Proof of radius of convergence for geometric series.

Question

I am just getting into proofs involving a radius of convergence and I am tasked with proving for every $b$ with $0<b<1$ the series $\sum_{n=0}^\infty x^n$ converges uniformly to $\frac{1}{1-x}$ on the interval $[-b,b]$ while working directly with the definition of uniform convergence. So far I have $\lvert f(x) - s_k (x) \rvert = \lvert \frac{1}{1-x}-\frac{1-x^{k+1}}{1-x}\rvert = \lvert \frac{x^{k+1}}{1-x}\rvert < \epsilon$ But now I am stuck as far choosing an $N$ and doing the rest of the proof. If anyone could help me that would be greatly appriciated thank you!

Answer 1

Since, $x\in[-b,b]$,$$\left\lvert\frac{x^{k+1}}{1-x}\right\rvert=\frac{\lvert x\rvert^{k+1}}{\lvert1-x\rvert}\leqslant\frac{\lvert b\rvert^{k+1}}{1-\lvert b\rvert}.\tag1$$Since $\lvert b\rvert<1$,$$\lim_{k\to\infty}\frac{\lvert b\rvert^{k+1}}{1-\lvert b\rvert}=0.$$So, given $\varepsilon>0$, there is a natural $N$ such that$$k\geqslant N\implies\frac{\lvert b\rvert^{k+1}}{1-\lvert b\rvert}<\varepsilon$$and it now follows from $(1)$ that$$k\geqslant N\implies\bigl(\forall x\in[-b,b]\bigr):\left\lvert\frac{x^{k+1}}{1-x}\right\rvert<\varepsilon.$$