Convergence of $\sum_{n=1}^\infty \frac{\mathrm{ln}(n)}{n^x}$ for $x>1$ 
Show $\displaystyle\sum_{n=1}^\infty \frac{\mathrm{ln}(n)}{n^x}$ is convergent for $x>1$.

I've tried ratio test and I get
$$\underset{n \rightarrow \infty}{\mathrm{lim}} \left\lvert \frac{\mathrm{ln}(n+1)n^x}{(n+1)^x\mathrm{ln}(n)} \right\rvert = 1$$
so I can't really conclude anything there.
So then I look at the limit comparison test:
$$\underset{n \rightarrow \infty}{\mathrm{lim}} \frac{\mathrm{ln}(n)}{n^x}=0$$
so this is not helpful.
The end result I am really trying to prove here is that the Riemann zeta function converges uniformly on $[a, +\infty)$ (which I have done already) where $a>1$ and that $\displaystyle\zeta'(x) = \sum_{n=1}^{\infty} \frac{- \mathrm{ln}(n)}{n^x}$ for $x>1$.
The way that I showed the Riemann zeta function converges uniformly on $[a,\infty)$ was by using the Weierstrass $M$-test and noting that $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^x}$ converges for $x>1$ by the $p$-test. 
 A: Using the integral convergence test with $f(n)=\frac{\ln(n)}{n^{x}}$, we have:
$$\int_{1}^{\infty}\frac{\ln(n)}{n^{x}}\:dn=\lim_{a\to\infty}\int_{1}^{a}\frac{\ln(n)}{n^{x}}\:dn=\lim_{a\to\infty}\left[-\frac{1+(x-1)\ln(n)}{n^{x-1}(x-1)^2}\right]_{1}^{a}=\left(-\frac{1}{(x-1)^{2}}\right)-\left(0\right)$$
Therefore, for $x>1$, the integral converges and thus so does the series.
A: The function $f(n) = \frac{\ln(n)}{n^x}$ is positive and becomes decreasing for $n > M$ ($M$ depends on $x$). Use the Cauchy condensation test:
$$
2^n\frac{\ln(2^n)}{2^{xn}} = \frac{n\ln 2}{2^{(x-1)n}}
$$
Since $x > 1$, we have $x - 1 > 0$, $2^{x-1} > 1$ and the series converges.
A: Convergence of this series is provable using 'not helpful' limit
$$\lim_{n\to\infty} \frac{\ln n}{n^x}=0$$
for all $x>0$.
Let $x>1$, and take $c$ such that $0<c<x-1$. By previous limit, there exists $N$ such that
$$n>N \implies \left| \frac{\ln n}{n^{x-1-c}} \right|<1$$
. So if $n>N$ then
$$\frac{\ln n}{n^x}<\frac{1}{n^{1+c}}.$$
So
$$\sum_{n=1}^\infty \frac{\ln n}{n^x}<\sum_{n=1}^N \frac{\ln n}{n^x}+\sum_{n>N} \frac{1}{n^{1+c}}$$
Since each term of series (on previous inequality) is non-negative, so by comparison test, this series converges.
A: Yet another way: Compare with $$\sum \frac{1}{n^a}$$
for some $1 < a < x$. (Limit comparison test is the easiest here.)
