limit comparison test for alternating series I am trying to understand why does the limit comprasion test doesn't work for alternating series, is it even true? or is there a counter example? I can't find one. can you help me please?
thanks.
benny

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.
