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Once again, note No use of the alternating series test!
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$\begingroup$Directly: write $A_n=\sum_{k=1}^n\frac{(-1)^k}{k}$. You want to show that $(A_n)_n$ is a converging sequence. It is enough to show that both $(A_{2n})_n$ and $(A_{2n+1})_n$ converge, and that $A_{2n+1}-A_{2n}\to 0$. The last part is trivial as $A_{2n+1}-A_{2n} = -\frac{1}{2n+1}$; the first two are done in a similar (almost identical) way: $$ A_{2n+1} = -1 + \sum_{k=1}^n \left( \frac{1}{2k} - \frac{1}{2k+1}\right) = -1 + \sum_{k=1}^n \frac{1}{2k(2k+1)} $$ Since $\frac{1}{2k(2k+1)}\operatorname*{\sim}_{k\to\infty}\frac{1}{4k^2}$, this last series converges.
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