# Is the limit of non-decreasing sequence that is bounded equal to the upper bound?

I'm playing with one of theorem in my notes:

Theorem

If a convergent sequence $\displaystyle\lim_{n\to\infty} = L$ that is bounded by $M$, then $|L| \leq M$

I wonder is there a connection between $L$ and $M$ if the sequence is non-decreasing? In other words, is it always true that $L = M$ for non-decreasing sequence that is bounded? I can't find a counterexampe, as well as a convincing arugment to connect $L$ and $M$. A hint would be greatly appreciated.

-
$M$ can be of course larger than $L$. Think about $a_n=1-1/n$ and $M=2$ –  ziyuang Oct 1 '12 at 0:00

The least upper bound is equal to the limit. To show this, let $\varepsilon > 0$. By the definition of the least upper bound, there is $N \in \mathbb{N}$ for which $M - \varepsilon < a_N$. Since the sequence is non-decreasing, we have:
$$\forall n > N : M - \varepsilon < a_N \le a_n$$
Given that $a_n \le M$, Rearrange to get the definition of the limit:
$$\forall n > N : |a_n - M| < \varepsilon$$