Is there a monotonically increasing sequence that is bounded? I just read the following theorem: 
If a sequence of real numbers is increasing and bounded above, then its supremum is the limit.
How is it possible to have an increasing but bounded above sequence? Can you give me an example please? 
 A: $\{ a_n \}=\left(1-\frac{1}{n}\right)$ is one such sequence. You can easily verify that it's monotone increasing and it is bounded above by $1$. It converges at $1$, which is indeed its supremum.
A: Yes - and in fact you've seen lots of them! 
For example:
3.
3.1.
3.14.
3.141.
.
.
.
This sequence is implicit anytime you talk about the decimal expansion of $\pi$.

In fact, given any increasing bounded sequence $S$, there is a unique real number $\alpha$ which is the supremum of $S$ (that is, $\alpha$ is greater than every element of $S$, and is the least such real number). This $\alpha$ is to $S$ exactly as $\pi$ is to the sequence above. And, in fact, considerations like these are how to rigorously define the real numbers.
A: Probably the simplest one is $$a_n=\frac n{1+n}$$
A: You can think of it as a 'controlled' increase, ala $\tan^{-1}n$ which goes to $\pi /2$ in the long term. This is bounded increase, as opposed to unbounded increase like $a_n = n$. 
I'm being informal with the terms to give you more intuition. 
