# Absolute convergence of a telescoping series in a Banach space

Let $$f_k:X \to B$$ be a sequence of functions defined on an arbitrary space $$X$$ and taking values in a Banach space $$B$$. Suppose there exist a set $$A_n \subset X$$ such that $$|f_{k+1}(x)-f_k(x)|<\frac{1}{2^k}$$ for all $$k\geq n$$ and $$x\in X\setminus A_n$$. The exercise asks to show that $$(f_k)$$ converges absolutely and uniformly on $$X\setminus A_n$$.

I did the following:

$$\sum_{k=n}^{\infty} |f_{k+1}(x)-f_k(x)| \leq \sum_{k=n}^{\infty} \frac{1}{2^k}=\frac{1}{2^{n-1}}$$ for all $$x\in X\setminus A_n$$, so the series $$\sum_{k=n}^{\infty} (f_{k+1}-f_k)$$ is absolutely convergent on $$X\setminus A_n$$. Since $$B$$ is complete, the series is also convergent on $$X\setminus A_n$$. It is also uniformly convergent because

$$\bigg|\sum_{k=n}^{m} (f_{k+1}-f_k)- \sum_{k=n}^{\infty} (f_{k+1}-f_k)\bigg|= \bigg|\sum_{k=m+1}^{\infty} (f_{k+1}-f_k)\bigg|\leq \sum_{k=m+1}^{\infty} |f_{k+1}-f_k|\leq \frac{1}{2^{m}}\to 0 \text{ as }m\to\infty$$

uniformly on $$X\setminus A_n$$. Then we note that $$m> n$$ implies $$f_m=f_m-f_n+f_n=\sum_{k=n}^{m-1} (f_{k+1}-f_k) +f_n\to \sum_{k=n}^{\infty} (f_{k+1}-f_k) + f_n$$ as $$m\to \infty$$ on $$X\setminus A_n$$, and so the uniform convergence of $$\sum_{k=n}^{m} (f_{k+1}-f_k)$$ on $$X\setminus A_n$$ implies the uniform convergence of $$(f_k)$$ on $$X\setminus A_n$$. Finally, since the norm $$|\cdot|$$ is continuous, we also have that $$|f_k|$$ converges on $$X\setminus A_n$$.

Is this proof ok? Am I missing something? Thanks a lot for your feedback.