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How can we show that the series $$\sum_{n=2}^\infty\frac{1}{(\ln n)^x}$$ diverges for $x>1$ ?

The book gives the following hint: consider $$\sum_{k=1}^\infty\sum_{n=M_{k-1}+1}^{M_k}\frac{1}{(\ln M_k)^x}$$ where $\ln M_k=k$ and note that $M_k-M_{k-1}=e^{-1}(e-1)M_k$; hence show that the series diverges.

But I really can't figure out what this hint means.

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Note that by l'Hôpital $$\lim_{n\to\infty} \frac{\ln n}{n^{\frac{1}{x}}} = \lim_{n\to\infty} \frac{\frac{1}{n}}{\frac{1}{x}n^{\frac{1}{x}-1}} = \lim_{n\to\infty} \frac{x}{n^{\frac{1}{x}}} = 0.$$ Thus, for every $x>1$ there is some $N\in\mathbb{N}$, such that for all $n>N$, $$\frac{\ln n}{n^{\frac{1}{x}}}<1$$ or equivalently $\ln n < n^{\frac{1}{x}}$.

Thus $$\sum_{n=2}^\infty \frac{1}{(\ln n)^x} = \sum_{n=2}^N\frac{1}{(\ln n)^x} + \sum_{n=N+1}^\infty\frac{1}{(\ln n)^x} > \sum_{n=2}^N\frac{1}{(\ln n)^x} + \sum_{n=N+1}^\infty\frac{1}{n}=\infty.$$

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  • $\begingroup$ Your answer is much more comprehensible than the hint given in the book. Thank you very much. $\endgroup$ – Gary May 21 '13 at 14:40
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Or; as the following limit: $$\lim_{n\to\infty} n^1\times\frac{1}{\ln^x(n)}=\infty,~~x>1$$ so the Comparison Test tells us the series diverges.

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  • $\begingroup$ $+! \land \ddot\smile$ $\endgroup$ – Namaste May 23 '13 at 0:12
  • $\begingroup$ Wow this is even simpler, but at first I found it hard to believe that $n > (\ln n)^x$ when $n\to\infty$. Anyway thanks! $\endgroup$ – Gary May 23 '13 at 12:57
  • $\begingroup$ @Gary: See this one math.stackexchange.com/a/398964/8581. :-) $\endgroup$ – mrs May 23 '13 at 18:46

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