Convergence with respect to the sup norm of the sequence $f_n(x) = x^{n+1} - x^n$ I'm trying to determine whether the sequence $f_n(x)=x^{n+1}-x^n$ converges in $C([0,1])$ with respect to the supremum norm, $||f||=\sup\{|f(x)|\,:\,x\in S\}$, where $f$ is defined on the set $S$, and if it is convergent find it's limit.  
I have proved that $f_n(x)=x^n$ is not convergent as it is not Cauchy, but I am unsure of how to proceed for this sequence.  Any help is appreciated.
 A: You can see that the limit of the sequence is $0$. So we have
$$ \sup_{0\leq x \leq 1}\bigg|x^n-x^{n+1} -0\bigg| \leq \frac{\left( \frac{n}{n+1}\right)^n }{n+1}< \frac{1}{n+1}< \epsilon.  $$ 
You can find the sup by using the derivative test to the function $x^n-x^{n+1}$.
A: To get an idea for what the limit could be, first calculate $x^{n+1} - x^n \to 0-0= 0$ for $x\in (0,1)$ and $x^{n+1} - x^n =1-1= 0$ for $x=1$. Hence, the limit function is $0$, we just have to check for uniform convergence.
When you considered the sequence $x^n$, you probably noted that the convergence is uniform on $[0,1-\varepsilon]$ for every $\varepsilon > 0$, but not on the whole interval.
The idea here is that $x^{n+1} - x^n = x^n (x-1)$ and the factor $x-1$ is small (in absolute value) for $x \in [1-\varepsilon, 1]$.
I suggest that you take these ideas to derive the proof yourself. If you get stuck, look at the rest of this answer.
Let $\varepsilon>0$ be arbitrary. For $x\in\left[0,1\right]$, there
are two cases:


*

*We have $x\in\left[0,1-\varepsilon\right]$. Then
$$
\left|x^{n-1}-x^{n}\right|=\left|x^{n}\left(1-x\right)\right|\leq\left|x^{n}\right|\leq\left(1-\varepsilon\right)^{n}.
$$

*We have $x\in\left[1-\varepsilon,1\right]$. Then
$$
\left|x^{n-1}-x^{n}\right|=\left|x^{n}\left(1-x\right)\right|\leq\left|1-x\right|=1-x\leq1-\left(1-\varepsilon\right)=\varepsilon.
$$


All in all, we conclude
$$
\left\Vert x^{n+1}-x^{n}\right\Vert _{\sup}\leq\max\left\{ \varepsilon,\left(1-\varepsilon\right)^{n}\right\} \xrightarrow[n\to\infty]{}\varepsilon
$$
and thus
$$
\limsup_{n\to\infty}\left\Vert x^{n+1}-x^{n}\right\Vert _{\sup}\leq\varepsilon
$$
for all $\varepsilon>0$. This yields $\left\Vert x^{n+1}-x^{n}\right\Vert _{\sup}\xrightarrow[n\to\infty]{}0$,
so that your sequence converges uniformly to $0$.
