If a subsequence of $(a_n)$ converges to a limit $L$, prove that $(a_n)$, which is non-decreasing, also converges to $L$ I know that if a sequence converges to $L$, its subsequences also converge to $L$. However, I'm not sure how to show the converse in this case, given $a_n$ is monotonic.
If I show somehow that $a_n$ is bounded, can I say that by the monotone convergence theorem $a_n$ converges to L because its subsequence does? If so, how would I begin in proving $a_n$ is bounded? Thanks for any advice
 A: This can proved by contradiction.  Suppose $(a_{n})$ is a real, non-decreasing sequence which has a subsequence converging to limit $L$.  We want to prove $a_{n}$ converges to $L$.
Suppose by contradiction that $a_{n}$ does not converge to $L$.  That means $\exists \epsilon > 0$ such that $\forall N \in \Bbb N$, $\exists m \geq N$ with $a_{m} \not \in (L - \epsilon, L + \epsilon)$.  Denote the subsequence that converges to $L$ by $(a_{n_{k}})$.  We know since $a_{n_{k}} \to L$, there is $K \in \Bbb N$ such that $k \geq K$ implies $a_{n_{k}} \in (L - \epsilon, L + \epsilon)$.  
Since the sequence in non-decreasing, and $a_{n_{k}} \in (L - \epsilon, L + \epsilon)$, and $(a_{n})$ doesn't converge to $L$ by assumption, then there is some $m'$ such that $a_{m'} \geq L + \epsilon$.  But we have $\forall k \geq K$, $a_{n_{k}} \in (L - \epsilon, L + \epsilon)$, and for some $k$, $n_{k} > m'$ )(which implies $a_{m'} \leq a_{n_{k}}$ since $(a_{n})$ is non-decreasing).  But then $a_{n_{k}} < L + \epsilon \leq a_{m'}$, which implies $a_{n_{k}} < a_{m'}$ even though $m' < n_{k}$, which contradicts the fact that the sequence is non-decreasing.  
So, the sequence $(a_{n})$ must converge to $L$.
