infinite series with natural logarithm

logistic_guy

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Determine the convergence or the divergence of the infinite series.

k=31ln(klnk)\displaystyle \sum_{k = 3}^{\infty}\frac{1}{\ln(k \ln k)}
 
Determine the convergence or the divergence of the infinite series.

k=31ln(klnk)\displaystyle \sum_{k = 3}^{\infty}\frac{1}{\ln(k \ln k)}
From experience I can calmly tell that the infinite series is divergent.
 
From experience I can calmly tell that the infinite series is divergent.
Say we want to show that k=31ln(klnk)>k=31k \displaystyle{\sum_{k=3}^\infty \dfrac{1}{\ln\left(k \ln k\right)}} > \sum_{k=3}^\infty \dfrac{1}{k} which would be sufficient to show divergence. Then we needed ln(klnk)<k \ln\left(k \ln k\right)< k or equivalently klnk<ek. k\ln k < e^k. I would be satisfied with the remark that klnk<k2<ek k\ln k<k^2<e^k but you can of course prove this.
 
Say we want to show that k=31ln(klnk)>k=31k \displaystyle{\sum_{k=3}^\infty \dfrac{1}{\ln\left(k \ln k\right)}} > \sum_{k=3}^\infty \dfrac{1}{k} which would be sufficient to show divergence. Then we needed ln(klnk)<k \ln\left(k \ln k\right)< k or equivalently klnk<ek. k\ln k < e^k. I would be satisfied with the remark that klnk<k2<ek k\ln k<k^2<e^k but you can of course prove this.
Thanks professor for passing by.

To be more precise:
I know that 1ln(klnk)>12k\displaystyle \frac{1}{\ln(k\ln k)} > \frac{1}{2k}, and from previous problem I have shown that k=31k\displaystyle \sum_{k=3}^{\infty}\frac{1}{k} is a divergent series,

therefore

k=31ln(klnk)\displaystyle \sum_{k = 3}^{\infty}\frac{1}{\ln(k \ln k)} is divergent too by comparison test.
 
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