The alternating series test is a sufficient condition for the convergence of a numerical series. I am searching for a counterexample for its inverse: i.e. a series (alternating, of course) which converges, but for which the hypothesis of the theorem are false.

In particular, if one writes the series as $\sum (-1)^n a_n$, then $a_n$ should not be monotonically decreasing (since it must be infinitesimal, for the series to converge).

  • 3
    $\begingroup$ Interlace the terms $1/n^2$ and $1/n^3$ (with appropriate signs). $\endgroup$ – David Mitra May 25 '13 at 9:58
  • 2
    $\begingroup$ did you mean to say that the absolute values of the general terms should not be monotinically decreasing (since if the series has alternating signs, surely the general terms can't be monotone). And you should really change the title. There are no counterexamples to Leibnitz's test! $\endgroup$ – Ittay Weiss May 25 '13 at 9:58
  • $\begingroup$ Yes, I was thinking of $\sum (-1)^n a_n$, where $a_n\ge0$. $\endgroup$ – zar May 25 '13 at 11:23
  • $\begingroup$ Do you want to stipulate that the series is still conditionally convergent? $\endgroup$ – Ted Shifrin May 25 '13 at 21:26
  • $\begingroup$ Yes (are you thinking of something else?) $\endgroup$ – zar May 26 '13 at 12:50

Put: $$ b_n = \begin{cases} n^{-2} &: n \text{ odd} \\ 2^{-n} &: n \text{ even} \end{cases} $$

$b_n$ is not monotonically decreasing. Still, $\sum (-1)^n b_n$ converges.


If you want a conditionally convergent series in which the signs alternate, but we do not have monotonicity, look at $$\frac{1}{2}-1+\frac{1}{4}-\frac{1}{3}+\frac{1}{6}-\frac{1}{5}+\frac{1}{8}-\frac{1}{7}+\cdots.$$ It is not hard to show that this converges to the same number as its more familiar sister.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.