Uniform convergence on compact subset Let there be two functional squences $$a_n(x)=\sqrt[n]{x} \quad \textrm{ for $x\in(0,\infty)$}$$ $$b_n(x)=\sum_{k=0}^{n}x^k(1-x)^k=\frac{1-x^{n+1}(1-x)^{n+1}}{x^2-x+1} \quad \textrm{ for $x\in (0,1)$}$$
The pointwise converge to the functions $a(x)=1$ and $b(x)=\frac{1}{x^2-x+1}$. How can I prove that $a_n$ and $b_n$ converge uniformly to $a$ and $b$ on any compact subinterval. My idea for $a_n$ would be if we have the subinterval $[p,q]\subset(0,\infty)$, then $\|a_n-a\|=sup\{ \lvert \sqrt[n]x-1\rvert, x\in[p,q]\}=\lvert\sqrt[n]q-1\rvert\rightarrow 0$, for $n\to\infty$. So it should be convergening uniformly on any compact subinterval. But does it also converges uniformly on the complete interval? I think not but I dont know how to prove it...
My idea for $b_n$: similar like up there let $[p,q]\subset(0,1)$, $\|b_n-b\|=\sup\{\lvert\frac{x^{n+1}(1-x)^{n+1}}{x^2-x+1}\rvert\, x\in[p,q]\}$. Now, we can calculate that he maximum value of $\frac{x^{n+1}(1-x)^{n+1}}{x^2-x+1}$ is at $x=\frac{1}{2}$, so if $\frac{1}{2}\notin[p,q]$ we just take the one closer to $\frac{1}{2}$, let it be $t$, and now we have that $\|b_n-b\|=\frac{t^{n+1}(1-t)^{n+1}}{t^2-t+1}\to 0$, for $n\to\infty$, so it too converges uniformly in the compact subinterval. But here comes also the question: does it also converge uniformly on the whole interval $(0,1)$? And is my proof up til now good?

For the first question:
If $x \in [p,q]$ with $p \geqslant 1$ then $h_n(x) = a_n(x) - 1 = x^{1/n} - 1 \geqslant 0,$ and using Bernoulli's inequality
$$x = (1 +h_n(x))^n \geqslant nh_n(x)\\ \implies |x^{1/n}-1| = x^{1/n} - 1 = h_n(x) \leqslant \frac{x}{n} \leqslant \frac{q}{n} \to 0$$
If $x \in [p,q]$ with $q < 1$ then $h_n(x) = x^{-1/n} - 1 > 0,$ and using Bernoulli's inequality
$$x = \frac{1}{(1 +h_n(x))^n} \leqslant \frac{1}{nh_n(x)}\\ \implies |x^{1/n}-1| = 1- x^{1/n} = \frac{h_n(x)}{1 + h_n(x)} \leqslant h_n(x) \leqslant\frac{1}{nx} \leqslant \frac{1}{pn} \to 0$$
If $p < 1 < q$ then we have uniform convergence on $[p,1]$ and $[1,q]$ -- hence, uniform convergence on $[p,q]$
For the second question, you are done. You found the maximum at $x = 1/2$ and 
$$\sup_{x \in (0,1)}|b_n(x) - b(x)| = \frac{4}{3\cdot2^{2n+2}} \to 0.$$
Note that $|b_n(0) - b(0)| =|b_n(1) - b(1)| = 0$ so the convergence is uniform on the open interval.
