this is a hard one I tried many times and I always failed here is it: $\displaystyle \sum_{n=2}^{\infty} \frac{1}{ln^e \, (n)}$
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Originally Posted by TWiX this is a hard one I tried many times and I always failed here is it: $\displaystyle \sum_{n=2}^{\infty} \frac{1}{ln^e \, (n)}$ There always exists $\displaystyle N_k\in\mathbb{N}\,\,\,s.t.\,\,\,n>N_k\Longrightarr ow \ln^k(n)<n$ , no matter what $\displaystyle k\in\mathbb{N}$ is, so... Tonio
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