proving divergence I understand how to prove if a sequence is convergent, however I am having trouble proving whether or not a sequence is divergent.  I was wondering what the general method or proof would be to determine convergence.  For example obviously the sequence {n} is divergent, but how would you formally prove this?
 A: If $a_n = n$ is convergent, then there is some limit $L$ to which it converges. This means that given $\epsilon > 0$, we can find a positive integer $N$ such that $|a_n - L| < \epsilon$ for all $n > N$. Plugging in $a_n = n$, the inequality becomes
$$|n - L| < \epsilon$$
which is true if and only if
$$-\epsilon < n-L < \epsilon$$
if and only if
$$-\epsilon + L < n < \epsilon + L$$
There is no way for this inequality to be satisfied for all sufficiently large $n$, because eventually $n$ will exceed $\epsilon + L$.
We conclude that $a_n$ cannot be convergent, hence it is divergent.
A: A sequence is divergent, if it is not convergent. 
This might be because the sequence tends to infinity or it has more than one limit point.
Your example $x_n = n$ is of the first type:
$$
\lim_{n \to \infty} x_n = \lim_{n \to \infty} n = \infty
$$
You prove it by showing that for any number $K$ you can response with some index $N$ such that from that index on, the sequence surpasses the challenge. (See here).
Let $K \in \mathbb{R}$ then choose $N = \max(0, \lceil K \rceil) \in \mathbb{N}$ then for any $n$ with $n \ge N$ we have 
$$
 n \ge \max(0, \lceil K \rceil) \ge K 
\Rightarrow \\
x_n \ge K
$$
The maximum is added to deal with negative $K$ which might otherwise lead to invalid index values, e.g. if your sequence elements are indexed by natural numbers including $0$.
A: A general criterion is the following: A sequence $(a_n)_{n\geq0}$ is divergent iff it is not a Cauchy sequence, that is to say: Iff there are an $\epsilon_0>0$ and  arbitrarily large $n$, $m$ satisfying $|a_n-a_m|\geq\epsilon_0$.
An important case is if you can spot two different accumulation points $\alpha\ne\beta$.
