Value of sum and uniform convergence of $\sum_{n=0}^{\infty} (1-\log x)\log^{n}x$?

Question

Can you prove that this series converges uniformly with $x>0$? And can you calculate the sum?

I found that the derivative is $0$ when $x \ge e^{\frac{n}{n+1}}$. So i think that $\forall n \ge e^{\frac{\delta}{\delta+1}}$, with $\delta > 0$, we have $\sup_{x\ge\delta} f_n (x) = f_n (\delta)$, but i can't prove that $\sum_{n=0}^{\infty} (1-\log\delta)\log^{n}\delta$ converges (it must).

Answer 1

The sum converges uniformly when $|\log x|<1$ or $1/e<x<e$.

If this holds, the sum is a ordinary geometric series with sum, ignoring the $1-\log x$, of $\frac1{1-\log x}$ so the whole expression is just 1.