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Check convergence of

$$\sum^{\infty}_{n=1}\frac{1}{(\ln \ln n)^{\ln n}}.$$

Please verify my solution below.

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4  
Please use \ln and \exp. –  Brian M. Scott Jan 9 '13 at 17:03
    
OK, don't know about this –  Steve Jan 9 '13 at 17:04
1  
Please note that the series is not defined for $n = 1, 2$. –  JavaMan Jan 9 '13 at 17:30
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This is a very nice question (+1) –  Chris's sis Jan 12 '13 at 15:23

3 Answers 3

up vote 15 down vote accepted
+250

$$\sum^{\infty}_{n=1}\frac{1}{(\ln\, \ln\, n)^{\ln\, n}}=\sum^{\infty}_{n=1}\frac{1}{\exp( \ln\, \ln\ln\, n\,*\ln\,, n)}=\sum^{\infty}_{n=1}\frac{1}{\exp(\ln\, n\,*\ln\, \ln\, \ln\, n)}=\sum^{\infty}_{n=1}\frac{1}{n^{\ln\, \ln\, \ln\, n}}$$

For a lagre n, $$\frac{1}{n^{\ln\, \ln\, \ln\, n}} < \frac{1}{n^2}$$

So the series converge by comparison test

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1  
Looks okay to me. –  Brian M. Scott Jan 9 '13 at 17:11
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+1. Obviously the way to go. –  Did Jan 9 '13 at 17:40
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@Steve +1 for answering your own question. –  Rustyn Jan 11 '13 at 17:52
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@Steve you should accept your own answer~! –  Rustyn Jan 11 '13 at 19:27
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Am I the only one who sees Elias's edit as a disimprovement? –  TonyK Jan 12 '13 at 16:51

There is $n_0$ such that

$$\sum^{\infty}_{n=n_0}\frac{1}{(\ln \ln n)^{\ln n}}<\sum^{\infty}_{n=n_0}\frac{1}{3^{\ln n}}=\sum^{\infty}_{n=n_0}\frac{1}{n^{\ln 3}}$$ and this converges since $\ln 3>1$

Q.E.D.

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Use the inequality $$ \frac{x-1}{x^p}<\ldots<\frac{x-1}{x^2}<\frac{x-1}{x}<\log x , \quad \forall \,x>1, \,\forall p\in\mathbb{N}\backslash\{1,0\}. $$ and the fact that $\ln$ is an increasing function. We have for $n$ large, \begin{align} \left|\frac{1}{\left[\log\big(\log n\big)\right]^{(\log n)}}\right|\leq & \frac{1}{\left|\log\circ\log n\right|^{\big(\frac{n-1}{n^p}\big)}} \\ \leq & \frac{1}{\left|\log\big(\frac{n-1}{n}\big)\right|^{\big(\frac{n-1}{n^p}\big)}} \\ \leq & \frac{1}{\left|\frac{\big(\frac{n-1}{n}\big)-1}{\big(\frac{n-1}{n}\big)}\right|^{\big(\frac{n-1}{n^p}\big)}} \\ = & \frac{1}{\left(\frac{n}{n-1}\right)^{\big(\frac{n-1}{n^p}\big)}} \\ = & \bigg(1-\frac{1}{n}\bigg)^{\frac{n-1}{n^p}} \\ = & \left[\bigg(1-\frac{1}{n}\bigg)^{n}\right]^{\frac{1}{n^p}} \\ \end{align} And by root test, if $a_n= \left[\bigg(1-\frac{1}{n}\bigg)^{n}\right]^{\frac{1}{n^p}}$ we have $ \lim_{n\to \infty}\sqrt[n]{a_n}<1 $ for $p$ enogh large. By comparation test the serie $$ \sum_{n=1}^{\infty}\frac{1}{\left[\log\big(\log n\big)\right]^{(\log n)}} $$ converge.

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