# Sequences in Banach spaces [duplicate]

I am very bad with proofs that ask you to show that "$\exists$ something..." because in most of them you have to explicitly show the something. The following question is one such. Any help will be greatly appreciated!

Let $\{u_n\}$ a sequence in a Banach space $X$. Suppose that $\sum_{n=1}^{\infty} \|u_n\| < \infty$. Prove that there exists some $x \in X$ such that

$$\lim_{n \to \infty} \sum_{k=1}^{n} u_k = x$$

Thank you very much in advance!

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## marked as duplicate by 23rd, egreg, Davide Giraudo, Lord_Farin, Johannes KloosNov 23 '13 at 23:13

You meant $\sum_{k=1}^nu_k$. –  1015 Feb 21 '13 at 3:45
Precisely! I will correct it. –  user44069 Feb 21 '13 at 3:47

Let $m,n \in \mathbb{N}$ satisfy $m \leq n$. By the Triangle Inequality for the norm $\| \cdot \|_{X}$, we have $$\left\| \sum_{k=m}^{n} u_{k} \right\|_{X} \leq \sum_{k=m}^{n} \| u_{k} \|_{X}.$$ As $\displaystyle \sum_{k=1}^{\infty} \| u_{k} \|_{X} < \infty$, it follows from the Cauchy Criterion for Series that for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ sufficiently large so that $$\forall m,n \in \mathbb{N}_{\geq N}: \quad m \leq n ~ \Longrightarrow ~ \sum_{k=m}^{n} \| u_{k} \|_{X} < \epsilon.$$ Consequently, $$\forall m,n \in \mathbb{N}_{\geq N}: \quad m \leq n ~ \Longrightarrow ~ \left\| \sum_{k=m}^{n} u_{k} \right\|_{X} < \epsilon.$$ We thus see that the sequence $\displaystyle \left( \sum_{k=1}^{n} u_{k} \right)_{n \in \mathbb{N}}$ is Cauchy in $X$. As $X$ is a Banach space, every Cauchy sequence in $X$ converges; in particular, the sequence $\displaystyle \left( \sum_{k=1}^{n} u_{k} \right)_{n \in \mathbb{N}}$ converges to some $x \in X$: $$\sum_{k=1}^{\infty} u_{k} ~ \stackrel{\text{def}}{=} ~ \lim_{n \to \infty} \sum_{k=1}^{n} u_{k} = x.$$