Xrfgtjtk

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

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

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

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

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

$$ left| sum_{k = 1}^m x_k - sum_{k = 1}^nx_kright| = left| sum_{k=n+1}^m x_kright| leq sum_{k = n+1}^m | x_k |.$$

So for any $epsilon > 0$, we have

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

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

No, your proof does not work because:

1. The conclusion that you get from the reverse triangle inequality is that$$leftlVertsum_{k=1}^nx_krightrVert<varepsilon+lvert Lrvert.$$
   


2. If you define $varepsilon'=varepsilon+lvert Lrvert$, 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.