How to use the Comparison Test to investigate the convergence of $\sum (\ln n)/n^\alpha$?

Question

Let $$\sum\limits_{n=1}^\infty \frac{\ln n}{n^\alpha}, \alpha\in\Bbb{R}$$

I need to investigate the convergence of this series.

I've read that since the series is positive for all $n$ then it converges if and only if $\alpha > 1$, but isn't it true only for a p-series ($\sum \frac{1}{n^p})$?

For $\alpha \le 1$ the auther mention we can use this equality:
$$\frac{1}{n} \le \frac{\ln n}{n^\alpha}. \forall n\in\Bbb{N}$$

And then compare it to the harmonic series and we're done. But this claim is non-trivial. I wish to know how to prove it before using it. I tried to do so by induction but got stuck somewhere.

I'd be glad if you could help me understand it properly.

Thanks.

Answer 1

• If $\alpha\le1$ then $$\frac1n\le\frac{\ln n}{n^\alpha},\quad n\ge3$$ so by comparison with the harmonic series the given series is divergent.
• If $\alpha>1$ then let $1<\alpha'<\alpha$ so $$\frac{\ln n}{n^\alpha}\le\frac1{n^{\alpha'}},\;\text{for}\; n \;\text{sufficently large}$$ so the given series is convergent by comparison with a convergent Riemann series.

Answer 2

Use the properties $\varepsilon^{-1}\log(n^{\varepsilon})=\log(n)$ and $\log(n)<n$ to get an upper bound in a simpler (power) form.