# Let $X$ be a Banach space then every absolutely convergent series in $X$ converges in $X$

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?

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.

• In the last two lines, the triangular inequality is in the norm of $X$ not in absolute value – Dreamer123 Dec 31 '18 at 18:47
• @Dreamer123 Thanks! – Kenny Wong Dec 31 '18 at 18:48

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.

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