# 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?

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

• You are right. If $0<\alpha<1,$ you can still apply the limit comparison test (with $1/n$) to show that the series diverges. Dec 4 '20 at 19:29
• Dec 4 '20 at 19:35

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 $$\frac{1+\alpha}{2}\le 1$$.