If $X$ is a Banach space, then every absolutely convergent series is convergent.

$\mathit{pf:}$ Suppose s_n is absolutely convergent. For any $m

Question

If $X$ is a Banach space, then every absolutely convergent series is convergent.

$\mathit{pf:}$ Suppose s_n is absolutely convergent. For any $m<n$.

$||s_n-s_m||=||\sum_{k=1}^nx_k-\sum_{k=1}^mx_k||=||\sum_{k={m+1}}^nx_k||$

$\le \sum_{k=m+1}^n||x_k||=|(\sum_{k=1}^n||x_k||-\sum_{k=1}^m||x_k||)|$

How does this show that $s_n$ is Cauchy?

zzr0ck3r · Answer

@eliassaab :)

zzr0ck3r · Answer

is it true that every convergent sequence is Cauchy in a complete normed space?

zzr0ck3r · Answer

if that is true then we choose n,m>N (the N for absolute convergence) and we are done...I think.....

zzr0ck3r · Answer

Ok I think that is it, but ill leave it up for you to confirm that is the point of the proof:)

anonymous · Answer

Since 
$$
 \sum_{k=0}^\infty||x_k|| 

$$
is convergent, for each $ \epsilon>0 $then there is N, such that for n  greater that N
$$

\sum_{k=n}^\infty||x_k|| \le \epsilon 
$$
That is what you need to finish your proof.

zzr0ck3r · Answer

and this is true because its Cauchy?

anonymous · Answer

Let $ \epsilon >0$ and let N> found above then
$$
m>n> N \implies || s_m-s_n|| =||\sum_{k=n+1}^m x_k||
\le \sum_{k=n+1}^m||x_k||  \le  \sum_{k=n+1}^\infty ||x_k|\le \epsilon
$$

anonymous · Answer

Now s_n is Cauchy, so it is convergent in a Banach space, since it is complete.

zzr0ck3r · Answer

right, great. tnx again.  Hibert spaces are next:)

anonymous · Answer

YW

zzr0ck3r · Answer

is the converse hard to prove?

zzr0ck3r · Answer

nm I found it.