# How to show $\ln(\ln(k+1))-\ln(\ln(k))<\frac{1}{k \ln(k)},\forall K\in \mathbb{N}, K\geq 2.$

How can I show the inequality $\ln(\ln(k+1))-\ln(\ln(k))<\dfrac{1}{k\ln(k)},\forall K\in \mathbb{N}, K\geq 2.$

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From Mean value theorem we have, $$\displaystyle\frac{\ln(\ln(k+1))-\ln(\ln(k))}{k+1-k}=\frac{1}{\ln c}.\frac{1}{c},c\in(k,k+1)$$ $$k<c\Rightarrow\frac{1}{\ln c}.\frac{1}{c}<\frac{1}{\ln k}.\frac{1}{k}$$ $$\Rightarrow\displaystyle{\ln(\ln(k+1))-\ln(\ln(k))}=\frac{1}{\ln c}.\frac{1}{c}<\frac{1}{\ln k}.\frac{1}{k}$$ $$\Rightarrow\displaystyle{\ln(\ln(k+1))-\ln(\ln(k))}<\frac{1}{\ln k}.\frac{1}{k}$$ We are done.
By the way is it possible to show $\sum_{k=2}^{\infty}(ln(ln(k+1))-ln(lnk))=\infty$ so that by comparison test $\sum_{K=2}^{\infty}\frac{1}{klnk}$ also divergent? – ftolessa Feb 9 '13 at 8:13
Another test which you can use $\sum a_i$ is convergent if and only if $\sum 2^ka_{2^k}$ is convergent. – Abhra Abir Kundu Feb 9 '13 at 9:02