Garrett Black

2022-06-21

How do I check whether the sequence converges uniformly?

${f}_{n}={x}^{n}-{x}^{n+1}={x}^{n}(1-x),\phantom{\rule{1em}{0ex}}x\in [0,1]$

${f}_{n}={x}^{n}-{x}^{n+1}={x}^{n}(1-x),\phantom{\rule{1em}{0ex}}x\in [0,1]$

Bruno Hughes

Beginner2022-06-22Added 25 answers

It is controlled by

$\begin{array}{r}{(1-{\displaystyle \frac{1}{n+1}})}^{n}(1-{\displaystyle \frac{n}{n+1}})={(1-{\displaystyle \frac{1}{n+1}})}^{n}{\displaystyle \frac{1}{n+1}}.\end{array}$

We know that

$\begin{array}{r}{(1-{\displaystyle \frac{1}{n+1}})}^{n}\to {e}^{-1},\end{array}$

so for large n,

which tends to zero as $n\to \mathrm{\infty}$

$\begin{array}{r}{(1-{\displaystyle \frac{1}{n+1}})}^{n}(1-{\displaystyle \frac{n}{n+1}})={(1-{\displaystyle \frac{1}{n+1}})}^{n}{\displaystyle \frac{1}{n+1}}.\end{array}$

We know that

$\begin{array}{r}{(1-{\displaystyle \frac{1}{n+1}})}^{n}\to {e}^{-1},\end{array}$

so for large n,

which tends to zero as $n\to \mathrm{\infty}$