analysis - Showing that $sum frac{log n}{n^x}$ converges for $x>1$

I'm trying to show that $\sum \frac{\log n}{n^x}$ converges for $x>1$ by the ratio test. Here's what I've got so far $$\frac{a_{n+1}}{a_n} = \frac{\log (n+1) n^x}{(n+1)^x \log n}$$
$$=\left(\frac{n}{n+1}\right)^x \frac{\log (n+1)}{\log n}$$ but I can't see how to manipulate the $\frac{\log (n+1)}{\log n}$ term to make this congerge to a limit less than 1, can anyone help?

Answer

The reason you can't do that is because the limit of those terms is $1,$ so the ratio test won't work here. Instead, note that $\displaystyle f(t) = \frac{ \log t}{t^x} $ is eventually decreasing, since $\displaystyle f'(t) = \frac{1- x \log t}{t^{x+1}}$. Then either the integral test or Cauchy condensation will finish this off.