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

• integral test against $\frac{1}{x\log(x)^{1+2\epsilon}}$ – achille hui May 21 at 6:22

$$\sum_{n=1}^\infty \frac{1}{n(\log(n))^{1+2\epsilon}} \sim \sum_{n=1}^\infty \frac{2^n}{2^n(\log(2^n))^{1+2\epsilon}} = \sum_{n=1}^\infty \frac{1}{(n\log(2))^{1+2\epsilon}}< \infty$$
$$\int_2^x \frac{dt}{t (\log t)^{1+2\epsilon}} = \frac{1}{(\log x)^{2 \epsilon}}-\frac{1}{(\log 2)^{2 \epsilon}}$$
Proving that $$\int_2^\infty \frac{dt}{t (\log t)^{1+2\epsilon}}$$ converges. You can then use a series / integral comparison.