Is there a convergent, alternating series that fails the AST?
The alternating series test (AST) says, briefly, that if
- $a_k>0$
- $a_k \geq a_{k+1}$, and
- $a_k \to 0$ as $k \to \infty$
then $\sum_k (-1)^k a_k$ converges.
This seems to be a one-way test (that is, if an alternating series fails the test, we don't know that it diverges).
This document says so explicitly. It then gives an example of an alternating series which fails the AST. That doesn't prove the series is divergent (but it turns out to be divergent, anyway, by $n$th term test).
Is there a convergent, alternating series that fails the AST?
Of course, if the series fails condition 3, then it also fails the $n$th term test, and must diverge. And if it fails the first condition, then it's not strictly alternating, anyway.
So it must be a series that is not getting smaller (condition 2), but still converges.
Solution 1:
Yes, there are such series. Consider, for example, a sequence such as
$$1, 2, \frac 1 2, 1, \frac 1 4, \frac 1 2, \frac 1 8, \frac 1 4, \dots$$
The series
$$1 - 2 + \frac 1 2 - 1 + \frac 1 4 - \frac 1 2 + \frac 1 8 - \frac 1 4 + \dots$$
is alternating and (absolutely) convergent, but it clearly fails to be monotonically decreasing.
Solution 2:
Let $a_n = \begin{cases}2^{-n} & n \equiv 0\pmod{2} \\ 3^{-n} & n \equiv 1\pmod{2} \end{cases}$. Then, $a_n$ is not strictly decreasing, since $a_3 = \dfrac{1}{27} < \dfrac{1}{16} = a_4$.
However $\displaystyle\sum_{n = 0}^{\infty}(-1)^na_n$ still converges to $\dfrac{1}{1-\tfrac{1}{4}} - \dfrac{\tfrac{1}{3}}{1-\tfrac{1}{9}} = \dfrac{23}{24}$.
Solution 3:
Pick any series $\sum (-1)^na_n$ that satisfies the AST, and pick any sequence $b_n$ that converges to $0$.
Define
$$c_{n}=a_n+b_{\lfloor \frac{n}{2} \rfloor}$$ that is $$c_{2n}=a_{2n}+b_n \\ c_{2n+1}=a_{2n+1}+b_n$$
Then, it is easy to prove that $\sum(-1)^n c_n$ is always convergent, but it is very easy to make examples where $c_n$ is not positive, or not decreasing. Or to fail both conditions.
Solution 4:
Here is an example where the terms keep on oscillating forever: that is, $a_1>a_2$, $a_2<a_3$, $a_3>a_4$ and so on. For all $n\ge1$ define $$a_{2n-1}=\frac{5}{n^2}\ ,\quad a_{2n}=\frac{1}{n^4}\ .$$ Checking what I asserted above, we have $$a_{2n-1}>\frac{1}{n^2}\ge\frac{1}{n^4}=a_{2n}$$ while $$a_{2n}\le\frac{1}{n^2}=\frac{4}{n^2+2n^2+n^2}<\frac{5}{n^2+2n+1}=a_{2n+1}\ .$$ It is not hard to see that it is ok to rearrange the terms of $$\sum_{k=1}^\infty (-1)^ka_k=-\frac{5}{1^2}+\frac{1}{1^4}-\frac{5}{2^2}+\frac{1}{2^4}-\frac{5}{3^2}+\frac{1}{3^4}-\cdots\ ,$$ so that its sum is $$-5\Bigl(\frac{1}{1^2}+\frac{1}{2^2}+\cdots\Bigr) +\Bigl(\frac{1}{1^4}+\frac{1}{2^4}+\cdots\Bigr)=-\frac{5\pi^2}{6}+\frac{\pi^4}{90}\ .$$
Solution 5:
Another construction to make lots of examples - take any series that converges absolutely, say $ 1 + 1/4 + 1/9 + 1/16 + \ldots$. Now rearrange the terms to make it as non-monotonic as you like. Since it converges absolutely, the rearrangement will still converge (to the same sum). Now make the rearranged sum alternating.