probability - Why does $sum_{n=1}^infty frac{1}{n(log(n))^{1+2epsilon}}$ converge?

I am looking through examples on convergences of random series, and in one of the proofs the following result is used: If $\epsilon > 0$ then

$$\sum_{n=1}^\infty \frac{1}{n(\log(n))^{1+2\epsilon}}<\infty$$

No explanation is given hence my confusion. I know that if $p>1$ then $\sum_{n} \frac{1}{n^p} < \infty$ but as this involes $\log(n)$, I do not understand how they reach this conclusion. Especially since $\sum_n \frac{1}{n(\log(n))}=\infty$.

Could someone explain how I would show this converges?

Answer

You may use Cauchy condensation test:

$$\sum_{n=1}^\infty \frac{1}{n(\log(n))^{1+2\epsilon}} \sim \sum_{n=1}^\infty \frac{2^n}{2^n(\log(2^n))^{1+2\epsilon}} = \sum_{n=1}^\infty \frac{1}{(n\log(2))^{1+2\epsilon}}< \infty$$