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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?

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    $\begingroup$ integral test against $\frac{1}{x\log(x)^{1+2\epsilon}}$ $\endgroup$ – achille hui May 21 at 6:22
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You may use Cauchy condensation test:

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

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Hint

You have

$$\int_2^x \frac{dt}{t (\log t)^{1+2\epsilon}} = \frac{1}{(\log x)^{2 \epsilon}}-\frac{1}{(\log 2)^{2 \epsilon}}$$

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

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