Prove $\sum_{n=1}^{\infty} \frac 1 {n(\ln n)^x}$ is congervent $x > 1$ and divergent $x \le 1$ using Cauchy theorem. I've proved the theorem due to Cauchy: $$\sum_{n=1}^{\infty}a_n \le \sum_{n=1}^{\infty}2^na_{2^n} \le 2 \sum_{n=1}^{\infty}a_n$$
and I've showed that $\sum_{n=1}^{\infty}\frac 1 {n^x}$ is convergent when $x>1$ and divergent when $x \le 1$ by rewriting as a geometric series and using the inequality above.
However I have trouble proving $\sum_{n=1}^{\infty} \frac 1 {n(\ln n)^x}$ is convergent when $x>1$ and divergent when $x \le 1$.
I know I can write $\sum_{n=1}^{\infty} 2^n\frac 1 {2^n(\ln 2^n)^x}=\frac 1 {(\ln 2^n)^x}$ but from here I cannot reduce to some know convergent series.
Could I get some advice ?

After you applied Cauchy theorem,
$$\sum_{n=1}^{\infty} 2^n\frac 1 {2^n(\ln 2^n)^x}=\sum_{n=1}^{\infty}\frac 1 {(\ln 2^n)^x}$$
But what is $(\ln 2^n)^x$?
$$(\ln 2^n)^x=(n\ln 2)^x=(\ln 2)^x\cdot n^x$$
Thus,
$$\sum_{n=1}^{\infty}\frac 1 {(\ln 2^n)^x}=\sum_{n=1}^{\infty}\frac 1 {(\ln 2)^x\cdot n^x}=\frac{1}{(\ln 2)^x}\sum_{n=1}^{\infty}\frac{1}{n^x}$$
Where the is last convergent when $x>1$ and divergent when $x\leq 1$, as I believe, you already shown.  
