Prove the uniform convergence of $\sum \frac{x^n(1-x)}{\log(n+1)}$ for $x \in [0,1]$

Question

As it's told on the title, need to prove the uniform convergence of $\sum_{n=1}^{n=\infty} \frac{x^n(1-x)}{\log(n+1)}$ for $ x \in I=[0,1]$.

I tried to find the $\sup_{x \in I}g_n(x)$ (where $g_n(x)$ it's the function $\frac{x^n(1-x)}{\log(n+1)}$) to prove the uniform convergence with the Weierstrass-M test. The maximum value it's for $x=\frac{n}{n+1}$. But then I get to prove $\sum\frac{n^n}{(n+1)^{n+1}\log(n+1)}$ is conditionally convergent

Much thanks.

Answer 1

Hint: We want to show

$$\tag 1 \sum_{n=M}^{N}\frac{x^n(1-x)}{\ln (n+1)}$$

is uniformly small if $M$ is large. Try summation by parts.