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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?
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.
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.
