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- Thread starter jack5322
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mathman

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Your question is incomplete. What variable (n???) is changing and what is it going to?

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n is changing, and its going to infinity

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HallsofIvy

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thank you

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mathman

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^{n}converges to 0 so 1- x^{n}converges to 1. If x> 1 or x< -1 then x^{n}does not converge at all so 1- x^{n}does not converge. If x= 1, 1- x^{n}= 0 for all x while if x= -1, 1- x^{n}= 0 for even n and 2 for odd n and so 1- (-1)^{n}does not converge. If a sequence of functions converges on a compact (closed and bounded) set, then it converges uniformly on that set so 1- x^{n}converges uniformly on [itex][\epsilon, 1][/itex] for any [itex]\epsilon> 0[/itex] but does NOT converge uniformly on (0, 1].

I believe the interval of convergence should be[itex][\epsilon-1, 1][/itex]

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