The alternating series test (AST) says, briefly, that if

  1. $a_k>0$
  2. $a_k \geq a_{k+1}$, and
  3. $a_k \to 0$ 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.

  • 1
    $\begingroup$ You can take any alternating series and append a few terms of arbitrary finite magnitudes in front of it, so that condition (2) is violated for small $k$... but I guess that's not what you had in mind? $\endgroup$
    – David Z
    Jun 17 '14 at 2:04
  • 3
    $\begingroup$ Take your favourite alternating series, say $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$. Interchange terms $1$ and $3$, also terms $5$ and $7$, also terms $9$ and $11$, and so on. Signs still alternate, series still converges, still to $\log 2$, but AST does not apply. $\endgroup$ Jun 17 '14 at 2:28
  • $\begingroup$ @DavidZ: Two good points. A good example and, yes, that is not quite what I had in mind. In fact, the complete AST says conditions 1 and 2 must hold after some point (that is, after $k>K$). $\endgroup$
    – Jeff
    Jun 17 '14 at 3:35

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.

  • 5
    $\begingroup$ What is the sequence's pattern here? $\endgroup$
    – Jeff
    Jun 17 '14 at 5:51
  • $\begingroup$ looks like subsequent terms are alternately multiplied by negative two and negative one-fourth. $\endgroup$
    – Dapianoman
    Sep 11 '20 at 8:57

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 $\dfrac{1}{1-\tfrac{1}{4}} - \dfrac{\tfrac{1}{3}}{1-\tfrac{1}{9}} = \dfrac{23}{24}$.


Pick any series $\sum (-1)^na_n$ that satisfies the AST, and pick any sequence $b_n$ that converges to $0$.


$$c_{n}=a_n+b_{\lfloor \frac{n}{2} \rfloor}$$ 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.


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 $$-5\Bigl(\frac{1}{1^2}+\frac{1}{2^2}+\cdots\Bigr) +\Bigl(\frac{1}{1^4}+\frac{1}{2^4}+\cdots\Bigr)=-\frac{5\pi^2}{6}+\frac{\pi^4}{90}\ .$$


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.


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