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my trial

Let $\sum x_k$ be absolutely convergent in $X$ $\implies$

$\sum \|x_k \|$ converges in $\mathbb{R}$ $\implies$

$\forall \epsilon >0, \exists N(\epsilon)$ st $\forall n>N(\epsilon)$. we have $| \| \sum_{k=1}^{k=n} x_k \| - L |$ < $\epsilon$ where $L$ is the limit in $\mathbb{R}$. Now fix an $\epsilon$>0 then there exist some $N$ st for all $n>N$

$| \| \sum_{k=1}^{k=n} x_k \| - L |$ < $\epsilon$

By reverse-triangular inequality $\implies$ $ \| \sum_{k=1}^{k=n} x_k \| $ < $\epsilon + |L| = \epsilon'$. Similarly for $m>n>N$ we have

$ \| \sum_{k=1}^{k=m} x_k \| $ < $\epsilon'$

By triangular inequality, we get $| \| \sum_{k=1}^{k=n} x_k \| - \| \sum_{k=1}^{k=m} x_k \| | <2 $$\epsilon$'

But $\| \sum_{k=1}^{n} x_k - \sum_{k=1}^{m} x_k\|$ =$\| \sum_{k=n+1}^{m} x_k \|=$$| \| \sum_{k=1}^{k=n} x_k \| - \| \sum_{k=1}^{k=m} x_k \| | <2 $$\epsilon$' Hence $\{ \sum_{k=1}^{n} x_k \}$ is a cauchy sequence in $X$ thus has a limit in $X$ so $\sum x_k $ converges in $X$.


Is my proof correct?

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You've more or less got the right idea, but we've got to put the norm signs in the right place, and there is a place near the end where we need to use the triangle inequality slightly differently.

The fact that $\sum_{k = 1}^\infty \| x_k \|$ converges and is equal to $L$ (say) means that for any $\epsilon > 0$, there exists an $N(\epsilon) \in \mathbb N$ such that

$$ n \geq N(\epsilon ) \implies \left| \sum_{k = 1}^{n}\| x_k \| - L\right| < \epsilon .$$

Following through with your original approach, we find that $$ m > n \geq N(\epsilon) \implies \sum_{k = n + 1}^{m} \| x_k\| < 2\epsilon .$$

[To spell it out, I'm using the triangle inequality like this: $$ \sum_{k = n+1}^m \| x_k \| = \left| \left( \sum_{k = 1}^m \| x_k \| - L\right)- \left( \sum_{k = 1}^n \| x_k \| - L \right)\right| \leq \left| \sum_{k = 1}^{m}\| x_k \| - L\right| + \left| \sum_{k = 1}^{n}\| x_k \| - L\right| < 2\epsilon$$ ]

To show that $n \mapsto \sum_{k = 1}^n x_k$ is a Cauchy sequence, we must use the triangle inequality like this:

$$ \left\| \sum_{k = 1}^m x_k - \sum_{k = 1}^nx_k\right\| = \left\| \sum_{k=n+1}^m x_k\right\| \leq \sum_{k = n+1}^m \| x_k \|.$$

So for any $\epsilon > 0$, we have

$$ m > n \geq N(\epsilon) \implies \left\| \sum_{k = 1}^m x_k - \sum_{k = 1}^nx_k\right\| < 2\epsilon$$

which implies that $n \mapsto \sum_{k = 1}^n x_k$ is Cauchy, and hence, convergent.

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  • $\begingroup$ In the last two lines, the triangular inequality is in the norm of $X$ not in absolute value $\endgroup$ – Dreamer123 Dec 31 '18 at 18:47
  • $\begingroup$ @Dreamer123 Thanks! $\endgroup$ – Kenny Wong Dec 31 '18 at 18:48
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No, your proof does not work because:

  1. The conclusion that you get from the reverse triangle inequality is that$$\left\lVert\sum_{k=1}^nx_k\right\rVert<\varepsilon+\lvert L\rvert.$$
  2. If you define $\varepsilon'=\varepsilon+\lvert L\rvert$, then $\varepsilon'$ is not an arbitrary number greater than $0$, as it should be.

In order to get a proof, apply the Cauchy criterion for series.

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  • $\begingroup$ Why is that? since $\epsilon$ is chosen arbitrarily and $L$ is fixed then $\epsilon$' is arbitrary $\endgroup$ – Dreamer123 Dec 31 '18 at 17:56
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    $\begingroup$ Not so, since $\varepsilon'\geqslant L$. $\endgroup$ – José Carlos Santos Dec 31 '18 at 17:57

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