approximation. It appears to me that since I use the deep result only to prove that something that appears negligible is indeed negligible, and I leave some degrees of freedom in that proof, I am optimistic that there is a more elementary way to do this part.

Lemma. There is only a finite number of solutions to the inequality $\left| \sin n \right| > (\frac 1 2) ^{1/\sqrt n}$.

Proof of Lemma. I will skip some (I think simple) details in the proof. Let $\epsilon > 0$ be very small and observe the inequality $|\sin n| > 1-\epsilon$. Since $\sin$ is Lipschitz continuous and equal to $\pm 1$ exactly in the points $\pi/2 + \pi k$ for integers $k$ (and is symmetric around these points), there is some $\delta = \Theta(\epsilon)$ such that the above inequality is equivalent to: $|n - \pi/2 - \pi k| < \delta$. (That is, solutions are now in both $n$ and $k$.)

Dividing by $\pi n$ we get:

$$\left|\frac 1 \pi - \frac {2k+1} {2n}\right| < \frac \delta {\pi n}$$

This is a Diophantine approximation problem - how well can we approximate $1/\pi$ by rationals? Before we answer the question, let's plug in $\delta = \Theta(\epsilon)$ with the $\epsilon$ from the lemma. So up to some constant, we are interested in

$$\left|\frac 1 \pi - \frac {2k+1} {2n}\right| < \frac 1 n \left(1 - \left(\frac 1 2\right)^{1/\sqrt n} \right)$$

We now quote Mahler's theorem [1953]: $1/\pi$ is not a Liouville number*, meaning rational approximations are only "polynomially good": Given a rational approximation $p/q \approx 1/\pi$, the difference $p/q - 1/\pi$ is bounded (below!) by some constant power of $q$. However our Diophantine inequality requires a superpolynomial approximation, so it cannot be solved infinitely many times. Edit: This is wrong.

_{* Actually, Mahler's theorem is about $\pi$, but Liouville numbers are closed under the reciprocal operation.}

Proposition. The series $\sum_{n=1}^\infty \frac {(\sin n)^n} n$ converges absolutely.

Proof. All except a finite amount of values of $n$ satisfy $|\sin n|^n \le \left( \frac 1 2 \right)^{\sqrt n}$, which decays at least as fast as $1/n^3$. QED.

Answer 3

I just fell by accident on this question, and figured I should give out my proof as it uses other (simpler in my opinion) ideas. I originally had to solve it during an oral exam for a competition.
It can always be useful for newcomers that want a different answer.

The key argument is to use the discrete equivalent of integration by parts: Abel’s summation which can so often save you from finding convenient paquets for almost alternating sequences. I’ll try to explain this thoroughly because it is a very convenient technique.

Now in the continuous version, we use $$\int_a^b f(x)g(x) \, dx = \left[f(x) \int_a^x g(t) \, dt \right]_a^b - \int_a^b f’(x) \int_a^x g(t) \, dt $$

Now in our setting we want discrete objects, so $a=1,\, b= \infty$, $x=n$ and the integral becomes a sum and the differential a difference.

What we want is to get a more convenient sum to study, we know that the differential of $1/x$ is $-1/x^2$ which is much more suited for convergence, so we set $d_n = \frac{1}{n} - \frac{1}{n+1}= \frac{1}{n(n+1)}$, the negative of the differential and $S_n= \sum_{k=1}^n \sin(k)^k$.

$$\begin{align*} \sum_{n=1}^N \frac{\sin(n)}{n} &= \sum_{n=1}^N (S_n-S_{n-1})\frac{1}{n} \\ &= \sum_{n=1}^N S_n\frac{1}{n} - \sum_{n=1}^N S_{n-1}\frac{1}{n} \\ &= \sum_{n=1}^N S_n\frac{1}{n} - \sum_{n=-1}^{N-1} S_{n}\frac{1}{n+1} \\ &= S_N\frac{1}{N} - S_{0}\frac{1}{1} + \sum_{n=1}^{N-1} S_n d_n &(\ast)\\ &= S_N\frac{1}{N} + \sum_{n=1}^{N-1} \frac{S_n}{n(n+1)} \\ \end{align*}$$

Now I hope you recognised the integration by parts formula at the step $(\ast)$. Now we only need to prove that $S_n$ is « bounded enough » to ensure convergence since $d_n$ is absolutely convergeant.

