Uniformly convergent and Weierstrass M-test Heres my question:
I used The Weierstrass M-test showing that $|x^n(1-x)|\le |x|^n(1-x)$ and $$\sum_{n=0}^\infty |x|^n(1-x) = (1-x)\sum_{n=0}^\infty |x|^n=(1-x)\frac{1}{1-x}=1 < \infty$$
So the series is uniformly convergent.
However, $sup|x^n(1-x)-1|=1\neq0$, showing that it is not pointwise convergent.
Is there something wrong? Thank you!
 A: First, note that a hypothesis of Weierstrass test is that the series of bounding constants converges.
Note that for $x = 1$ we have $\sum_{n \geq 1}x^{n}(1-x) = 0$; for $-1 < x < 1$ we have $\sum_{n \geq 1}x^{n}(1-x) = x$; hence the series does not converge uniformly on $]-1,1]$. 
A: The statement of the Weierstrass M-Test goes something like this:
Let $\sum_{n=1}^{\infty} f_{n}(x)$ be a series of functions. Suppose there exists a convergent series $\sum_{n=1}^{\infty} M_{n} < \infty$ and that for all $n \in \mathbb{N}$ we have $|f_{n}(x)| \leq M_{n}$, for any $x \in A$. Then the series of functions $\sum_{n=1}^{\infty} f_{n}(x)$ converges uniformly on the domain $A$.
In short, you have not shown that yet. 
The series of functions $\sum_{n=1}^{\infty} |x|^{n}(1-x)$ converges pointwise on $(-1,1]$ (hint: use geometric series for each $x$, except $x = 1$). 
However, it is probably worthy to note that the series is not uniformly convergent. This can be shown using some basic results in Real Analysis. Consider the function $F: (-1,1]: \to \mathbb{R}$ defined by $F(x) = \sum_{n=1}^{\infty} |x|^{n}(1-x)$. Can you find the value of $F(x)$ for each $x \in(-1,1]$? If the series was uniformly convergent, what can you say about $F(x)$?
