I'm stuck in the next Analysis problem:
Let $(f_n)_{n\in \mathbb{N}}$ with $f:[0,1]\rightarrow \mathbb{R}$, $f_n(x)=x^n$ Prove that the sequence is not uniformly Cauchy.
Attempts:
So first of all the negation of the definition of being Cauchy is:
$\exists \varepsilon >0$ such that for all $N\in \mathbb{N}$ with $m,n\geq N$ we have $|f_n(x)-f_m(x)|\geq \varepsilon$
So to work this out, there are two obvious inequalities
$|f_n(x)-f_m(x)|=|x^n-x^m|\geq |x^n|-|x^m|$
$|f_n(x)-f_m(x)|=|x|^m |x^{n-m}-1|$
But I can't manage to find a lower bound for the difference $|f_n-f_m|$. Should I try for specific values of $\varepsilon$?
If you know what is the limit of $(1-1/m)^m$, try with $n=2m$, $x=1-1/m$ and your second idea, ie you have $|f_n(x)-f_m(x)|=(1-\frac{1}{m})^m (1-(1-\frac{1}{m})^m)$, and choose $\varepsilon$ sufficiently small.