Finding supremum, infimum, limsup and liminf of the sequence $a_{2n}=(-1)^n+\frac1{2n}$, $a_{2n+1}=2(-1)^n-\frac1{2n+1}$ $$\begin{align*}
a_{2n}&=(-1)^n+\frac1{2n}\\
a_{2n+1}&=2(-1)^n-\frac1{2n+1}
\end{align*}$$
The supremum is $2$ (when $\frac1{2n+1}$ in $a_{2n+1}$ tends to $0$).
The infimum is when $n=1$, $-2-\frac13=-\frac73$.
As $m\to\infty$, the supremum over $n\ge m$ is $2-\frac1{4k+1}$, where $m=2k$ or $m+1=2k$; this tends to $2$, so the limit superior is $2$.
As $m\to\infty$, the infimum over $n\ge m$ is $-2-\frac1{4k+3}$, where $m=2k+1$ or $m+1=2k+1$; this tends to $-2$, so the limit inferior is $-2$.
Could someone give me an alternative solution to this exercise? Or, will exercise could improve to make it formally?
 A: You might begin by breaking the sequence down into four subsequence:
$$\begin{align*}
a_{4k}&=1+\frac1{4k}\tag{1}\\
a_{4k+1}&=2-\frac1{4k+1}\tag{2}\\
a_{4k+2}&=-1+\frac1{4k+2}\tag{3}\\
a_{4k+3}&=-2-\frac1{4k+3}\tag{4}
\end{align*}$$
It’s very clear how each of these is behaving: 


*

*$(1)$ is monotonically decreasing from a maximum of $\frac54$ to a limit of $1$;  

*$(2)$ is monotonically increasing from a minimum of $1$ to a limit of $2$;  

*$(3)$ is monotonically decreasing from a maximum of $-\frac12$ to a limit of $-1$; and  

*$(4)$ is monotonically increasing from a minimum of $-\frac73$ to a limit of $-2$.


From this information alone you can already see that the sequence attains its minimum at $a_3=-\frac73$, which must therefore by its infimum. You can also see that $a_n<2$ for all $n$, while $\lim\limits_{k\to\infty}a_{4k+1}=2$, so $2$ must be the supremum of the sequence.
How you proceed from there depends on what you know about limits superior and inferior. It’s clear from $(1)-(4)$ that if a subsequence of $\langle a_n:n\in\Bbb N\rangle$ converges at all, it must converge to $1,2,-1$, or $-2$. 

If the subsequence has infinitely many terms in two or more of the subsequences $(1)-(4)$, it does not converge. If it has infinitely many terms in only one of them, it converges to the same limit as that subsequence.

If you know that $\liminf_na_n$ is the smallest $x$ such that $\langle a_n:n\in\Bbb N\rangle$ has a subsequence converging to $x$, then you can immediately conclude that $\liminf_na_n=-2$, the smallest of the limits $-2,-1,1$, and $1$. Similarly, if you know that $\limsup_na_n$ is the largest $x$ such that $\langle a_n:n\in\Bbb N\rangle$ has a subsequence converging to $x$, then you can immediately conclude that $\liminf_na_n=2$, the largest of the four subsequential limits.
If you have to work directly from the definitions of $\liminf$ and $\limsup$, you can’t do much better than what you actually did, though you might find it easier to state in terms of the four-way split into subsequences.
