1. The problem statement, all variables and given/known data
Show that the sequence of functions ##x(1-x), x^2(1-x),...## converges uniformly on ##[0,1].##
2. The attempt at a solution
I have a quick question. For the following proof why is ##\left ( \frac{n}{n+1}\right )^n < 1##?
Proof:
We need to prove that, given ##\epsilon > 0##, there exists an ##N## such that for every ##n > N## and for every ##x \in[0, 1]##, we have ##|x^n(1 - x)-0| < \epsilon.##
##x^n## and ##(1-x)## are both continuous functions. Now ##x^n(1 - x)## has a maximum on ##[0, 1]## at $x=\frac{n}{1+n}$ since ##\frac{d}{dx}[x^n(1-x)] = -x^n +nx^{n-1}-nx^n = -x-nx+n## thus ##x=\frac{n}{1+n}##.
Then ##|x^n(1-x)|<(\frac{n}{n+1})^n(\frac{1}{n+1})<\frac{1}{n+1}<\epsilon.## Choose ##N = \frac{1-\epsilon}{\epsilon}## therefore for ##n>N## we have ##|x^n(1-x)|<\epsilon.##
Why is ##\left ( \frac{n}{n+1}\right )^n < 1##?
Show that the sequence of functions ##x(1-x), x^2(1-x),...## converges uniformly on ##[0,1].##
2. The attempt at a solution
I have a quick question. For the following proof why is ##\left ( \frac{n}{n+1}\right )^n < 1##?
Proof:
We need to prove that, given ##\epsilon > 0##, there exists an ##N## such that for every ##n > N## and for every ##x \in[0, 1]##, we have ##|x^n(1 - x)-0| < \epsilon.##
##x^n## and ##(1-x)## are both continuous functions. Now ##x^n(1 - x)## has a maximum on ##[0, 1]## at $x=\frac{n}{1+n}$ since ##\frac{d}{dx}[x^n(1-x)] = -x^n +nx^{n-1}-nx^n = -x-nx+n## thus ##x=\frac{n}{1+n}##.
Then ##|x^n(1-x)|<(\frac{n}{n+1})^n(\frac{1}{n+1})<\frac{1}{n+1}<\epsilon.## Choose ##N = \frac{1-\epsilon}{\epsilon}## therefore for ##n>N## we have ##|x^n(1-x)|<\epsilon.##
Why is ##\left ( \frac{n}{n+1}\right )^n < 1##?
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