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.
 A: 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 $Q_+$.
A: 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 $n\to\infty$.
