Test for convergence of the series $\sum_{n=2}^\infty\frac{1}{(\ln n)^{\ln n}}$

Question

Could I have a hint for testing the convergence of the following series please?

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

Edit

The integral test does not work because $\int_1^n\frac{1}{(\ln x)^{\ln x}}dx$ has not an elementary primitive.

Thank You.

@Alex I'm trying with the integral test, but this integral is really tough for me... I'll check your link now! Thank you for your help man; I really appreciate it. — Charlie, Jan 26 '14 at 20:03
@Alex Am I on the right way? Can this integral be reduced to a simple one? — Charlie, Jan 26 '14 at 20:41

Antonio Vargas · Accepted Answer · 2014-01-26 20:06:49Z

15

Alternate hint:

$$ (\ln n)^{\ln n} = n^{\ln \ln n}. $$

answered Jan 26 '14 at 20:06

Antonio Vargas

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$\begingroup$ Thank you. I did it with the comparison test. $\endgroup$ – Charlie Jan 26 '14 at 22:18
$\begingroup$ You're welcome :) $\endgroup$ – Antonio Vargas Jan 26 '14 at 23:31
$\begingroup$ I have never seen this before, how do you get to that equation? $\endgroup$ – Nhat Jan 27 '14 at 17:20
3

$\begingroup$ @KitKat, take $\ln$ of $(\ln n)^{\ln n}$ to get $(\ln n)\ln\ln n$, rewrite this as $\ln(n^{\ln\ln n})$, then exponentiate this to get $n^{\ln \ln n}$. $\endgroup$ – Antonio Vargas Jan 27 '14 at 17:38
$\begingroup$ @AntonioVargas Nice, thanks. $\endgroup$ – Nhat Jan 27 '14 at 19:52

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Ma Ming · Accepted Answer · 2014-01-26 20:01:50Z

0

Hint:

Consider

$$ 2^k \frac{1}{(k\ln e)^{k\ln 2}}>\sum_{2^k\le n< 2^{k+1}} \frac{1}{(\ln n)^{\ln n}}> 2^k\frac{1}{((k+1)\ln e)^{(k+1)\ln 2}}, $$

answered Jan 26 '14 at 20:01

Ma Ming

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2 Answers 2