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$$\sum_{n = 2}^\infty \frac{1}{(\log n)^2}.$$

I think there's the ratio test and when I tried it got stuck

****Important, while this question was answwered very good, just had a concern with one of the answers so I JUST put a comment in and would appreciate if someone would be able to address it

Thanks!***

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  • $\begingroup$ Use $\sum_{n = 2}^\infty \frac{1}{(\log n)^2}$ to show $\sum_{n = 2}^\infty \frac{1}{(\log n)^2}.$ Formatting tips here. $\endgroup$
    – Em.
    Feb 17, 2016 at 2:02
  • $\begingroup$ In the long run, $(\log n)^2\lt n$. $\endgroup$ Feb 17, 2016 at 2:09

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Use Cauchy Condensation Test: you want to test the convergence of $$ \sum_{n=2}^{\infty}a_n $$ where $a_n=1/(\log n)^2$ which is a positive and decreasing sequence; thus the above series converges iff $\sum_{n=2}^{+\infty}2^na_{2^n}$ does.

But $$ 2^na_{2^n}=\frac{2^n}{(\log2^n)^2}=\frac{2^n}{n^2\log^22}\stackrel{n\to+\infty}{\longrightarrow}+\infty $$ thus $\sum_{n=2}^{+\infty}2^na_{2^n}$ diverge.

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Comparison test and Integral test also can be used. Since $n > \log n$ For all $n \ge 2$, Following inequality holds: $$ 0\le\frac{1}{n\log n} < \frac{1}{(\log n)^2} $$ Then $f(x)=\frac{1}{x\log x}$ monotonely decreases on $[2,\infty)$ and \begin{align} \int_2^{\infty} \frac{dx}{x \log x}&=\lim_{N\to\infty}\int_{\ln 2}^{\ln N}\frac{1}{t}dt\\ &=\lim_{N\to\infty}\left[\ln t\right]_{\ln 2}^N\\ &=\lim_{N\to\infty}(\ln(\ln N)-\ln(\ln 2))\\ &=\infty \end{align} Thus, $\sum_{n=2}^{\infty} \frac{1}{n\log n}$ diverges by integral test. Therefore $\sum_{n=2}^{\infty} \frac{1}{(\log n)^2}$ diverges by comparison test.

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  • $\begingroup$ I am having trouble understanding why you compared the series to 1/(n log(n)) WHy did you not compare this to 1/n(log(n)^2 and just use the p series test to test if it was convergent or divergent $\endgroup$
    – mary
    Feb 18, 2016 at 9:56
  • $\begingroup$ Because $\frac{1}{n(\log n)^2}\le \frac{1}{(\log n)^2}$ and so we can't apply comparison test to show given series converges. p-series is $\sum \frac{1}{n^p}$, $\log$ is not included. $\endgroup$ Feb 18, 2016 at 12:44
  • $\begingroup$ still do not get why you cannot compare it since I even looked at the comparison test and this didnt make sense $\endgroup$
    – mary
    Feb 19, 2016 at 7:18
  • $\begingroup$ Remark the statement of comparison test: Given two sequence $(a_n)$ and $(b_n)$, If $0\le a_n \le b_n$ and $\sum b_n$ converges, then $\sum a_n$ converges, and if $0\le b_n \le a_n$ and $\sum b_n$ diverges, then $\sum a_n$ diverges. If you want to use $\frac{1}{n(\log n)^2}$ to show that given series converges, then you have to show that $\frac{1}{n(\log n)^2}$ is bigger than $\frac{1}{(\log n)^2}$. However, it is false. $\endgroup$ Feb 19, 2016 at 7:23

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