Is f_n(x)=(1-x^n)/(1-x) uniformly convergent?

[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!

 

[mathman]

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

 

[jack5322]

n is changing, and its going to infinity

 

[HallsofIvy]

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].

 

[jack5322]

thank you

 

[mathman]

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$[\epsilon-1, 1]$

 

[jack5322]

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?