Basically, $S_n$ should have the same behaviour as the integral $\int_{0}^n \sin(t)^t \, dt$, but the deviation from it increases over time, so we will use a much more basic method. Let $\epsilon >0$. We know that $\cos(x) \leq 1-\frac{x^2}{2}$, so for $x \in [-\pi/2,\pi/2]$, $|x| \leq \sqrt{2\epsilon} \implies \cos(x) \leq 1-\epsilon$.
By translation, and $\pi$-periodicity of $|\sin(x)|$, we deduce that $$\begin{align*} &x \notin \bigcup_{k \in \mathbb{Z}} ](k+1/2)\pi - \sqrt{2\epsilon}, (k+1/2)\pi + \sqrt{2\epsilon}[ \\ \iff & x \mod \pi \notin ]\pi/2 -\sqrt{2\epsilon},\pi/2 +\sqrt{2\epsilon}[ \\ \implies & |\sin(x)| \leq 1-\epsilon. \end{align*}$$

Now since $(n \mod \pi)_{n \in \mathbb{N}^*}$ is well equidistributed, if we set $$\begin{align*} {}^\epsilon \! A_{m}^{n} &= \{m+1,\dots,n\} \setminus {}^\epsilon \! B_n^m \\ {}^\epsilon \! B_{m}^{n} &= \left\{k \in \{m+1,\dots,n\} \mid \pi/2- \sqrt{2\epsilon} \le k \mod \pi \le \sqrt{2\epsilon}\right\} \end{align*}$$ then $$\begin{eqnarray} \frac{\sharp {}^\epsilon \! A_{m}^{n}}{n} &\to& \frac{\pi - 2 \sqrt{2\epsilon}}{\pi}\\ \frac{\sharp {}^\epsilon \! B_{m}^{n}}{n} &\to& \frac{2 \sqrt{2\epsilon}}{\pi}\\ \end{eqnarray}$$

And furthermore it has uniform linear speed, i.e. $$\begin{eqnarray} \left| \frac{ \sharp {}^\epsilon \! B_{m}^{n}}{n-m} - \frac{ 2 \sqrt{2\epsilon}}{\pi} \right| & \le & \frac{C \sqrt{\epsilon}}{n-m} \end{eqnarray}$$ since $\pi$ is badly approximated by rationals, otherwise we could have gotten some $1/n\ln(n)$.

Only the second bound interests us: Let $N = \frac{P(P+1)}{2}$, $$\begin{align*} \sum_{n=1}^{N} |\sin(n)|^n &= \sum_{p=0}^P \sum_{k=1}^{p} |\sin(p(p+1)/2 + k)|^{p(p+1)/2 + k}\\ &= \sum_{p=0}^P \sum_{k\in {}^\epsilon \!A_{p(p+1)/2}^{(p+1)(p+2)/2}} |\sin(p(p+1)/2 + k)|^{p(p+1)/2 + k}\\ &+ \sum_{p=0}^P \sum_{k\in {}^\epsilon \!B_{p(p+1)/2}^{(p+1)(p+2)/2}} |\sin(p(p+1)/2 + k)|^{p(p+1)/2 + k} \\ &\le \sum_{p=0}^P \sum_{k\in {}^\epsilon \!A_{p(p+1)/2}^{(p+1)(p+2)/2}} (1-\epsilon)^{p(p+1)/2 + k} + \sum_{p=0}^P \sum_{k\in {}^\epsilon \!B_{p(p+1)/2}^{(p+1)(p+2)/2}} 1 \\ &\le \sum_{p=0}^P \sum_{k = 1}^{p} (1-\epsilon)^{p(p+1)/2 + k} + \sum_{p=0}^P \sharp {}^\epsilon \! B_{p(p+1)/2}^{(p+1)(p+2)/2} \\ &\le \sum_{n=1}^N (1-\epsilon)^n + \sum_{p=0}^P \left( \frac{2p\sqrt{2\epsilon}}{\pi} + C\sqrt{\epsilon} \right) \\ &\le (1-\epsilon)\frac{1-(1-\epsilon)^{N}}{\epsilon} + \frac{2N\sqrt{2\epsilon}}{\pi} + C \sqrt{2N\epsilon}\\ &\le \frac{1}{\epsilon} + \frac{2N\sqrt{2\epsilon}}{\pi} + C \sqrt{2N\epsilon} \end{align*}$$

Now by setting $\epsilon = N^{-2/3}$, we get $S_N = O(N^{2/3})$ (which is the best bound of the form $N^\alpha$), now from our previous result form Abel’s summation, we get the convergence and $$\sum_{n=1}^\infty \frac{\sin(n)^n}{n} = \sum_{n = 1}^\infty \frac{S_n}{n(n+1)}$$ And the convergence rate is at least in $O(1/\sqrt[3]{n})$, which is a better bound than the one given by the elegant technique of @Jade-Vanadium. But beware to next section, we can't ensure absolute convergence with this method, which is possible with @Jade-Vanadium's method!!

