Divergence of $\sum_{n=2}^\infty\frac{1}{(\ln n)^x}$ for $x>1$

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.

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 $$\sum_{n=2}^\infty \frac{1}{(\ln n)^x} = \sum_{n=2}^N\frac{1}{(\ln n)^x} + \sum_{n=N+1}^\infty\frac{1}{(\ln n)^x} > \sum_{n=2}^N\frac{1}{(\ln n)^x} + \sum_{n=N+1}^\infty\frac{1}{n}=\infty.$$
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.
• $+! \land \ddot\smile$ – Namaste May 23 '13 at 0:12
• 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! – Gary May 23 '13 at 12:57