I'm having trouble showing rigorously what is the limit of $x_n=|\sin(n)|^{1/n}$ in a rigorous manner. What I have shown is that, $x_n$ cannot converge to $0$ and is bounded by $1$, and that should suffice to show that $x_n$ effectively converges to $1$.

However, I can't figure out how to formalize this proof, and show it in a rigorous manner. My guess would be to try and show that the limit of $|a_n|^{1/n}$ can be $1$ if $|a_n|$ is bounded by $1$ and does not converge to $0$. I don't know if this more general statement holds, and if it would simplify or complexify the problem.

  • $\begingroup$ Let $a_n = (1/2)^n$. $\endgroup$ – Giovanni De Gaetano Feb 14 '16 at 17:46
  • $\begingroup$ Mmh. Right, I should have added the condition that $|a_n|$ does not converge to 0. $\endgroup$ – Frotaur Feb 14 '16 at 17:49
  • 1
    $\begingroup$ One of the subsequential limits of $|\sin n|^{1/n}$ is $1$. I do not see any obvious ways to rule out others. $\endgroup$ – André Nicolas Feb 14 '16 at 18:02
  • $\begingroup$ Your idea to formalize the proof doesn't work. A counterexample in this case would be $a_n = |\sin(\pi n)|$, which doesn't have limit. The point is that the function $|\sin(x)|^{1/x}$ from $\mathbb{R}$ to $\mathbb{R}$ doesn't admit a limit for $x \rightarrow \infty$. The case $|\sin(n)|^{1/n}$ for $n \in \mathbb{N}$ doesn't seem easy to me, for the reason already given by André Nicolas. $\endgroup$ – Giovanni De Gaetano Feb 14 '16 at 18:04
  • $\begingroup$ You are indeed right, we have also the fact that $sin(n)$ is never $0$. I agree, I should abandon the more general method to concentrate on the actual limit. My understanding is that even though sin(n) will get arbitralily close to zero, it will happen so scarcely that the square root will always be "powerful enough" to bring up the value close to one. I don't know if I am clear. $\endgroup$ – Frotaur Feb 14 '16 at 20:16


the idea is to bound $\sin(n)$ for $n\in \mathbb{N}$ from below in such a way that you see that $\sin(n)$ is so far away from $0$ that $\left|\sin(n)\right|^\frac{1}{n}$ goes to $1$. Therefore we have to show that natural numbers have a certain distance to multiples of $\pi$.

For this, you can use the fact that $\pi$ is not a Liouville number (see http://mathworld.wolfram.com/LiouvilleNumber.html).

So, there is an $n_o\in\mathbb{N}$ such that $\left|\pi-\frac{p}{q}\right|\geq \frac{1}{q^{n_o}}$ for all $p,q\in \mathbb{N}$, or, equivalently $\left|q\pi-p\right|\geq \frac{1}{q^{n_o-1}}$.

Now choose $p=n$, and $q$ in a way that $q\pi$ is close to $n$, i.e. $q\in[\frac{n-\frac{\pi}{2}}{\pi}, \frac{n+\frac{\pi}{2}}{\pi}]$.

As now $q\leq \frac{n+\frac{\pi}{2}}{\pi}$ and $\left|q\pi-p\right|\leq\frac{\pi}{2}$, and as for $x\in[0,\frac{\pi}{2}]$ there is the estimate $\sin(x)\geq \frac{x}{2}$, we get the following series of inequalities: $$ |\sin(n)|=|\sin(q\pi-n)|=\sin|q\pi-n|\geq \frac{1}{2}|q\pi-n|\geq\frac{1}{2q^{n_o-1}}\geq \frac{1}{2}\cdot\left(\frac{\pi}{(n+\frac{\pi}{2})}\right)^{n_o-1}. $$ Taking the $n$-th root, we obtain $$ |\sin(n)|^{\frac{1}{n}}\geq \frac{1}{2^{\frac{1}{n}}}\cdot\left(\frac{\pi^{\frac{1}{n}}}{(n+\frac{\pi}{2})^{\frac{1}{n}}}\right)^{n_o-1}. $$ As the limit of $n^{\frac{1}{n}}$ for $n\rightarrow\infty$ is $1$ and $n_o$ is fixed, the right hand side goes to $1$ for $n\rightarrow\infty$. As the left hand side is bounded from above by $1$ aswell, it has to converge to $1$.


I am not quite sure that this sequence converges.

Let $ a_n = |\sin(n)|^{1/n}$. First I would like to give a reference to the following paper http://www.jstor.org/stable/2688681?seq=1#page_scan_tab_contents . This paper asserts that the set of limit points of $\sin(n)$ is equal to $[-1,1]$. It would be apparent from here that the set of limit points of the absolute value of this function would be equal to $[0,1]$.

Consider $b_n = \log a_n = \frac{\log(|\sin(n)|)}{n}$. If we first look at the subsequences of $b_{n_k}$ of $b_n$ for which $a_{n_k}$ has a limit point of $y\in (0,1]$, then $b_{n_k}\rightarrow 0$ which would mean that the corresponding $a_{n_k}$'s would converge to $1$.

On the other hand, when we consider the subsequences of $b_n$ for which $a_{n_k}\rightarrow 0$, then we would need to apply L'Hopital rule to figure out the limiting value of $b_{n_k}$.

Let us abstract ourselves from the problems with absolute value as they would only correspond to minor technical details, and let us consider $b_{n_k}$ without the absolute values and take derivatives with respect to $n_k$ of both numerator and the denominator. This would get us $$ b_{n_k}\rightarrow -\frac{\cos(n_k)}{\sin(n_k)}. $$ Since $\sin(n_k)$ converge to $0$, we would have $\cos(n_k)$ coverging to $1$. This would imply that $b_{n_k}\rightarrow -\infty$. This in turn allows us to conclude that $a_{n_k} = e^{b_{n_k}} \rightarrow 0$.

As a result, $a_n$ has two limit points $0$ and $1$, thus it is not convergent.

  • $\begingroup$ Indeed, I did assume that the limit was 1. However, I checked with wolframalpha, and it yielded that the limit is in fact 1 (although I don't know what it's worth). I have also some remarks on your considerations. Firstly, I don't know if it is allowed to consider the problem "without" absolute value since the logarithm of a negative number isn't defined. Secondly, I am not sure that de l'Hopital rule can be applied to compute limits of sequences. Usually, it is used to compute a limit in the broader sense, in which we do not restrict how "$n_k$" tends to infinity. $\endgroup$ – Frotaur Feb 14 '16 at 22:02

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.