# $\sum_{n=2}^\infty \frac{1}{(\ln\, n)^2}$ c0nvergence

$$\sum_{n=2}^\infty \frac{1}{(\ln\, n)^2}$$

The series converge?

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Surely because $\ln 1 = 0$, the series won't converge because at $n=1$ we have an indeterminate form? Also, even if we were to change the bottom limit to $n=2$, the series won't converge for the reasons given in the comments below. – Andrew D Jan 9 '13 at 16:52
Should be n=2, fixed. – Steve Jan 9 '13 at 16:56

It diverges. See that $$\lim_{n\to\infty}\frac{\sqrt{n}}{\ln^2(n)}=\infty$$ Infact the power of $n$ is $\frac{1}{2}\leq1$ and $\lim_{n\to\infty}n^{1/2}u_n=\infty$

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$\frac{1}{n}\leq\frac{1}{(ln\, n)^{2}}\iff n\geq(ln\, n)^{2}\iff ln\, n\leq\sqrt{n}\iff n\leq e^{\sqrt{n}}$

$\begin{cases}(e^{\sqrt{n}})'>(n)'\\\text{for } n=1\,\,\,\, e^{1}>1\end{cases}$

so from comparison test, the series diverge

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Does $\sum 1/n$ converge? – user27126 Jan 9 '13 at 16:52
Perfect until the last word: the series $\sum\limits_n\frac1n$ diverges hence by comparison... – Did Jan 9 '13 at 16:52
That's a typo of course. Fixed. Thanks! – Steve Jan 9 '13 at 16:53

Use the Cauchy Condensation Test

$$\sum_{k=2}^{\infty} \frac{1}{(\ln k)^{2}} \text{converges} \iff \sum_{k=2}^{\infty} \frac{2^k}{(\ln 2^k)^{2}} \text{converges}$$

$$\sum_{k=2}^{\infty} \frac{2^k}{(\ln 2^k)^{2}} = \sum_{k=2}^{\infty} \frac{2^k}{(\ln 2^k)(\ln 2^k)}=\sum_{k=2}^{\infty} \frac{2^k}{k^2(\ln 2)(\ln 2)}=\sum_{k=2}^{\infty} \frac{2^k}{k^2(\ln 2)^2} \text{(diverges)}$$

So the original series diverges.

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:very nice proof – derivative Nov 21 '13 at 10:51