# Uniform Convergence: Explaining f_n(x)=1-x^n all over 1-x

• jack5322
In summary, the question is asking about the uniform convergence of the function f_n(x)= 1-x^n/(1-x) and its independence on the variable x. If the function is not uniformly convergent, a proof and explanation is requested. The discussion also includes the convergence of a sequence of functions on a compact set and its uniform convergence on a closed interval. The question also asks for a rigorous proof of the convergence of (1+z/n)^n on any closed interval containing zero and a specific number R. Suggestions for approaching this proof are also requested.

#### jack5322

could someone please explain if f_n(x)= 1-x^n all over 1-x is uniformly convergent? can someone show why its independent of x if it is and on what intervals? Also, can someone explain if it isn't uniform convergent then can they show a proof and explain. Thanks any help will be greatly appreciated!

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

n is changing, and its going to infinity

If x lies between -1 and 1, then xn converges to 0 so 1- xn converges to 1. If x> 1 or x< -1 then xn does not converge at all so 1- xn does not converge. If x= 1, 1- xn= 0 for all x while if x= -1, 1- xn= 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- xn converges uniformly on $[\epsilon, 1]$ for any $\epsilon> 0$ but does NOT converge uniformly on (0, 1].

thank you

HallsofIvy said:
If x lies between -1 and 1, then xn converges to 0 so 1- xn converges to 1. If x> 1 or x< -1 then xn does not converge at all so 1- xn does not converge. If x= 1, 1- xn= 0 for all x while if x= -1, 1- xn= 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- xn converges uniformly on $[\epsilon, 1]$ for any $\epsilon> 0$ but does NOT converge uniformly on (0, 1].

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

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how would we rigorously prove that (1+z/n)^n converges uniformly on any closed interval containing zero and R where R is a number between zero and positive infinity? I was thinking of using cauchy's convergence criterion but I always get stuck. Any suggestions?