# In the alternating series test, what is the condition for convergence? [closed]

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?

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

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

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 $-|u_n| \le u_n \le |u_n|$.
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 $0$.