Proving logarithmic property for series convergence

Question

Given a series $\sum_{n=1}^\infty a_n$, $a_n>0$. I need to prove that if $\alpha>0$ and $n_0 \in N$ exist, such that for every $n \ge n_0$, it is true that $\frac{\ln(\frac{1}{a_n})}{\ln(n)} \ge 1 + \alpha$ then the series converges. Also if $n_0 \in N$, such that for every $n \ge n_0$ it is true that $\frac{\ln(\frac{1}{a_n})}{\ln(n)} \le 1$, the series is divergent.

Answer 1

Hint:

Take the exponential of both sides:

\begin{align} \frac{\log(\frac{1}{a_n})}{\log(n)} \ge 1 + \alpha&\iff-\log a_n\ge( 1 + \alpha)\log n\iff\log a_n\le-( 1 + \alpha)\log n)\\ &\iff a_n\le\frac1{n^{1+\alpha}}, \end{align} which is a convergent Riemann series.

Similarly, the second hypothesis is equivalent to $\;a_n\ge \dfrac 1n$.