Limits of the comparison test http://en.wikipedia.org/wiki/Direct_comparison_test
I don't understand how this works, I understand the idea behind it. Similar to the squeeze theorem it is pretty logical and easy to visually see. 
So then what is happening when I want to evaluate
$$\sum_{n \geq 1} \frac{1}{n}.$$
So my $b_n$ is then $\frac{1}{n^2}$
My $\{b_n\}$ converges but my $\{a_n\}$ doesn't which I already know because it is drilled into you. I guesss I could prove it with a $p$-series test or something like that, but why does the comparison test fail here and where else does it fail that isn't in specified?
 A: The comparison test does not fail here as the hypothesis is not satisfied. Note that
$$ \frac{1}{n} > \frac{1}{n^{2}}$$ and not the other way round. Hence, the convergence of the series
$$\sum\limits_{n=1}^{\infty} \frac{1}{n^{2}}$$
gives no information about the convergence/divergence of $\sum\frac{1}{n}$.
By the way, to prove that $\sum\frac{1}{n}$, you can show that the sequence of partial sums is not Cauchy.
EDIT: The comparison test says that if $(a_{n})$ and $(b_{n})$ are two sequences such that $a_{n} \leq b_{n}$ and that $\sum b_{n}$ converges, then $\sum a_{n}$ converges.
A: What is written below is intended to help a bit with the intuition. All series discussed below have non-negative terms. 
Call a series $\sum x_n$ bad if it diverges, and good if it converges.
Comparison Test 1 says that if $\sum b_n$ is good, and $0\le a_n\le b_n$ for all $n$, then $\sum a_n$ is good.  
In cruder terms, if $\sum b_n$ doesn't blow up, and the $a_n$ are smaller, then $\sum a_n$ doesn't blow up.
Now suppose $\sum a_n$ is bad, and $a_n\le b_n$ for all $n$. Then $\sum b_n$ cannot be good. For if it were, then by Comparison Test 1, $\sum a_n$ would be good. But it isn't. So we have obtained:
Comparison Test 2: If $\sum a_n$ is bad, and $0\le a_n\le b_n$ for all $n$, then $\sum b_n$ is bad. If $\sum a_n$ 
blows up, and the $b_n$ are bigger than the $a_n$, then $\sum b_n$ blows up.
There are useful variants of the comparison tests. For one thing, what happens for "small" $n$ doesn't matter. If $a_n\le b_n$ for large enough $n$, and $\sum b_n$ is good, then $\sum a_n$ is good. Also, if $\sum b_n$ is good, and there is a positive constant $c$ such that $0\le a_n\le cb_n$, then $\sum a_n$ is good. 
In order to apply the Comparison Tests successfully, we need to have a basic stock of series whose convergence/divergence we know about.  
That basic stock need not be large. In practice, most comparisons are done with geometric series, or with $p$-series $\sum\frac{1}{n^p}$.
We do have to be careful about the direction of the inequalities. Suppose we know already that $\sum \frac{1}{n^2}$ is good. That doesn't  help at all with $\sum \frac{1}{n}$. For we have $\frac{1}{n^2}\le \frac{1}{n}$. So the terms of $\sum \frac{1}{n}$ are bigger than the terms of a good series. Not useful: A series bigger than a good series can easily blow up.
Very informally, a series of positive terms converges (is good) if the terms approach $0$ "fast enough."  It turns out that $\frac{1}{n^2}$ approaches $0$ fast enough, but $\frac{1}{n}$ does not approach $0$ fast enough. 
Interestingly, $\frac{1}{n}$ approaches $0$ almost fast enough. For example, $\frac{1}{n^{1.01}}$, which approaches $0$ only a little faster than $\frac{1}{n}$, is "fast enough."
