TITLE: $f_n(x):=nx(1-x)^n$ Determine whether the sequence $(f_n)$ converges uniformly on $[0,1]$
QUESTION [6 upvotes]: I am having a bit of trouble on this revision question.
To determine pointwise convergence: $\lim_{n\rightarrow\infty} = nx(1-x)^n $. For $x=0, x=1$, it's clear that the limit is $0$. How can I determine the limit for $0 \le x \le 1$? (My limit finding skills are rusty). I'm quite sure it is $0$ as well, but how do I go about showing this properly?
Supposing this is true, then to determine uniform convergence, let $d_n(x):= |f_n(x)-f(x)|= nx(1-x)^n - 0$
Then $$ d_n'(x)= n(1-x)^n - n^2(1-x)x = 0$$ if $x=1$. So the maximum of $f_n(x)$ occurs at $x=1$. It follows that:
$$0 \le d_n(x)= |f_n(x)-f(x)|< d_n(1) = 0$$ So $(f_n)$ converges uniformly 
on $[0,1]$. Would this be correct?
Thanks for the help in advance!

REPLY [4 votes]: To answer your first question about pointwise convergence for $0<x<1$, note that for a fixed value of $x$ there exists $y>0$ such that
$$(1-x) = \frac{1}{1+y}.$$
Hence,
$$0<nx(1-x)^n = \frac{nx}{(1+y)^n}.$$
Applying the binomial theorem we see that
$$(1+y)^n = 1 + ny + \frac{n(n-1)}{2}y^2 + \ ... \ >\frac{n(n-1)}{2}y^2,$$
and
$$0<nx(1-x)^n < \frac{2x}{(n-1)y^2}.$$
The RHS of the above inequality can be made smaller than any $\epsilon>0$ for fixed values of x and y by choosing a sufficiently large value of $n$.  
In particular, let $N$ be an integer such that
$$N > 1+\frac{2x}{\epsilon y^2}.$$
Then for all $n \geq N$ we have
$$0<nx(1-x)^n < \epsilon,$$
and the pointwise limit of the sequence is $0$.