It is asked to prove that the sequence $$f_n(x) = 1 + x + ... + x^n$$
Uniformly converges in $[0, r] (0<r<1)$ to
$$f(x)=\frac{1}{1-x}$$
That is, i need to show that
$$\lim_{n \to \infty} || f_n - f|| = 0$$
where $ ||f_n - f||= \sup_{x \in [0,r]}\{|f_n(x)-f(x)|; n \in N\}$
We can see that
$$g_n(x)=|f_n(x) - f(x) | = \frac{x^{n+1}}{1-x} $$
and $$\lim_{n\to \infty} g_n(x) =0, \forall x \in [0,r]$$
Does that proves that the limit of the norm goes to 0?
Also, Ive tried to find the maximum of g_n using the derivative and I've obtained that the derivative is zero in $x=0$ and $ x = 1+\frac{1}{n}$. Clearly, $1+1/n $is not convenient, so I've tried to shpw that x=0 is the maximum point. But I couldnt do this. If I could, then $g_n(0)=0$ is the supremum and it is proved.
Anyone can give me a little help to finish this? Thanks in advanced!
@Edit
$$g_n'(x) = \frac{x^n(n-nx+1)}{(x-1)^2} >0$$
Since x<1. Then, tg is increasing and its maximum occurs when x=r. Now, just applies limit in $g_n(r)$ and we are done.
You're on the right track, but what you really want to show is that, if we define $$M_n=\sup\{g_n(x):x\in[0,r]\}$$ that $\lim_{n\rightarrow\infty} M_n = 0$. Essentially, what you've proven is that the sequence converges pointwise, but you need this stronger condition on the maximum difference to show that it converges uniformly - for instance,in the interval $[0,1)$, the series converges pointwise, but not uniformly, since it takes a very long time to converge near $1$.
In particular, from the math you already have, noting that $g_n$ is an increasing function, it should be clear that $M_n=\frac{r^{n+1}}{1-r}$, which clearly goes to $0$ as $n$ goes to $\infty$.