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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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Welcome to MSE! If this is homework, please tag it as such, people will help you anyway. And please try and tell us what you have tried so far, and where you got stuck. Also, please take a look at the revision of the formatting I did on your post. –  Andreas Caranti Feb 9 '13 at 6:57
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Actually, it is not homework question. The question is taken from one of the previous mathematics entrance examination. Thank you –  ftolessa Feb 9 '13 at 7:39

1 Answer 1

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.

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Thank for your you valuable time. Now it is okay. –  ftolessa Feb 9 '13 at 8:02
    
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
    
I guess yes@user61455 –  Abhra Abir Kundu Feb 9 '13 at 8:57
    
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

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