I have made some numerical computations and it seems the convergence rate is exactly $1/n^{1/3}$, if you want here is my code (that I asked chatgpt to write for me since I am lazy):

import numpy as np
import matplotlib.pyplot as plt
Définir la fonction principale
def sequence(N, d):
    # Calculer les termes de sin(k)^k / k pour k dans [1, N^3]
    max_k = N3
    terms = np.array([np.sin(k)k / k for k in range(1, max_k+1)])
# Calculer la suite pour chaque n dans [1, N]
results = []
for n in range(1, N+1, d):
    # Somme des termes de n à n^3
    sum_terms = np.sum(terms[n-1:max_k])
    result = (n**(1/3)) * sum_terms
    results.append(result)

return results


Paramètres
N = 200
d = 5
Calculer les résultats
results = sequence(N, d)
Créer le graphique
n_values = np.arange(1, N+1, d)
plt.plot(n_values, results, label=r'$\sqrt[3]{n} \cdot \sum_{k=n}^{n^3} \frac{\sin(k)^k}{k}$')
plt.xlabel('n')
plt.ylabel('Valeur de la suite')
plt.title('Graphique de la suite $\sqrt[3]{n} \cdot \sum_{k=n}^{n^3} \frac{\sin(k)^k}{k}$')
plt.grid(True)
plt.legend()
plt.show()

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Remark

Note that the Abel summation was necessary: if we directly use the technique used in the second part, we can't get the absolute convergence, so although I have a better convergence speed, my technique does not ensure the absolute convergence of the sequence.

Indeed you get:

$$\begin{align*} \sum_{n=1}^{N} \frac{|\sin(n)|^n}{n} &= \sum_{p=0}^P \sum_{k=1}^{p} \frac{|\sin(p(p+1)/2 + k)|^{p(p+1)/2 + k}}{p(p+1)/2 + k}\\ &= \sum_{p=0}^P \sum_{k\in {}^\epsilon \!A_{p(p+1)/2}^{(p+1)(p+2)/2}} \frac{|\sin(p(p+1)/2 + k)|^{p(p+1)/2 + k}}{p(p+1)/2 + k}\\ &+ \sum_{p=0}^P \sum_{k\in {}^\epsilon \!B_{p(p+1)/2}^{(p+1)(p+2)/2}} \frac{|\sin(p(p+1)/2 + k)|^{p(p+1)/2 + k}}{p(p+1)/2 + k}\\ &\le \sum_{p=0}^P \sum_{k\in {}^\epsilon \!A_{p(p+1)/2}^{(p+1)(p+2)/2}} \frac{(1-\epsilon)^{p(p+1)/2 + k}}{p(p+1)/2 +k} + \sum_{p=0}^P \sum_{k\in {}^\epsilon \!B_{p(p+1)/2}^{(p+1)(p+2)/2}} \frac{1}{p(p+1)/2+k} \\ &\le \sum_{p=0}^P \sum_{k\in {}^\epsilon \!A_{p(p+1)/2}^{(p+1)(p+2)/2}} \frac{(1-\epsilon)^{p(p+1)/2 + k}}{p(p+1)/2 +k} + \sum_{p=1}^P \sum_{k\in {}^\epsilon \!B_{p(p+1)/2}^{(p+1)(p+2)/2}} \frac{1}{p(p+1)/2} + \sum_{k\in {}^\epsilon \!B_{0}^{1}} \frac{1}{k}\\ &\le \sum_{n=1}^N \frac{(1-\epsilon)^{n}}{n} + \sum_{p=1}^P \sharp {}^\epsilon \! B_{p(p+1)/2}^{(p+1)(p+2)/2} \frac{1}{p(p+1)/2} +1\\ &\le \sum_{n=1}^N \frac{(1-\epsilon)^{n}}{n} + \sum_{p=1}^P \left( \frac{2p\sqrt{2\epsilon}}{\pi p(p+1)/2} + \frac{C\sqrt{\epsilon}}{p(p+1)/2} \right) +1\\ &\le \sum_{n=1}^N \frac{(1-\epsilon)^{n}}{n} + \frac{4\sqrt{\epsilon}}{\pi} \sum_{p=1}^P \frac{1}{p+1} + 2C \sqrt \epsilon \sum_{p=1}^P \left(\frac{1}{p}- \frac{1}{p+1}\right) +1\\ &\le \sum_{n=1}^N \frac{(1-\epsilon)^{n}}{n} + \frac{4\sqrt{\epsilon}}{\pi}\ln(P+1) +2 C \sqrt \epsilon \left(1 - \frac{1}{P+1} \right) +1\\ &\leq \sum_{n=1}^N \frac{(1-\epsilon)^{n}}{n} + \frac{2\sqrt{\epsilon}}{\pi}\ln(2N) +2 C \sqrt \epsilon +1 \end{align*}$$

But if you set $\epsilon = f(N)$, you need to ensure $\sqrt{\epsilon} = O(\ln(2N)^{-1})$ so at least, $\epsilon = O(\ln(2N)^{-2})$, but in this case $\sum_{n=1}^N \frac{(1-f(N))^{n}}{n} =-\ln (O(\ln(2N)^{-2})) = \Omega(\ln(\ln(N)))$ which diverges.