Prove the sequence converges uniformly and satisfies f(x)=1/(1-x)

Question

Let $[a,b]\subset (-1,1)$, and define the sequence $f_n:[a,b] \to \mathbb{R}$ by $f_n(x)=x^n$. Prove that $\sum_nf_n$ converges uniformly, and that $f= \sum_nf_n$ satisfies $f(x)=1/(1-x)$.

Here's what I have: Let c=max{|a|,|b|}. Since $[a,b]\subset (-1,1)$, $0≤c<1$. Moreover, $|x^n|≤c^n$ for all n∈N and all x∈[a,b]. Since $\sum_n c^n=1/(1-c)<\infty$, the Weierstrass M-test implies the uniform convergence of $\sum_n x^n=1/(1-x)$ on any closed interval $[a,b]\subset (-1,1)$.

Will this work for the first part? And how do I show it satisfies for the second part?

Answer 1

You can write the partial sums, so that $F_n(x)=\sum_{i=1}^n f_i(x)$. This defines a function pointwise. Next, you write $|f(x)-F_n(x)|$ and find an upper bound which goes down to zero, independently of $x$.

Answer 2

Hint: $\sum_{n=0}^N f_n = (x^{N+1} - 1)/(x-1)$, which you can show directly by multiplying both sides by $(x-1)$ and grouping together terms.