infinite series with natural logarithm

[logistic_guy]

Determine the convergence or the divergence of the infinite series.

\(\displaystyle \sum_{k = 3}^{\infty}\frac{1}{\ln(k \ln k)}\)

 

[nessmell]

Dating for Sex.

 

[logistic_guy]

Determine the convergence or the divergence of the infinite series.

\(\displaystyle \sum_{k = 3}^{\infty}\frac{1}{\ln(k \ln k)}\)

From experience I can calmly tell that the infinite series is divergent.

 

[fresh_42]

From experience I can calmly tell that the infinite series is divergent.

Say we want to show that [imath] \displaystyle{\sum_{k=3}^\infty \dfrac{1}{\ln\left(k \ln k\right)}} > \sum_{k=3}^\infty \dfrac{1}{k}[/imath] which would be sufficient to show divergence. Then we needed [imath] \ln\left(k \ln k\right)< k [/imath] or equivalently [imath] k\ln k < e^k. [/imath] I would be satisfied with the remark that [imath] k\ln k<k^2<e^k [/imath] but you can of course prove this.

 

[logistic_guy]

Say we want to show that [imath] \displaystyle{\sum_{k=3}^\infty \dfrac{1}{\ln\left(k \ln k\right)}} > \sum_{k=3}^\infty \dfrac{1}{k}[/imath] which would be sufficient to show divergence. Then we needed [imath] \ln\left(k \ln k\right)< k [/imath] or equivalently [imath] k\ln k < e^k. [/imath] I would be satisfied with the remark that [imath] k\ln k<k^2<e^k [/imath] but you can of course prove this.

Thanks professor for passing by.

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

therefore

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