# Convergence of $(\sin x)^x$

This showed up on a forum and had no clear answer so I'm asking this here to see if anyone can give some light.

Define the sequence $$a_n := (\sin n)^n \ \forall n \in \mathbb{N}$$

How do you proof the existence of $$\lim_{n\to \infty} a_n$$? Thanks.

• I'd guess that there is a constant $q<1$ such that $|\sin n|<q$ for all $n$, since sine is equal to $\pm 1$ on multiples of $\pi$. Then it follows by squeeze theorem. – I was suspended for talking Jun 14 at 10:34
• @Sisyphus There is none because $\sin(\mathbb{N})$ is dense in $[-1,1]$. – Spenser Jun 14 at 10:42
• Maybe you can find some help here – Tomás Jun 14 at 11:01
• Your title is misleading. $(\sin x)^x$ does not converge. – Yves Daoust Jun 14 at 12:11
• I believe my answer is complete now. Would you check it out please? – K. Sadri Jun 14 at 12:30

My answer: It does not converge to anything! It must be clear that this does not converge to any non-zero $$L$$. So the question is: "Does it converge to zero." Let the proposition $$P$$, be the "converges to zero!" statement. Then $$P^c="\exists\varepsilon>0 \textrm{ s.t. } \#\{n\in\mathbb{N}\Big||\sin n|>\varepsilon^\frac{1}{n}\}=\infty"$$ But using the approximation $$\varepsilon^\frac{1}{n}\approx\frac{-\log\varepsilon}{n}$$ This becomes $$P^c="\exists a<\infty \textrm{ s.t. } \#\{n\in\mathbb{N}\Big||\sin n|>1-\frac{a}{n}\}=\infty"$$ Then it is easy to use the properties of the sine function to see that $$P^c=Q_-\cup Q_+$$ With $$Q_\pm="\#\Big\{n\in\mathbb{N}\Big||\{\frac{n}{2\pi}\mp\frac{1}{4}\}|<\frac{a}{\sqrt n}\Big\}=\infty"$$ From now on we focus only on the proposition $$Q_+$$ and try to prove it. Let $$A$$ be the conjecture

$$A$$: For any sequence $$0\leq e_n\leq 1$$ with infinite sum $$\sum_0^\infty e_n$$, any irrational number $$\alpha\in(0,1)$$ and any sequence $$t_n\in[0, 1]$$ for which the limit $$\lim_{n\rightarrow\infty}t_n$$ exists, the following holds

$$\#\Big\{(n, \{n\alpha\})\Big|n\in\mathbb{N}, |\{n\alpha\}-t_n|\leq e_n\Big\}=\infty$$

Then with $$e_n=\min(1, \frac{a}{\sqrt n})$$, $$\alpha=\frac{1}{2\pi}$$, $$t_n=.25$$ we get $$A\Rightarrow Q_+\Rightarrow P^c$$ And

I believe $$A$$ holds. But to solve the recent problem it is easier to use Dirichlet's Approximation Theorem. It states that that for every irrational $$\alpha\in(0,1)$$ there is an infinite sequence $$n_k$$ for which $$\{n_k\alpha\}\leq\frac{1}{n_k}$$. Now consider the infinite sequence $$m_k:=n_k\Big\lfloor\frac{1}{4\{n_k\alpha\}}\Big\rfloor$$ For this we have $$m_k\{n_k\alpha\}^2=n_k\{n_k\alpha\}^2\Big\lfloor\frac{1}{4\{n_k\alpha\}}\Big\rfloor\leq.25n_k\{n_k\alpha\}\leq.25$$ Or equivalently $$\{n_k\alpha\}\leq\frac{.5}{\sqrt{m_k}}$$ This guarantees $$|\{m_k\alpha\}-.25|\leq\{n_k\alpha\}\leq\frac{.5}{\sqrt{m_k}}$$ Which proves $$Q_+$$.

• Thanks a lot for your effort, this is what I was searching for! – Null Jun 14 at 14:17
• Good, as good as your comment to my hasty and wrong answer. – Piquito Jun 14 at 14:30

This is a long comment, not a full answer!

