# convergence of $\sum_{n = 2}^\infty \frac{1}{(\log n)^2}$in the log test and conditions for success

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

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!***

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

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.

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.

• 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
– mary
Feb 18, 2016 at 9:56
• 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. Feb 18, 2016 at 12:44
• still do not get why you cannot compare it since I even looked at the comparison test and this didnt make sense
– mary
Feb 19, 2016 at 7:18
• 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. Feb 19, 2016 at 7:23