Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero? If $a_n$ is a sequence such that
  $$a_1 \leq a_2 \leq a_3 \leq \dotsb$$
  and has the property that $a_{n+1}-a_n \to 0$, then can we conclude that $a_n$ is convergent?


I know that without the condition that the sequence is increasing, this is not true, as we could consider the sequence given in this answer to a similar question that does not require the sequence to be increasing.

$$0, 1, \frac12, 0, \frac13, \frac23, 1, \frac34, \frac12, \frac14, 0, \frac15, \frac25, \frac35, \frac45, 1, \dotsc$$

This oscillates between $0$ and $1$, while the difference of consecutive terms approaches $0$ since the difference is always of the form $\pm\frac1m$ and $m$ increases the further we go in this sequence.  

So how can we use the condition that $a_n$ is increasing to show that $a_n$ must converge?  Or is this still not sufficient?
An easy way to visualize why this can't be true is to try putting some points on a number line.

Start with 1 point in [0, 1):



2 points in [1, 2):



And so on:



Now you have a sequence that grows to infinity but keeps getting closer together.
The condition $a_{n+1}-a_n \to 0$ is not sufficient, as José Carlos Santos pointed out. But, a necessary and sufficient condition, that doesn't require the series to be increasing, is that $\lim\limits_{n\to\infty}(a_{n+m(n)}-a_n)=0$ for all $m(n)\in \mathbb{N}$, where $m$ is a function of $n$. Sequences which satisfy this property are called Cauchy sequences.

Also, if you show that a sequence is monotonically increasing and bounded from above, then it converges. The same applies for monotonically decreasing sequences which are bounded from below.