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?


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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.