Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

the definition of this test is:

if $a_n$ decreases monotonically and goes to 0 in the limit then the alternating series $\sum_{n=1}^{\infty}(-1)^na_n$ converges

my question is: why does the series $a_n$ has to be monotonically descending, isn't it enough for $\lim_{n\to\infty}a_n = 0$ ? can someone give me an example for that?

share|cite|improve this question
Also see earlier thread Monotonicity in alternating Series. – Jeppe Stig Nielsen Jun 10 '15 at 14:29
up vote 4 down vote accepted

Counter example:

Check the following alternating series diverges:

$$\sum_{n=1}^\infty(-1)^na_n\;\;,\;\;a_n=\begin{cases}0&,\;\;n\;\;\text{is odd}\\{}\\\frac1n&,\;\;n\;\;\text{is even}\end{cases}$$

share|cite|improve this answer
thanks! a great example that can be easily understood! – guynaa Dec 8 '13 at 19:48

Let $$a_n=\frac{1}{\sqrt {n+(-1)^n}}$$ then by taylor series we have $$(-1)^na_n=\frac{(-1)^n}{\sqrt n}\left(1-\frac{(-1)^n}{2\sqrt n}+O\left(\frac{1}{n}\right)\right)$$ The first and the last term give a convergent series by Leibniz theorem and by comparison with the Riemann series respectively and the term $\frac{1}{2n}$ gives a divergent series hence the series $\sum (-1)^n a_n$ is divergent.

share|cite|improve this answer

Consider $a_n = 1/n$ when $n$ is even, and $a_n=1/n^2$ when $n$ is odd. Then $a_n \to 0$ but $\sum (-1)^n a_n$ diverges.

share|cite|improve this answer
why does it diverge? – guynaa Dec 8 '13 at 19:43
Perhaps you can prove that yourself! – GEdgar Dec 8 '13 at 19:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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