# $\sum (-1)^n/n$ fails the p-series test, but passes the alternating series test?

P-Series Reference

Alternating Series Test Reference

$$\sum_{i=0}^\infty \frac{(-1)^n}{n}$$

This alternating series fails the p-series test because the exponent of n = 1.

Yet it seems to pass the alternating series test.

1 - $a_n$ must be positive. True.

2 - Terms must be decreasing. $\frac{d}{dn} 1/n = -n^{-2}$, which is < 1. True.

3 - $\lim_{n\rightarrow\infty} 1/n = 0$ True.

$(-1)^n/n$ is clearly a divergent series, so why does it pass the AST?

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What you are noting is that the series $$\sum_{i=0}^\infty \,\frac{(-1)^n}{n}\quad {\bf {converges},}$$ as you found by the alternating series test, but does not converge absolutely: $$\sum_{i=0}^\infty \,\left|\frac{(-1)^n}{n}\right| \quad = \quad \sum_{i = 0}^\infty\,\frac 1n\quad\bf{does\; not\; converge.}$$

Note: the $p$-series test is applicable for sums of the form: $\displaystyle\sum \frac 1{n^{p}},$ and your "given" series does not "fit" that form for odd $n$; indeed, the most appropriate test to use here, as you used in the end, is the alternating series test.

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+1 from me, but I let my answer stand as a cautionary tale of sorts. –  Harald Hanche-Olsen Apr 21 '13 at 16:19

What you seem to be missing is this: What you call the power series test merely fails to prove that the series converges. It does not prove that the series diverges, absolutely or otherwise. The alternating series test, on the other hand, proves that the series does converge.

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That's not true: the p-series test does prove divergence for series of the form $\sum \frac 1{n^p}$ when $p\leq 1$. –  amWhy Apr 21 '13 at 16:11
@amWhy: Sure, but how did the p-series test enter into the discussion? –  Harald Hanche-Olsen Apr 21 '13 at 16:13
Look at the title for the question. –  amWhy Apr 21 '13 at 16:14
@amWhy: Oh, the title, eh? Well, there should be a law against mentioning relevant information only in the title. He clearly said “power series test” in the question body – a term which could be taken to mean any test which yields the radious of convergence, but not by any stretch of the imagination to include the p-series test. –  Harald Hanche-Olsen Apr 21 '13 at 16:18
I understand completely...Argh!! mismatching questions/titles! –  amWhy Apr 21 '13 at 16:19