# $\sum (-1)^n/n$ 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
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