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I've spent a good week and half manipulating and trying different tests to find the convergence or divergence of this series:

$$\sum_{n=0}^\infty \frac{1}{(\ln n)^{\ln n}}$$

I've tried all the Convergence Tests I know, and have tried to exhaust the Ratio Test for many different manipulations of the above, including $\ln n^{-\ln n}$, $\ln n^{\ln(1/n)}$, and, using the relationship $a^x = e^{x\ln a}$, $e^{\ln(1/n)*\ln(\ln(n))}\implies(1/n)^{\ln(\ln n)}$.

Basically, I'm just burnt out trying to solve this, and came for any help someone on here can offer me.

Cheers.

EDIT: the only hint we're given is to note that $n > e^2$.

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    $\begingroup$ Perhaps that you meant $\displaystyle\sum_{n=2}^\infty\frac1{(\log n)^{\log n}}$. $\endgroup$ – José Carlos Santos Aug 16 '18 at 22:25
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    $\begingroup$ No, the question uses natural logarithms. $\endgroup$ – anyone Aug 16 '18 at 22:26
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    $\begingroup$ @anyone The base of the logarithm is (probably) immaterial. Compare the comment with the sum in the question: they are entirely different. $\endgroup$ – Umberto P. Aug 16 '18 at 22:27
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    $\begingroup$ Hint: try Cauchy-condensation test. It is very good for series with logarithms. $\endgroup$ – Mark Aug 16 '18 at 22:33
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    $\begingroup$ I believe @JoséCarlosSantos is pointing out that at $n=0$ and $n=1$, the summand is undefined. (Also $\log$ is a common notation for natural log) $\endgroup$ – Simply Beautiful Art Aug 17 '18 at 0:10
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HINT:

$$(\ln n)^{\ln n} = n ^{\ln \ln n} > n^2$$ for all $n > n_0$. So $$\frac{1}{n ^{\ln \ln n}} < \frac{1}{n^2}$$

Then use comparison test

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