Proving that a given bounded sequence converges Let $(a_n)$ be a bounded real sequence. If $a_n \leq a_{n+2}$ and $a_n \leq a_{n+3}$ for all $n\in \mathbb{N}$ prove that $(a_n)$ converges. I think I should start by stating that there are at least two subsequences of $(a_n)$ that converge to $sup\{a_n\}$ but I'm not sure which direction to go from here. Thank you for your help.
 A: A bounded sequence need not necessarily converge. I proceed to provide a counter example:
Define the sequence $\left(a_n\right)$, s.t. $a_n = 1$ for $n = 2k+1$, $a_n = -1$ for $n = 2k$, for all $k \in \mathbb{N}$. The sequence is bounded by $M = 1$ i.e. $\left| a_n \right| \leq M, \forall n \in \mathbb{N}$. Moreover the sequence does not converge since $\left(a_{2k+1}\right)$ is a subsequence converging to $1$ while $\left(a_{2k}\right)$ is a subsequence converging to $-1$.
Take note however of the Bolzano-Weierstrass Theorem, which states that every real bounded sequence has a convergent subsequence. 
A: I think it should actually converge to it's infimum. If you observe the sequences
$\{ a_{2k} \}_{k=1}^\infty$, $\{ a_{1+2k} \}_{k=1}^\infty$,  and $\{ a_{3k} \}_{k=1}^\infty$, by the assumption they are both monotonically decreasing, and therefore converge to their infimums which will be denoted $L_1$,$L_2$ and $L_3$ accordingly. However $\{ a_{2k} \}_{k=1}^\infty$  and $\{ a_{3k} \}_{k=1}^\infty$  both have the subsequence $\{ a_{6k} \}_{k=1}^\infty$ which must converge to the limit of the original sequence. This gives you that:
$L_1=L_3$
Similarly $\{ a_{1+2k} \}_{k=1}^\infty$  and $\{ a_{3k} \}_{k=1}^\infty$ have a subsequence $\{ a_{3+6k} \}_{k=1}^\infty$, which gives us that:
$L_2=L_3$
Denote $L:=L_1$. Then for all $\epsilon>0$ there exists $N_\epsilon$ such that:
$\vert a_{2k}-L\vert< \epsilon \quad$ and $\quad \vert a_{2k+1}-L\vert< \epsilon \quad$  if $\quad k>N_\epsilon$. 
And since every $n \in \mathbb{N}$ can be written as either $2k$ or $2k+1$, this shows you the sequence converges.
