Here is the problem:

Determine whether the series $$\sum_{n=2}^{\infty} \frac{(-1)^n}{n(\ln(n))^{2/3}}$$ is absolutely convergent, conditionally convergent, or divergent.

I tried the Ratio Test and the Root Test and they both came out to $L=1$ or inconclusive, so now what do I do?

  • 1
    $\begingroup$ Integral Test. Or Cauchy Condensation, but that is not taught as routinely. It will turn out to be not absolutely convergent. But convergent, sure. $\endgroup$ – André Nicolas Dec 14 '11 at 21:16
  • $\begingroup$ You could consider using a combination of the $n^{th}$ root test and the comparison test possibly. $\endgroup$ – analysisj Dec 14 '11 at 21:16

It converges (alternating series) and you can compare to an integral to test for absolute convergence $$ \int_2^{\infty}\frac{1}{x}(\log x)^{-2/3}dx=\int_{\log 2}^{\infty} u^{-2/3}du=3u^{1/3}\Bigg|_{\log 2}^{\infty}=\infty $$

  • $\begingroup$ Quick question: Why is the 1/x on the far left no longer (-1)/x? $\endgroup$ – Kyle V. Dec 14 '11 at 21:27
  • $\begingroup$ Also, so since the integral test came out to $\infty$ which is not < 1 it means it is not absolutely convergent? $\endgroup$ – Kyle V. Dec 14 '11 at 21:34
  • $\begingroup$ @StickFigs the integral test (look in your book if you have one) is a way of determining convergence/divergence of a series of positive terms $\sum a_n$ by comparing the series to the area under the graph of a function $f$ with $f(n)=a_n$ (draw a picture thinking of the series geometrically as a bunch of columns with height $a_n$ and width 1 with the base of the column on the interval $[n,n+1]$ or $[n-1,n]$) $\endgroup$ – yoyo Dec 14 '11 at 21:39

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