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I am trying to prove that in a normed space, if absolute convergence of any series implies the convergence of that series, then the normed space is complete.

Below is my attempt, but it concludes that every Cauchy sequence converges to $0$. Where have I gone wrong?

Proof. Let $(x_n)$ be a Cauchy sequence. Define the sequence $(S_n)$ by $S_n=\sum_{k=1}^{n}\|x_k\|$. For $n> m$ we have

$|S_n-S_m|=\sum_{k=m+1}^{n}\|x_k\|\leq\sum_{k=m+1}^{\infty}\|x_k\|\to 0$ as $m\to\infty$.

Thus $(S_n)$ is a Cauchy sequence of reals. As $\mathbb{R}$ is complete, it converges.

By supposition, absolute convergence implies convergence. Hence $\sum_{k=1}^{n}x_k$ converges as $n\to\infty$.

So $x_n=\sum_{k=1}^{n}x_k-\sum_{k=1}^{n-1}x_k\to\sum_{k=1}^{\infty}x_k-\sum_{k=1}^{\infty}x_k=0$ as $n\to\infty$.

So $(x_n)$ converges to $0$?

I've clearly made some error here. Thanks for your time.

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    $\begingroup$ How did you conclude that $\sum_{k=m+1}^{\infty}\|x_k\|\to 0$? This would not be true even if we knew $x_k\to 0$, but you seem to have concluded it for an arbitrary Cauchy sequence $x_k$, where even the individual terms may not converge to $0$. $\endgroup$
    – M W
    Oct 10, 2023 at 16:18
  • $\begingroup$ @MW as $m\to\infty$, the sum decreases. Then by Monotone Convergence Theorem, it converges to $0$. $\endgroup$
    – jet
    Oct 10, 2023 at 16:20
  • $\begingroup$ you need to know it's finite before you can talk about it decreasing. $\endgroup$
    – M W
    Oct 10, 2023 at 16:21
  • $\begingroup$ @MW True, I think that's the error here. Supposing it is finite, then the rest of the argument is sound? $\endgroup$
    – jet
    Oct 10, 2023 at 16:22
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    $\begingroup$ @jet the only time a Cauchy sequence $x_n$ will ever result in a finite sum $\sum x_n$ is on the occasion that $x_n\to 0$, and even then, not always. So "the rest of the argument is sound" is sort of like "other than that, Mrs. Lincoln, how did you like the play?" It is soundly proving that When a Cauchy sequence results in a convergent sum, it must be converging to $0$. But that is not what you want to prove. $\endgroup$
    – M W
    Oct 10, 2023 at 16:34

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Adding the terms of the Cauchy sequence makes no sense. Here is a Cauchy sequence: take the normed space to be $\mathbb R$, and $x_n=\frac1n$. Then $(x_n)$ is Cauchy, in fact it converges to $0$, while your sequence $(S_n)$ are the partial sums of the harmonic series which diverges.

You definitely need series somewhere, since your hypothesis is stated in terms of series. One way you can express the limit of a sequence as a series, is by telescoping. Indeed, if $x_n\to x$ you have $$\tag1 \lim_N\sum_{n=1}^Nx_{n+1}-x_n=\lim_N x_{N+1}-x_1=x-x_1. $$ So we want to say that $x=-x_1+\sum_{n=1}^\infty x_{n+1}-x_n$. The problem is that this series will fail to converge in general. To get around this problem we can note that since our original sequence is already Cauchy, we can work with a subsequence; if a subsequence converges, then the whole thing converges.

Hence we choose a subsequence so that the series in $(1)$ converges. A typical choice is as follows. Since $(x_n)$ is Cauchy, there exists $n_1$ such that $\|x_n-x_m\|\leq\frac12$ for all $n,m\geq n_1$. Again because the sequence is Cauchy, there exists $n_2\geq n_1$ such that $\|x_n-x_m\|\leq\frac14$ whenever $n,m\geq n_2$. Continuing this way we obtain a subsequence $(x_{n_k})$ with the property that $\|x_{n_{k+1}}-x_{n_k}\|\leq2^{-k}$.

Now consider $$\tag2 S_N=x_{n_1}+\sum_{k=1}^Nx_{n_{k+1}}-x_{n_k}. $$ We have $$ \sum_{k=1}^\infty\|x_{n_{k+1}}-x_{n_k}\|\leq\sum_{k=1}^\infty2^{-k}=1. $$ This shows that $(S_N)$ converges absolutely, and then by hypothesis it converges. This allows us to define $$ x=x_{n_1}+\sum_{k=1}^\infty x_{n_{k+1}}-x_{n_k} $$ We want to show that $\lim_kx_{n_k}=x$. A remarkable feature of the series in $(2)$, being telescopic, is that given any $j$ we have $$ x=x_{n_j}+\sum_{k=j}^\infty x_{n_{k+1}}-x_{n_k}. $$ This gives us $$ \|x-x_{n_j}\|\leq\sum_{k=j}^\infty\|x_{n_{k+1}}-x_{n_k}\|\leq\sum_{k=j}^\infty2^{-j}=2^{-j+1}. $$ Thus $\|x-x_{n_j}\|\to0$ with $j$, showing that $\lim_kx_{n_k}=x$. And since $(x_{n_k})$ is a subsequence of $(x_n)$ and $(x_n)$ is Cauchy, it follows that $\lim_nx_n=x$.

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