<s>[INST] prove that $\sum_{n=1}^\infty \frac{(-1)^n}{n}$ convergies using Cauchy criterion Prove using Cauchy criterion that the sum of the series $$\sum_{n=1}^\infty \frac{(-1)^n}{n}$$
converges.
The cauchy criterion says that $\sum_{i=0}^\infty a_{i}$ converges if and only if for every $\epsilon>0$ exists $N$ such that $\forall m>n>N$ implies $\vert\sum_{k=n+1}^m a_{k}\vert<\epsilon$
Thats what I got so far $\vert\sum_{k=n+1}^m \frac{(-1)^k}{k}\vert=\vert\frac{(-1)^{n+1}}{n+1}+\frac{(-1)^{n+2}}{n+2}+...+\frac{(-1)^{m}}{m}\vert\le\vert\frac{(-1)^{n+1}}{n+1}\vert+\vert\frac{(-1)^{n+2}}{n+2}\vert+\vert\frac{(-1)^{n+3}}{n+3}\vert...+\vert\frac{(-1)^{m}}{m}\vert$ but this is the harmonic series and we know that she is divergent so what am I missing?
 [/INST] Hint:
$$\frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} - \frac{1}{n+4} \pm \ldots \\= \frac{1}{n+1} - \left(\frac{1}{n+2} - \frac{1}{n+3}\right) - \left(\frac{1}{n+4} - \frac{1}{n+5}\right) - \ldots \\\leqslant \frac{1}{n+1}$$
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