0
$\begingroup$

When testing for convergence of a series $a_n$ using the Alternating series test, we need to satisfy the condition that $$|a_{n+1}| \leq |a_n|$$ and that $$\lim\limits_{n\rightarrow\infty} a_n = 0$$.

But I've seen other sources that state that the second condition (the limit) is another series that is just positive terms and leaves out the alternating term (eg, $(-1)^n$). That is, that we need to test $$\lim\limits_{n\rightarrow\infty} b_n = 0$$, where, for example, $$a_n = (-1)^nb_n $$

Which one is it?

$\endgroup$

closed as unclear what you're asking by Jack, Namaste, Olivier Bégassat, José Carlos Santos, Claude Leibovici Jul 20 '17 at 5:46

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ These are two equivalent conditions. $\endgroup$ – Parcly Taxel Jul 19 '17 at 16:28
  • $\begingroup$ $a_n$ is not a series. $\sum a_n$ is $\endgroup$ – Jack Jul 19 '17 at 16:29
  • $\begingroup$ If multiple sources claim non-identical definitions for the alternating test, likely, the two definitions result in the same thing. $\endgroup$ – Simply Beautiful Art Jul 19 '17 at 16:41
1
$\begingroup$

It doesn't matter if you use the terms with the (alternating) sign, or just their absolute values since: $$|u_n| \to 0 \iff u_n \to 0$$


  • if $u_n \to 0$, then clearly $|u_n| \to 0$;
  • if $|u_n| \to 0$, note that $-|u_n| \le u_n \le |u_n|$.
$\endgroup$
0
$\begingroup$

If we call the sequence in question $a_n = (-1)^nb_n$, then you need to verify that $b_1\ge b_2 \ge b_3\ge\dots \ge 0$ and $b_n\to 0$.

The first condition is that $b_n$ is monotonically decreasing. The second says that $b_n$ converges to $0$.

$\endgroup$

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