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Is \begin{align}\sum_{n=1}^\infty \frac{\left(-1\right)^{n+1}\log\left(n\right)}{n}\tag{1}\end{align} divergent? I think it is because by comparison \begin{align} \sum_{n=1}^\infty \frac{1}{n}<\sum_{n=1}^\infty \frac{\log\left(n\right)}{n},\tag{2} \end{align} but can I compare the absolute value of these two summations or would I have to compare the following instead: \begin{align} \sum_{n=1}^\infty \frac{\left(-1\right)^{n+1}}{n}<\sum_{n=1}^\infty \frac{\left(-1\right)^{n}\log\left(n\right)}{n}.\tag{3} \end{align} Thank you for your time,

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    $\begingroup$ Hint: Look up the Alternating Series Test $\endgroup$ – Simon S Feb 7 '15 at 2:40
  • $\begingroup$ Okay so via the alternating series test I see that $$ \lim_{n\rightarrow\infty}\frac{\log\left(n\right)}{n}=0, $$ and therefore this appears to be convergent. But because this series is not absolutely convergent, why is it necessarily convergent in this way? After all, for the harmonic series $$ \lim_{n\rightarrow\infty}\frac{1}{n}=0 $$ also, yet it is not convergent via the integration test... $\endgroup$ – jm324354 Feb 7 '15 at 2:46
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It is convergent. Since $\frac{\log n}{n}$ is decreasing for $n\ge 3$, and it decreases to zero, the sum converges by the alternating series test. However, as you point out, the series is not absolutely convergent (that is, it does not converge if you remove the $(-1)^{n+1}$), since then it dominates the harmonic series, which diverges. So the given series is conditionally convergent only.

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