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

Question

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 :(

Answer 1

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

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

Answer 2

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 .