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Test the following series for absolute and conditional convergence:

$$\sum_{k=3}^\infty \frac {(-1)^k \log k}{k\log (\log k)}. \tag{1} $$

What test do I have to use to test for convergence of this series? Use of the double log always confuses me.

I have two more series:

$$\sum_{k=2}^\infty \frac{(-1)^k}{k^\alpha+(-1)^k}, \alpha \gt 0 \tag{2} $$

$$\sum_{k=1}^\infty (-1)^kk^{-\alpha}, \quad \alpha \in \mathbb{R} \tag{3}$$


$(3)$ is easy.

Convergence:

By alternating series, so $\lim_{k\rightarrow \infty} k^{-\alpha}$ if $\alpha \gt 0 $. In this case, this series converges conditionally.

However the series does not converge absolutely unless $\alpha \gt 1$ by p-series.

Divergence:

If $\alpha \le 0$ this series does not converge absolutely or conditionally. This is because $\lim_{k\rightarrow \infty} k^{-\alpha}$ is not equal to zero.


(Reference: Elementary Classical Anaylsis, Marsden, pg. 323)

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  • $\begingroup$ For the first, you are probably expected to use the alternating series test to prove convergence. That absolute convergence fails could come via comparison with $\sum \frac{1}{k\log(\log k)}$, which your course may already have shown is divergent. $\endgroup$ Jun 7 '15 at 16:16
  • $\begingroup$ @Dr.MV Your answer until now seems so clear to me. Maybe third one also converge conditionally, but not absolutely. I will set up my idea for third one via your answer. Wait for a moment. $\endgroup$ Jun 7 '15 at 16:41
  • $\begingroup$ @Dr.MV How about my third one solution? Is that OK to you? $\endgroup$ Jun 7 '15 at 16:55
  • $\begingroup$ @André Nicolas Thank you for your comment. $\endgroup$ Jun 7 '15 at 16:55
  • $\begingroup$ You are welcome. Indeed the third is the easiest. You probably should have said why non-convergence if $\alpha\le 0$: the terms do not have limit $0$. $\endgroup$ Jun 7 '15 at 17:01
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We have

$$\int_3^{L}\frac{\log x}{x\log (\log x)}dx=\int_{\log (\log 3)}^{\log (\log L)}\frac{e^{2u}}{u}du$$

which clearly diverges as $L \to \infty$. Thus there series does not absolutely converge by the integral test. Inasmuch as $\lim_{k\to \infty}\frac{\log k}{k\log(\log k)}=0$, the series conditionally coverges.


For the second series

$$\sum_{k=2}^{\infty}\frac{(-1)^k}{k^{\alpha}+(-1)^k} \tag 1$$

we have

$$\lim_{k\to \infty }\frac{1}{k^{\alpha}+(-1)^k}= 0$$

for $\alpha >0$ only. Thus, the series $(1)$ converges for $\alpha >0$. Now, to test for absolute convergence, we have

$$\left|\frac{(-1)^k}{k^{\alpha}+(-1)^k}\right|<\frac{1}{k^{\alpha}-1}$$

and thus, the series converges absolutely for $\alpha >1$.

NOTE: The third series is effectively the same as the second one.

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  • $\begingroup$ You mean this series converges absolutely, but not conditionally? $\endgroup$ Jun 7 '15 at 16:21
  • $\begingroup$ You used substitution method for integration. How about just setting logx=u? $\endgroup$ Jun 7 '15 at 16:26
  • $\begingroup$ @MathforyourLife The first series converges, but does not converge absolutely. The substitution used was $u=\log (\log x)$. If one uses instead, $u=\log x$, then the transformed integral is $\int_{\log 3}^{\infty}\frac{udu}{\log u}$ which diverges also. $\endgroup$
    – Mark Viola
    Jun 7 '15 at 16:36
  • $\begingroup$ The reason is that $u$ is larger than $\log u$. So you mean $\log x=u$ is alright to show the absolute convergence of this problem by integral test. $\endgroup$ Jun 7 '15 at 16:39
  • $\begingroup$ @MathforyourLife That substitution in the integral test shows that the convergence is NOT absolute since the integral, and hence the series, diverges. The original series with the alternative sign converges, but the series of absolute value of the terms diverges. That implies conditional convergence only. $\endgroup$
    – Mark Viola
    Jun 7 '15 at 16:41
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The series is conditionally convergent by Dirichlet's test, since the partial sums of $(-1)^k$ are bounded and $\frac{\log k}{k \log(\log k)}$ is eventually decreasing to zero. It is not absolutely convergent by comparison with the harmonic series, since for any $k$ big enough: $$\frac{\log k}{k\,\log\log k} > \frac{\log k}{k\,\log k} = \frac{1}{k}.$$

You may use similar arguments for dealing with the series in the second part, too.

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