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I could not find how I can show that the following series is convergent or not.

$\sum_{n=2}^\infty(-1)^n\frac{\ln(n)}{\sqrt{n}}$

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    Note that the series is not absolutely converges.2012-12-25

3 Answers 3

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$\log(n)=3\log(n^{\tfrac{1}{3}}) \leq 3(n^{\tfrac{1}{3}}-1)$ so the terms of this alternating series converge to zero and

$\frac{\log(n+1)}{\sqrt{n+1}}\leq\frac{\log(n)+\frac{1}{n}}{\sqrt{n}+\frac{1}{2\sqrt{n+1}}} = \frac{\log(n)}{\sqrt{n}} - \frac{\log(n)-2\sqrt{1+1/n}}{\textrm{something positive}}$

which shows that the terms are eventually decreasing. Therefore the series converges.

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    @DavidMitra Excellent point. :-)2012-12-25
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Hint: Use Dirichlet test for series convergence.

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    Quote: *A particular case of Dirichlet's test is the more commonly used alternating series test for the case...*2012-12-25
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WimC has shown that indeed $\frac{\log n}{\sqrt{n}}$ goes to $0$ as $n\to \infty$. But it's also necessary to show that the sequence $\frac{\log n}{\sqrt{n}}$ is decreasing, or at least decreasing after a certain value of $n$. To do this, you can take the derivative of $\frac{\log n}{\sqrt{n}}$ with respect to $x$, which is: $\frac{2-\log x}{2x^{3/2}}$ after some simplification. The denominator is clearly positive, but after $\log x > 2$ the numerator is less than 0, so the derivative is clearly negative. This means that when $n > e^2$, the sequence will be decreasing. Hence the series, by the alternating series test, is convergent.