latentcorpse
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Does [itex]\sum_{n=0}^{\infty} x^n(1-x)[/itex] converge uniformly on [itex](0,1)[/itex]?
[itex]S_n(x)=\sum_{k=0}^{\infty} x^k(1-x)=\frac{1-x^{n+1}}{1-x}(1-x)=1-x^{n+1}} \rightarrow 1[/itex] as [itex]n \rightarrow \infty[/itex]
so we get pointwise convergence to 1
now we test for uniform convergence
[itex]d_{\infty}(S_n,1) = sup |S_n(x)-1|= sup x^{n+1}[/itex] now remembering that x is in (0,1), i said that this should tend to 0 as n goes to infinity giving uniform convergence but the answers say that it doesn't go to 0 and so convergence is not uniform, merely pointwise - why is this?
[itex]S_n(x)=\sum_{k=0}^{\infty} x^k(1-x)=\frac{1-x^{n+1}}{1-x}(1-x)=1-x^{n+1}} \rightarrow 1[/itex] as [itex]n \rightarrow \infty[/itex]
so we get pointwise convergence to 1
now we test for uniform convergence
[itex]d_{\infty}(S_n,1) = sup |S_n(x)-1|= sup x^{n+1}[/itex] now remembering that x is in (0,1), i said that this should tend to 0 as n goes to infinity giving uniform convergence but the answers say that it doesn't go to 0 and so convergence is not uniform, merely pointwise - why is this?