# Why is $\sum_{n=1}^{\infty} x^n-x^{n+1}$ not uniformly convergent in $[0,1)$?

So, I needed to determine whether $$\sum_{n=1}^{\infty} x^n-x^{n+1}$$ is uniformly convergent in $$[0,1)$$. This is what I tried to do:

First, this a series of functions where $$f_n(x)=x^n-x^{n+1}$$ for all $$n\in \mathbb{N}$$.

This seuqence of functions is pointwise convergent $$0$$ for all $$x\in [0,1)$$.

I will try to check whether the series is uniformly convergent in $$[0,1)$$ by using:

$$f_n(x)$$ uniformly converges to $$f(x)$$ in the domain $$D$$ $$\iff$$ $$Sup|f_n(x)-f(x)|$$ converges to $$0$$ for all $$x\in D$$

$$f_n(x)$$ is differentable in $$[0,1)$$ for all $$n\in \mathbb{N}$$, and the derivative of $$f_n(x)$$ is: $$-x^{n-1}\left(\left(n+1\right)x-n\right)$$

$$-x^{n-1}\left(\left(n+1\right)x-n\right)$$=$$0$$ if $$x=0$$ or $$x=\frac{n}{n+1}$$, and $$f_n(x)$$ is decreasing for $$x>\frac{n}{n+1}$$

Now: $$f_n(0)=0$$, $$f_n(\frac{n}{n+1})=(\frac{n}{n+1})^n-(\frac{n}{n+1})^{n+1}$$

So, now that I have the maximum for all $$n\in \mathbb{N}$$, what I tried to say is that because $$(\frac{n}{n+1})^n-(\frac{n}{n+1})^{n+1}$$ converges to $$0$$ then $$Sup|f_n(x)-f(x)|$$ converges to $$0$$ for all $$x\in [0,1)$$

However, in the answers it says that the function is not uniformly convergent in this domain.

Using Desmos to graph this sequence, it really seems like it is indeed uniformly convergent.

Where is my mistake?

EDIT: I understand what my mistake was now, however I can't understand how to prove that this is not uniformly convergent.

• Hint: you can simplify $f_N(x)=\sum_{n=1}^N(x^n - x^{n+1})$. The limit function $f(x)$ should be easy to evaluate, and it should also be easy to show that $\sup_{x\in[0,1)}|f_N(x) - f(x)| = 1$ for all $N$. – Jason May 17 at 21:20

One way to do this is to compute the sum explicitly and work from there: set

$$S_n(x)=\sum_{k=1}^{n} (x^n-x^{n+1})=(1-x)\sum_{k=1}^{n} x^n$$

Then,

$$S_n(x)=(1-x)\left(\frac{1-x^{n+1}}{1-x}-1\right)=x-x^{n+1}$$

and this converges pointwise to $$S(x)=x$$ on $$[0,1).$$

But

$$\sup \{|S_n(x)-x|:x\in [0,1)\}= \sup \{x^{n+1}:x\in [0,1)\}=1$$

so $$S_n$$ does not converge uniformly.