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I try to prove that $\ell_1$, the space of absolutely convergent sequences in $\mathbb{C}$ with norm $\| x \| = \sum_{k=1}^{\infty} |x_k|$, is complete. I already proved that, if $\{ x_n \}$ is a Cauchy sequence (with elements in $\ell_1$), then, for a fixed $k \in \mathbb{N}$, the sequence $\{ x_{n,k} \}$ is Cauchy in $\mathbb{C}$.

The obvious way to define a limit of a Cauchy sequence $\{ x_n \}$ is to take the limits of $\{ x_{n,k}\}$, which exist since $\mathbb{C}$ is complete. Let's call this sequence $x$. But now we only know $x_n \to x$ pointwise.

Problem: how do I prove $\lim_{n \to \infty} \| x_n - x \| = \lim_{n \to \infty} \sum_{k=1}^{\infty} |x_{n,k} - x_k| = 0$?

I suppose I need to use the absolute convergence of $x_n$, but I do not see how. If I can prove the above, it follows that $x \in \ell_1$ therefore $\ell_1$ is complete.

Edit: Knowing $x \in \ell_1$, to establish convergence, I know (for $n,m>N$) $\sum_{k=1}^K |x_{m,k} − x_{n,k} | \leq \|x_m−x_n\| < \epsilon$. I let $m \to \infty$, can I conclude $\sum_{k=1}^K |x_k − x_{n,k}| \leq \| x−x_n \| < \epsilon$? That seems odd. But so does $\sum^K_{k=1} | x_k − x_{n,k} | \leq \| x_m − x_n \|$. What can I say when $m \to \infty$?

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    $\begingroup$ This is a special case of the Fischer-Riesz theorem. The trick is to pick a subsequence $(n_k)$ such that $\|x_{n_{k-1}}-x_{n_k}\|\le 1/2^k$. Then the crucial estimate comes down to a geometric series. $\endgroup$ – J.R. Feb 9 '14 at 14:15
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    $\begingroup$ You can use the fact that the norm is a continuous function to introduce the limit into the norm. $\endgroup$ – user99680 Feb 9 '14 at 14:15
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    $\begingroup$ See this. $\endgroup$ – David Mitra Feb 9 '14 at 14:16
  • $\begingroup$ Knowing $x \in \ell_1$, to establish convergence, I know (for $n, m > N$) $\sum_{k=1}^K |x_{m,k} - x_{n,k}| \leq \| x_m - x_n \| < \epsilon$. I let $m \to \infty$, can I conclude $\sum_{k=1}^K |x_k - x_{n,k}| \leq \| x - x_n \| < \epsilon$? That seems odd. But so does $\sum_{k=1}^K | x_k - x_{n,k} | \leq \| x_m - x_n \|$. What can I say when $m \to \infty$? $\endgroup$ – Math Student 020 Feb 9 '14 at 15:31

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