I'm given the succession $$a_n= [1+\sin n]$$ and I should find the superior limit $ \limsup_{n \to \infty}a_n$ and the inferior limit $ \liminf_{n \to \infty}a_n$
$-1<\sin n<1$ and then $0< [1+\sin n]<2$
In my opinion it should be $ \limsup_{n \to \infty}a_n=2$
$ \liminf_{n \to \infty}a_n=0$
but in the book the suggested solution is 1 for the superior limit (the inferior is =0). Where am I making mistakes?
I guess we are supposed to interpret $[t]$ as $\lfloor t\rfloor$, i.e., as the largest integer $\leq t$.
Given that all real numbers $x$ where $\sin x=1$, i.e. the numbers $\bigl(2j+{1\over2}\bigr)\pi$ with integer $j$, are irrational, it follows that all values $\sin n$ $\>(n\in{\mathbb N})$ are in fact $<1$. Therefore $\lfloor 1+\sin n\rfloor\leq1$ for natural $n$, and this implies $\limsup_{n\to\infty}\lfloor1+\sin n\rfloor\leq1$.
On the other hand $\sin x>0$ in infinitely many disjoint intervals of length $>3$. This implies that there are infinitely many $n$ with $\sin n>0$, hence $\lfloor1+\sin n\rfloor\geq1$, so that $\limsup_{n\to\infty}\lfloor1+\sin n\rfloor\geq1$.
These two paragraphs show that $\limsup_{n\to\infty}\lfloor1+\sin n\rfloor=1$.