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Stumped on this absolute convergence problem! (Converge conditionally, absolutely, or diverges)

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

First, I try to do the absolute convergence test: $$\sum_{n=2}^\infty \left\lvert \frac{-1^n}{\ln(n)}\right\rvert = \sum_{n=2}^\infty \frac{1}{\ln(n)} $$

Not sure how to evaluate this series. Which test to use?

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The series fulfills the conditions of Leibniz alternating series and then it converges, but it doesn't converge absolutely since by comparison

$$\frac1{\log n}\ge\frac1n$$

and the harmonic series diverges.

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  • $\begingroup$ What's the most intuitive way to know that $ln(n) < n$ ? Or directly that $\frac{1}{n} \le \frac{1}{ln(n)}$ Exponent for e to get 10 is way smaller than 10. So, ln(10) < 10. ie: ln(n) < n $\endgroup$ – JackOfAll Mar 16 '16 at 21:11
  • $\begingroup$ @JackOfAll Evaluate, for example with l'Hospital, the limit $\;\lim_{x\to\infty}\frac{\log x}{x^\epsilon}\;$ , for any $\;\epsilon>0\;$ . You'll see at once that the limit is zero, so that for $\;x\;$ big enough you have that inequality. $\endgroup$ – DonAntonio Mar 16 '16 at 21:13
  • $\begingroup$ How do you algebraically go from $ln(n) < n $ to $\frac{1}{ln(n)} > \frac{1}{n}$ $\endgroup$ – JackOfAll Mar 16 '16 at 21:16
  • $\begingroup$ For $\;n>2\;,\;\;\log n>0\;$ , so $\;\log n<n\iff\frac1{\log n}>\frac1n\;$ . This is simply inequalities and fractions. $\endgroup$ – DonAntonio Mar 16 '16 at 21:17
  • $\begingroup$ Just because $log(n )> 0$, why can you assume $log( n) < n$ ? $\endgroup$ – JackOfAll Mar 16 '16 at 21:18
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You know that for $n\ge 1$, $\ln(n) \le n$ which in turn says that $\frac{1}{n}\le \frac{1}{\ln(n)}$. What does this tell you about the absolute convergence?

As for conditional, think about the alternating series test.

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  • $\begingroup$ Well since you're doing series, I would think you've done derivatives. You can easily show that $x-\ln(x)$ is increasing by taking a derivative and doing some basic manipulations. $\endgroup$ – Cameron Williams Mar 16 '16 at 21:13
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do you know alternated series ?

you can also group the terms by two, and find that $$\frac{1}{\ln (2 n)} - \frac{1}{\ln (2 n+1)} = \frac{\ln (2 n+1) - \ln (2 n)}{\ln (2n) \ln (2n+1)} = \frac{\ln \left(1+\frac{1}{2n}\right)}{\ln (2n) \ln (2n+1)} \sim \frac{1}{2n \ln^2 n}$$

and $\sum_{n=2}^\infty \frac{1}{n \ln^2 n}$ is by the Cauchy condensation test, or the integral test, an absolutely convergent Bertrand series.

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