# Uniform convergence and sequence of functions

I was asked to prove that $f_n(x) = (1-x)^{\alpha}x^n$ is uniformly convergent on $[0,1]$ where $\alpha>0$ I did this part then I was asked if wether the corresponding series of functions converges uniformly on the given interval ? And if it does for what alpha it should be ? Hints are really appreciated ? Thanks

• What do you mean by corresponding series of functions? You mean $\sum f_n$? Commented Dec 12, 2015 at 5:16

Let $S_n(x)=\sum _{k=0}^n(1-x)^\alpha x^k=(1-x)^\alpha \sum_{k=0}^nx^k=(1-x)^\alpha\dfrac{1-x^n}{1-x}=(1-x)^{\alpha-1}(1-x^n)$
$S(x)=\lim_{n\to\infty}S_n(x)=(1-x)^{\alpha-1}(1-x^n)$ which can be checked to be discontinuous irrespective of $\alpha$
• The limit of $S_n(x)$ for $0\leq x <1$ is $(1-x)^{(\alpha -1)}$ and every $S_n(1)=0$ so $S(1)=0$....$S(x)$ is discontinuous at $x=1$ iff $\alpha\leq 1$. Commented Dec 12, 2015 at 9:55
• how did you get " iff "; take $\alpha=2$ then $S_n(0)=1;S_n(1)=0$ @user254665 Commented Dec 12, 2015 at 13:26