This problem comes from Royden & Fitzpatrick Real Analysis.
Let $\{ u_n \}$ be a sequence in a Banach space $X$. Suppose that $\sum_{k=1}^{\infty} || u_k || < \infty$. Show that there is an $x \in X$ for which $\lim_{n \to \infty} \sum_{k=1}^n u_k = x$.
My approach: Since $X$ is a Banach space, it is complete. Notice that $\sum_{k=1}^{\infty} ||u_k|| < \infty$ implies that for every $\epsilon>0$ there is some index $N$ such that $\sum_{k=N}^{\infty} ||u_k|| < \epsilon$.
Let $x_n = \sum_{k=1}^n u_k$. Then for any $m \ge n \ge N$, we have: $||x_n - x_m|| = ||\sum_{k=1}^n u_k - \sum_{k=1}^m u_k|| = ||\sum_{k=n+1}^m u_k|| \le \sum_{k=n+1}^m ||u_k|| \le \sum_{k=N}^{\infty} ||u_k|| < \epsilon$. Therefore, this is a Cauchy sequence and since $X$ is complete there exists an $x$ that is the limit of the series.
This chapter introduced the definition of Banach space a few sections ago, and this particular section introduced The Open Mapping Theorem, The Closed Graph Theorem, as well as linear complements of spaces. It concerns me that my approach doesn't use any of the appropriate topics of the section so I just want to check its validity. Have I made any mistakes? Is there another approach that would use these topics?
Your proof is correct. Here is a tiny simplification. Since the sequence of partial sums $S_n=\sum_{k=1}^n||u_i||$converges in $\mathbb R$, it is a Cauchy sequence. You must show that the seuence $$T_n=\sum_{k=1}^nu_i$$ in the Banach space converges. Let $\epsilon>0.$ There exists $N$ such that $$\sum_{k=n}^m||u_i|| \text { for }m \ge n>N.$$ Thus $$||\sum_{k=n}^mu_i||\le\sum_{k=n}^m||u_i||<\epsilon$$ Thus the sequence $T_n$ is a Cauchy sequence and hence converges.