# 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?

• For the harmonic series, use the condensation test. – Hagen von Eitzen Jul 21 '13 at 14:40
• But why, it isn't specified as a short coming of the comparison test. – Paul the Pirate Jul 21 '13 at 14:43

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.

• I thought the $a_n$ and $b_n$ were referring to the part being squared so I was just replacing the function part of it. Like that weird function replacement stuff. I don't know what "Cauchy" means. – Paul the Pirate Jul 21 '13 at 14:46
• See this for the test of whether the partials sums are a Cauchy sequence. In particular, if the series were to converges, then $\sum_n^{2n} a_n$ should go to $0$ (taking $p=n$); for the harmonic series, you have $$\sum_n^{2n} \frac{1}{n} \geq n\cdot \frac{1}{2n} = \frac12$$ for all $n$. – Clement C. Jul 21 '13 at 15:00

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."

• So how is this limited? If I have $\sum \frac{1}{47^n - 44^n$ can I compare it to the larger function $\frac{1}{2^n}$ and state it converges? – Paul the Pirate Jul 29 '13 at 19:54
• This one is somewhat harder. Maybe I would do this. The denominator is $44^n((47/44)^n -1)$. So the denominator is $\ge 44^n (3/44)$. Now by comparison with a geometric series, we have convergence. Probably easier for most students is the limit comparison test, using $\sum \frac{1}{47^n}$. – André Nicolas Jul 29 '13 at 20:05
• I suppose what I am getting at is why can't I just do what I did above? What specifically is wrong with that claim. – Paul the Pirate Jul 29 '13 at 23:09
• It is not enough to have a good intuition (though that is enormously useful). One also needs to be able to give detailed verifications. – André Nicolas Jul 29 '13 at 23:25
• I guess the idea of "good" is kind of fuzzy. How good is good enough is what I don't get. – Paul the Pirate Jul 30 '13 at 1:13