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Consider $f_n(x) = \sum_{k=0}^{n} {x^k}$. Does $f_n$ converge pointwise on $[0,1]$? Does it converge uniformly on $[0,1]$?

Well, my approach would be, first of all to notice that if $x \neq 1$, then by simple induction we get $$f_n(x) = \frac{1 - x^{n+1}}{1-x}$$

So, $(f_n) \rightarrow \frac{1}{1-x}$. But at $x = 1$, the $f_n \rightarrow \infty$, hence $(f_n)$ does not converge pointwise. Therefore, it does not converge uniformly. Is this correct? Hope to get feedback


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Yes, you are correct (since uniform convergence would imply pointwise convergence). Note that for any $0<b<1$, $f_n$ converges uniformly (and hence pointwise) on $[0,b]$. – copper.hat Nov 22 '12 at 6:48
May be good to show uniform -> pointwise as an exercise for yourself. – Frudrururu Mar 31 '13 at 21:18
up vote 3 down vote accepted

What you have looks good. I thought I'd chime in with something a little different.

$f_n(x)\to f(x)$ pointwise on $[0,1]$ iff for each $x\in[0,1]$, $|f_n(x)-f(x)|\to 0$ as $n\to\infty$. This clearly fails at $x=1$ for any proposed function $f(x)$ since $f_n(1)=n+1\to\infty$ as $n\to\infty$. And since pointwise convergence on $[0,1]$ fails, then uniform convergence on $[0,1]$ fails as well.

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