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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.

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  • $\begingroup$ Pointwise convergence ($g_n(x) \to 0$) does not imply uniform convergence ($\|g_n\|_\infty \to 0$) in general. Hint: The $r<1$ condition is there for a reason, $f_n \not\to f$ uniformly on $(0,1)$. Only pointwise convergence is guaranteed there. $\endgroup$ – AlexR Oct 16 '14 at 22:29
  • $\begingroup$ Can you see my edit? Now I think it is correct. $\endgroup$ – Giiovanna Oct 16 '14 at 22:33
  • $\begingroup$ Thanks for all the answers, Ive figured out this by myself after a while. $\endgroup$ – Giiovanna Oct 16 '14 at 22:34
  • $\begingroup$ Yes it is correct this way. You should put that edit into a complete answer of the question and mark it as solved to help future readers ;) Alternatively pick the best answer and add a comment to it pointing out the details you found. $\endgroup$ – AlexR Oct 16 '14 at 22:38
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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 $\infty$.

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The problem is that $g_n$ has a minimum at $0$: $g_n(0) \equiv 0 \ \forall\ n\in\mathbb N$. The maximum is indeed attained at $\min(r, 1+\frac1n)=r$ as can be seen from monotonicity on $0<1+\frac1n$.

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Hint: You must show that $\|g_n\|\to0$. And $$\|g_n\|\leq r^{n+1}$$

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