Weierstrass-M test find $a_n$

Question

We have $\sum_{n=2}^{\infty} \frac{log(n)^x}{n}$ for $x\in$R. We let r<-1 and I have to show that $\sum_{n=2}^{\infty} \frac{log(n)^x}{n}$ is uniform convergent for all x $\leq$r. I think I can use Weierstrass-M test? So I was searching for $a_n$ so $\mid$ $f_n(x)\mid$ $\leq$$a_n$. But it has not been possible for me to find. Can anyone help me?

Answer 1

hint

$$x\le r \implies -x\ge -r>1$$

for $n $ great,

$$\frac{1}{n(\ln(n))^{-x}}\le \frac{1}{n(\ln(n))^{-r}}$$

and by comparison with the improper integral,

$$\int^{+\infty}\frac{(\ln(u))^{r}du}{u}$$

whose antiderivative is

$$\frac{(\ln(u))^{r+1}}{r+1}$$

with $ r+1<0$

we conclude that the convergence is uniform.