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I'm being asked to determine if $\displaystyle\sum\limits_{n=3}^\infty \frac1{n (\ln n)\ln(\ln n)}$ converges. So, using Cauchy's Condensation Test, I reduced the problem to one of determining the convergence of $\displaystyle\sum\limits_{n=3}^\infty\frac 1{n\ln (n\ln 2)}$. Am I on the right path, and how do I proceed from here?

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4 Answers 4

up vote 3 down vote accepted

If $a_n$ is eventually nonincreasing, then $\sum a_n$ converges iff $\sum 2^na_{2^n}$ converges (Cauchy condensation test). This takes us from $$\sum\frac 1{n\ln n \ln\ln n}$$ to $$\sum\frac {2^n}{2^n\cdot n\ln 2 \cdot (\ln n + \ln\ln 2)}=\sum\frac {1}{n\ln 2 \cdot (\ln n + \ln\ln 2)}$$ and then to $$\sum\frac {2^n}{2^n\ln 2 \cdot (n \ln 2 + \ln\ln 2)}=\sum\frac {1}{n\ln^2 2 + \ln2\cdot\ln\ln 2}$$ which is essentially the harmonic series.

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I love your idea of the repeated application of the Condensation Test-- I guess I hadn't thought hard enough! Thank you. –  Ryan Jan 16 '13 at 20:26
    
Contemplating about it, you will surely note that $n$ times any "log-gy stuff" in the denominator can be treated like this. Actually, it is enough to play with $\sum\frac1{n\ln^k n}$. –  Hagen von Eitzen Jan 16 '13 at 20:42

Apply directly integral test

$$\int_3^{\infty}\frac{(\ln(\ln x))'}{ \ln(\ln x)}\mathrm{dx}= \left[\ln(\ln(\ln(x)))\right]_3^{\infty}\longrightarrow \infty$$

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Since $\ln 2 $ is constant $\frac 1{n\ln (n\ln 2)} $ is comparable to $\frac 1{n\ln (n)}$ you can use integral test here.

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Observe that

$$\frac{d}{dx}(\ln\ln x)=\frac1{\ln x}\cdot\frac{d}{dx}(\ln x)=\frac1{x\ln x}$$

and use the integral test.

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