TITLE: limit of a sequence $\ln n\cdot(\sin n)^2$
QUESTION [6 upvotes]: It's easy to see that $\lim\limits_{x\to \infty}\ln x\cdot(\sin x)^2$ does not exist (sin can take 0 and 1 as values, so liminf is $0$, limsup is $\infty$).
how about the limit of a sequence:
$\lim\limits_{n\to \infty}\ln n\cdot(\sin n)^2$ ?

REPLY [5 votes]: If $\frac mn$ is a good approximation for $\frac1\pi$, that is $$\left|\frac mn-\frac1\pi\right|<\frac 1{n^2}$$
(and infinitely many such approximations do exist!),
then $n$ differs from $m\pi$ by less than $\frac \pi n$, which makes $\sin^2n<\frac{\pi^2}{n^2}$. Then for a sequence of such $n$, $\ln n\sin^2n\to 0$. On the other hand, $|\sin(n+1)|\approx \sin 1$  for such $n$, so that $\ln n\sin^2n$ is not bounded from above. Hence again $\liminf = 0$, $\limsup=\infty$.