To use the Alternating Series Test on a series: $$ \sum_{n=k}^{\infty} (-1)^nb_n, \quad b_n\geq 0 $$ I have been told that I need to check that

  1. $b_{n+1} \leq b_n$ for all $n$
  2. $\lim_{n\to \infty} b_n= 0$.

But I can't understand why it isn't enough to check the limit part. IF the limit goes to zero, don't the $b_n$'s have to be decreasing?

  • 1
    $\begingroup$ Well, your last claim is certainly not true. Consider for example the following sequence: $b_n = \frac{1}{n}$ if $n$ odd and $b_n = \frac{2}{n}$ if $n$ even. $\endgroup$ – Jef Apr 7 '15 at 15:38
  • $\begingroup$ $s_ {n} =\displaystyle \Big( \frac{(-1)^n}{n} \Big)_{n \geq 1}$ is an example of sequence going to $0$ without being either increasing or decreasing. $\endgroup$ – pitchounet Apr 7 '15 at 15:38
  • $\begingroup$ @jibounet This doesn't work as a counterexample because $b_{n} = 1/n$ in your case. $\endgroup$ – JessicaK Apr 7 '15 at 15:40
  • $\begingroup$ @JessicaK You're right. I just wanted to give an example of sequence going to $0$ without being decreasing. It is just to show that what John Doe had in mind is not right. Even though, the sequence $(b_{n})$ in the alternating test must be a sequence of non-negative real numbers, so my counterexample does not fit in this theorem. $\endgroup$ – pitchounet Apr 7 '15 at 15:43
  • $\begingroup$ See earlier threads Monotonicity in alternating Series and Leibniz's alternating series test as well. $\endgroup$ – Jeppe Stig Nielsen Jun 10 '15 at 14:28

Consider the sequence $$1,0,\frac{1}{2},0,\frac{1}{3}, ...$$, or if you want a more clearly alternating sequence then consider $$1, -1, \frac{1}{2}, -\frac{1}{4}, \frac{1}{3}, -\frac{1}{16}, ..., \frac{1}{n}, -\frac{1}{4^n},...$$

Both of these sequences satisfy the limit $\to 0$ (and the second explicitly alternates). But the series associated to either example diverges. The reason here is that the positive terms are much bigger than the negative terms. This is very visible in the second example, but distilled in the first.


No. For example, if $b_n$ is $1/n$ for even $n$ and $0$ for odd $n$, then $\lim\limits_{n\to \infty} b_n = 0$ but $b_n$ is not decreasing.

  • $\begingroup$ So this series is actually divergent? $\endgroup$ – John Doe Apr 7 '15 at 15:41
  • $\begingroup$ Yes, that's right. $\endgroup$ – kobe Apr 7 '15 at 15:42

As another example, you could consider the series

$\displaystyle \frac{2}{1}-\frac{1}{1}+\frac{2}{2}-\frac{1}{2}+\frac{2}{3}-\frac{1}{3}+\cdots+\frac{2}{n}-\frac{1}{n}+\cdots$.

This is an alternating series whose terms approach 0, but the terms are not decreasing in absolute value

(since $\displaystyle\frac{1}{n}<\frac{2}{n+1}$ for $n>1$).


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.