# Proving the uniform convergence of $\sum_{i=1}^{\infty}nx^n$ where $x \in [0,1)$ using Weirstrass M test.

I want to prove the uniform convergence of $$\sum_{i=1}^{\infty}nx^n$$ where $$x \in [0,1)$$. I know this sum has been asked a lot of times before but Im supposed to prove this with tools I have seen and this means not using any Theorem involving derivatives. Im trying to prove this using M Weierstrass test, so I want to bound each term of the sum. This means finding a sequence $$\lbrace M_{n} \rbrace \geq 0$$ such $$|nx^n| \leq M_{n}$$ for every $$n \in \mathbb{x}$$ and then proving $$\sum_{n=1}^{\infty} M_{n}< \infty$$. Im stuck finding a sequence which each term bound the original sum terms. I know that for every $$x \in [0,1)$$ we have$$0 \leq x^{n} \leq 1$$ then $$0 \leq n x^{n} \leq n$$ but the series $$\sum_{i=1}^{\infty} n$$ is not convergent :(

Hints:

Consider separately $$[0,a]$$ with $$a < 1$$ where convergence is uniform and $$[0,1)$$ where it is not.

For $$x \in [0,a]$$ with $$a < 1$$ take $$x = \frac{1}{1+y}$$ where $$y \geqslant \frac{1}{a}-1$$

We have $$nx^n = \frac{n}{(1+y)^n}$$ and by the binomial theorem $$(1+y)^n > \frac{n(n-1)(n-2)}{6}y^3,$$

so for $$n > 2$$, $$nx^n < \ldots$$

For $$x \in [0,1)$$ note that

$$\sum_{k=n}^{2n} k x^k > (n+1)(n)x^{2n}$$

The series is not uniformly convergent on $$[0,1)$$. If it is, then $$nx^{n}$$ must tend to $$0$$ uniformly as $$n \to \infty$$. In particular there exists $$n_0$$ such that $$|nx^{n}| <1$$ for all $$n \geq n_0$$ for all $$x \in [0,1)$$. Take $$x=\frac 1 {n^{1/n}}$$ with $$n=n_0$$ to get a contradiction .