How to use a convergent subsequence to prove the full increasing sequence converges If we let $(a_n)_{n\in N_+}$ be an increasing sequence which contains a subsequence that converges to $\ell \in R$. How can I prove that the full subsequence always converges to $\ell$?
It is pretty clear to me, as if the subsequence converges then surely the sequence is bounded above by something, but I am not sure how best to construct my argument.
Thanks!
 A: Use proof by contradiction:
(1) Assume that $(a_n)$ does not converge to $\ell$. Then, there is $\varepsilon_0>0$ satisfying the following condition: for each $m\in\mathbb{N}$ there is $M\in\mathbb N$ such that
$$M>m\quad\text{and}\quad |a_{M}-\ell|\geq \varepsilon_0.$$
(2) Let $(a_{n_k})$ be the convergent subsequence. Then, there is $k_0\in\mathbb N$ such that
$$n_k\geq n_{k_0}\quad\Longrightarrow\quad |a_{n_k}-\ell|<\varepsilon_0.$$
(3) From (1) there is $M_0\in\mathbb N$ such that $M_0>n_{k_0}$ and
$$a_{M_0}\notin(\ell-\varepsilon_0,\ell+\varepsilon_0).\tag{A}$$
Take $k_1\in\mathbb N$ such that $a_{n_{k_1}}>M_0$. From (2), 
$$ a_{n_{k_0}},a_{n_{k_1}}\in(\ell-\varepsilon_0,\ell+\varepsilon_0).\tag{B}$$
As $(a_n)$ is increasing, it follows from $(A)$ and $(B)$ that
$$\ell+\varepsilon\leq a_{M_0}\leq a_{n_{k_1}}<\ell+\varepsilon_0$$
which is a contradiction.
A: If the subsequence is $(a_{n_k})$, then for $n_k \le n \le n_{k+1}$ we have 
$a_{n_k} \le a_n \le a_{n_{k+1}}$.   Given $\epsilon > 0$, take $k$ large enough that $a_{n_k}$ and $a_{n_{k+1}}$ are within $\epsilon$ of $\ell$.
