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For what values of $\alpha >0$ and $\beta >0$ does the following series $$\sum_{n=2}^\infty \frac{1}{n^{\alpha}(\log n)^{\beta}}$$ converge?

if $\alpha >1$ ,then limit comparison test suggest that given series converge for any $\beta >0$
Also when $\alpha=1$ given series converge iff $\beta >1$ by integral test.
Is it correct ? Also what will happen when $0 <\alpha <1$

Any suggestions?Thanks

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  • $\begingroup$ You are right. If $0<\alpha<1,$ you can still apply the limit comparison test (with $1/n$) to show that the series diverges. $\endgroup$ Dec 4 '20 at 19:29
  • $\begingroup$ Use Cauchy Condensation test $\endgroup$ Dec 4 '20 at 19:35
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What you said until using integral test is correct.

If $0 \lt \alpha \lt 1$, then as

$(\log n)^\beta \lt n^{\frac{1-\alpha}{2}}$ for $n $ large enough, you get

$$\frac{1}{n^\alpha(\log n )^\beta} \ge \frac{1}{n^{\frac{1+\alpha}{2}}}.$$

Hence the series diverges as $\frac{1+\alpha}{2}\le 1$.

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