Uniform convergence of complex sequence ''Does the sequence ${[4x(1-x)]^n}$ converge uniformly on the open interval where 0 < x < 1?''
I dont know what to do with the open interval information. Is the Wierstrass-M-Test appropriate here to determine uniform convergence? How would things be different if the interval was closed?
 A: Hint: The Uniform Convergence Theorem says that if a sequence of continuous functions converge uniformly to a limit, then that limit is continuous.
A: For this particular problem, the interval being open has no bearing on the answer. But you will run into examples of sequences of functions that converge uniformly on an open interval, but not on that same interval if it is closed. To prove uniform convergence, you have to figure out if there is a continuous function $f$ such that for all $\varepsilon>0$, there exists $N \in \Bbb{N}$ where $$\left|[4x(1-x)]^n-f(x) \right|<\varepsilon$$ for all $n \geq N$ and for all $x\in (0,1)$. To show a function does not converge uniformly, it suffices to show the sequence converges to a non-continuous function. In this train of thought, it may be useful to show that $[4(1/2)(1-(1/2))]^n=1$ for all $n \in \Bbb{N}$, and for any $x \neq 1/2$ that $0\leq [4(1/2)(1-(1/2))]^n <1$. Then, compare $$\lim_{n \to \infty} [4x(1-x)]^n$$ for the case of $x = 1/2$ versus the case of $x \neq 1/2$.
