Prove that $\sum_{n=1}^{\infty} (\frac{\ln x}{x})^n$ converge uniformly

Question

How to prove that $\sum_{n=1}^{\infty} (\frac{\ln x}{x})^n$ is uniformly convergent for $ x \in [1,+\infty)$, I tried the Weierstrass M-test but it doesn't work. Is there another way to prove this?

Answer 1

Observe \begin{align} \log x \leq \begin{cases} x-1 & \text{ if } 1 \leq x \leq 2\\ \frac{1}{2}x & \text{ if } x> 2 \end{cases} \end{align} which means \begin{align} \frac{\log x}{x} \leq \frac{1}{2} \end{align} for all $x\geq 1$. Then it follows \begin{align} \sum^\infty_{n=1}\left(\frac{\log x}{x}\right)^n \leq \sum^\infty_{n=1}\frac{1}{2^n}. \end{align}

Answer 2

If you consider $$f(x)=\frac{\log(x)} x\implies f'(x)=\frac{1-\log (x)}{x^2}\implies f''(x)= \frac{2 \log (x)-3}{x^3}$$ So,$f'(x)=0$ if $x=e$ and this a maximum (by the second derivative test). So, for $x \in [1,+\infty)$, $f(x)\leq \frac 1 e$.

Then, for any $x \in [1,+\infty)$, $$\begin{align} \sum^\infty_{n=1}\left(\frac{\log x}{x}\right)^n \lt \sum^\infty_{n=1}\frac{1}{e^n}=\frac{1}{e-1} \end{align}$$