Note that: $$(\sin n)^n=\left(-i\sinh(-in)\right)^n=\frac{(-i)^n}{2^n}\left(e^{-in}-e^{in}\right)^n.$$ By the binomial expansion, we get:

\begin{align*} s_n=(e^{-in}-e^{in})^n=\sum_{k=0}^n\binom{n}{k}e^{-ikn}(-1)^{n-k}e^{i(n-k)n}=\sum_{k=0}^n\binom{n}{k}(-1)^{n-k}e^{in(n-2k)}. \end{align*}

Consider the following cases:

• If $$n$$ is even, then, we have: $$s_n=\sum_{k=0}^{n/2}\binom{n}{k}(-1)^{k}e^{in(n-2k)}+\sum_{k=n/2+1}^n\binom{n}{k}(-1)^{k}e^{in(n-2k)}.$$ Let $$l=n-k$$ in the second sum: $$\sum_{k=n/2+1}^n\binom{n}{k}(-1)^{k}e^{in(n-2k)}=\sum_{l=0}^{n/2-1}\binom{n}{n-l}(-1)^{n-l}e^{in(2l-n)}=\sum_{l=0}^{n/2-1}\binom{n}{l}(-1)^{l}e^{in(2l-n)}.$$ So, we get: \begin{align*}s_n&=\sum_{k=0}^{n/2}\binom{n}{k}(-1)^{k}e^{in(n-2k)}+\sum_{l=0}^{n/2-1}\binom{n}{l}(-1)^{l}e^{in(2l-n)}=\\ &=\sum_{k=0}^{n/2}\binom{n}{k}(-1)^{k}e^{in(n-2k)}+\sum_{l=0}^{n/2}\binom{n}{l}(-1)^{l}e^{in(2l-n)}-\binom{n}{n/2}(-1)^{n/2}=\\ &=\sum_{k=0}^{n/2}\binom{n}{k}(-1)^{k}\left(e^{in(n-2k)}+e^{in(2k-n)}\right)-\binom{n}{n/2}(-1)^{n/2}=\\ &=2\sum_{k=0}^{n/2}\binom{n}{k}(-1)^{k}\frac{e^{in(n-2k)}+e^{-in(n-2k)}}{2}-\binom{n}{n/2}(-1)^{n/2}=\\ &=2\sum_{k=0}^{n/2}\binom{n}{k}(-1)^{k}\cosh(in(n-2k))-\binom{n}{n/2}(-1)^{n/2}=\\ &=2\sum_{k=0}^{n/2}\binom{n}{k}(-1)^{k}\cos(n(n-2k))-\binom{n}{n/2}(-1)^{n/2}. \end{align*}
• If $$n$$ is odd, then, we have: $$s_n=\sum_{k=0}^{(n-1)/2}\binom{n}{k}(-1)^{k}e^{in(n-2k)}+\sum_{k=(n+1)/2}^n\binom{n}{k}(-1)^{k}e^{in(n-2k)}.$$ Let $$l=n-k$$ in the second sum: $$\sum_{k=n/2+1}^n\binom{n}{k}(-1)^{k}e^{in(n-2k)}=\sum_{l=0}^{(n-1)/2}\binom{n}{n-l}(-1)^{n-l}e^{in(2l-n)}=\sum_{l=0}^{(n-1)/2}\binom{n}{l}(-1)^{l}e^{in(2l-n)}.$$ So, we get: \begin{align*}s_n&=\sum_{k=0}^{(n-1)/2}\binom{n}{k}(-1)^{k}e^{in(n-2k)}+\sum_{l=0}^{(n-1)/2}\binom{n}{l}(-1)^{l}e^{in(2l-n)}=\\ &=\sum_{k=0}^{(n-1)/2}\binom{n}{k}(-1)^{k}\left(e^{in(n-2k)}+e^{in(2k-n)}\right)=\\ &=2\sum_{k=0}^{(n-1)/2}\binom{n}{k}(-1)^{k}\frac{e^{in(n-2k)}+e^{-in(n-2k)}}{2}=\\ &=2\sum_{k=0}^{(n-1)/2}\binom{n}{k}(-1)^{k}\cosh(in(n-2k))=\\ &=2\sum_{k=0}^{(n-1)/2}\binom{n}{k}(-1)^{k}\cos(n(n-2k)). \end{align*}

It remains to study this as $$n\to\infty$$.

• This looks much much harder to do :)) – K. Sadri Jun 14 at 13:36
• Good insight thought, I will take it a carefully look later. Thanks both. – Null Jun 14 at 14:19
• I think this could somehow be transformed into a quasi Legendre polynomial expression, however, I cannot see it right now. – Βασίλης Μάρκος Jun 14 at 16:22