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The question is, whether the following series is convergent, absolutely convergent or divergent $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}.$$

  1. The term $\dfrac{(-1)^{n+1}}{2n-1}$ = $ \dfrac{(-1)\cdot(-1)^{n}}{2n-1}$ = $(-1)^{n} \dfrac{(-1)}{2n-1}$.
    So according to Leibnitz-Criterion, if the series $ \dfrac{(-1)}{2n-1}$ is a Zero-sequence, then the alternating series converges, too.
  2. The series $\sum_{n=1}^\infty \left|\dfrac{(-1)^{n+1}}{2n-1}\right|$ = $\sum_{n=1}^\infty \dfrac{1}{2n-1}$ converges. Therefore the series is absolutely convergent.

Is this correct?

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$\sum_{n=1}^\infty \frac{1}{2n-1}$ is not convergent as

$\sum_{n=1}^\infty \frac{1}{2n-1} \ge \sum_{n=1}^\infty \frac{1}{2n}$ and the series $\sum_{n=1}^\infty \frac{1}{n}$ is not convergent

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