limit comparison test for alternating series

Solution 1:

The limit comparison test works because if $a_n$ and $b_n$ are positive, then if the ratio $a_n/b_n$ fluctuates around its limit, these fluctuations affect $a_n$ and $b_n$ alternatingly and can't systematically change the convergence behaviour of one but not the other. That's not the case if $a_n$ and $b_n$ alternate. You can see this in the following counterexample: Let

$$\frac{a_n}{b_n}=1+\frac{(-1)^n}{n^p}$$

with $p>0$. Then

$$\lim_{n\to\infty}\frac{a_n}{b_n}=1\;,$$

and the limit comparison test would say that the series either both converge or both don't. But now consider

$$b_n=\frac{(-1)^n}{n^q}\;$$

with $q>0$. The corresponding series converges by the alternating series test. But then

$$ \begin{eqnarray} a_n&=&\frac{(-1)^n}{n^q}\left(1+\frac{(-1)^n}{n^p}\right)\\ &=&\frac{(-1)^n}{n^q}+\frac{1}{n^{p+q}}\;, \end{eqnarray} $$

and the corresponding series diverges for $p+q\le1$. You can see how the fluctuations around the limit of the ratio "interfere constructively" to let the series for $a_n$ diverge; this couldn't happen if $a_n$ and $b_n$ were both always positive.

Solution 2:

Usually, the Limit Comparison Test is stated as follows:

Limit Comparison Test. Let $\sum a_n$ and $\sum b_n$ be two series of positive terms. If $$\lim_{n\to\infty}\frac{a_n}{b_n}$$ exists and is positive, then both $\sum a_n$ and $\sum b_n$ converge, or both diverge.

In fact, it can be extended slightly to include the following two cases:

• If $\lim\frac{a_n}{b_n} = 0$ and $\sum b_n$ converges, then $\sum a_n$ converges.
• If $\lim\frac{a_n}{b_n} =\infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.

The extended test, at any rate, fails if you try to compare alternating series with positive series. Take for example $\sum\frac{1}{\sqrt[4]{n}}$ and $\sum \frac{(-1)^n}{\sqrt{n}}$. The first series diverges ($p$-series with $p\lt 1$), and second converges (alternating series, and the terms go to zero and are decreasing in absolute value). However, $$\lim_{n\to\infty}\frac{\quad\frac{1}{\sqrt[4]{n}}\quad}{\frac{(-1)^n}{\sqrt{n}}} = \lim_{n\to\infty}\frac{\sqrt{n}}{(-1)^n\sqrt[4]{n}} = \lim_{n\to\infty}\frac{\sqrt[4]{n}}{(-1)^n} = 0.$$ If the extended test worked here, then you would have to conclude, since the sequence $\sum b_n = \sum\frac{(-1)^n}{\sqrt{n}}$ converges, that the series $\sum a_n = \sum\frac{1}{\sqrt[4]{n}}$ also converges, which it does not.

What about the regular test, which requires the limit to exist and be finite? If we try to compare an alternating series with a series of positive terms, then we cannot have a limit that is both positive and exists (the terms alternate between positive and negative, so if the limit exists it has to be zero). So it would have to involve comparing two alternating series, and as I write this I see joriki has posted an example, so I'll leave it here.