Find the domain of $x$ for the convergence of the series $\sum_{n=1}^{\infty} \left(\frac{\ln(n)}{n}\right)^{x}$

Question

Question: Find the domain of $x$ for the convergence of the series $$\sum_{n=1}^{\infty} \left(\frac{\ln(n)}{n}\right)^{x}.$$

My Approach: The series $\sum_{n=1}^{\infty}$$\left(\frac{\ln(n)}{n}\right)^{x}$ diverges for $x\leq0$ by $n$th term divergence test.

The series diverges for $x=1$ by Investigate convergence of $\sum_{n=1}^\infty \frac{\ln(n)}{n}$. I can not prove that series diverges for 0 $< x < 1$.

The Book mentions the answer $(1,\infty)$.

Answer 1

Hint. Note that for $0< x\leq 1$, and $n\geq 3$, $$\frac{1}{n^x}\leq \left(\dfrac{\log n}{n}\right)^{x}.$$ Moreover, for $x> 1$, $$\lim_{n\to +\infty} \frac{\left(\dfrac{\log n}{n}\right)^{x}}{\dfrac{1}{n^{(x+1)/2}}}=\lim_{n\to +\infty} \frac{(\log n)^{x}}{n^{(x-1)/2}}=0$$ implies that for $n$ sufficiently large, $$\left(\dfrac{\log n}{n}\right)^{x}\leq \frac{1}{n^{(x+1)/2}}.$$

Answer 2

For large $n$, $0<\dfrac{\log n}{n}<1$, so $\dfrac{\log n}{n}\leq\left(\dfrac{\log n}{n}\right)^{x}$ for $0<x<1$.