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Here is my question:

Let $X$ be a Banach space with norm $\|·\|$. Prove that, for any sequence $\{x_n\}$ in $X$, if $\sum_{n=1}^\infty \|x_n\|\lt\infty$, then $\lim_{k\to\infty}\sum_{n=1}^k x_n$ exists.

Here is what I got:

Given that $\sum_{n=1}^\infty \|x_n\|\lt\infty$, we know that $\|x_n\|$ is convergent. So let $\epsilon >0$ be given, then there exists $N>0$ such that for all $m\geq n\geq N$, $\sum_{j=n}^m \|x_j\|\lt\epsilon$. So by the triangle inequality, we have:

$$\|(\sum_{j=n}^m x_j)\|\leq\sum_{j=n}^m \|x_j\|\lt\epsilon$$

So $\sum_{j=n}^m x_j$ is Cauchy, and as $X$ is a Banach space, $\sum_{j=n}^m x_j$ is convergent, and therefore $\lim_{k\to\infty}\sum_{n=1}^k x_n$ exists.

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  • $\begingroup$ Show that $\sum_{k=1}^n x_k$ is a Cauchy sequence. $\endgroup$ – mfl Oct 30 '14 at 23:44
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Hint: Use the triangle inequality $\Big\|\sum_{k=n}^{m}x_k\Big\|\le\sum_{k=n}^{m}\|x_k\|$ to prove it's a Cauchy sequnce.

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  • $\begingroup$ You have the terms reversed in the inequality. $\endgroup$ – mfl Oct 30 '14 at 23:50
  • $\begingroup$ @mfl Oops, thanks :) $\endgroup$ – user2345215 Oct 30 '14 at 23:50
  • $\begingroup$ @user2345215 I have edited my solution above, based on what you and Kim provided. Does that look correct? $\endgroup$ – user3784030 Oct 31 '14 at 0:18
  • $\begingroup$ @user3784030 I don't really understand how can you compare a vector with a real number (the first inequality), but otherwise it's correct if you leave that out. Also I don't understand "So $\sum_{j=n}^m x_n$ is Cauchy". Did you mean to say "So $\sum_{n=1}^m x_n$ is Cauchy" (sequence in $m$)? $\endgroup$ – user2345215 Oct 31 '14 at 6:52
  • $\begingroup$ @user2345215 I have corrected those two mistakes - should have been $x_j$ instead of $x_n$ in the sums. $\endgroup$ – user3784030 Oct 31 '14 at 7:46
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When I see Banach, I think of Cauchy. $\sum|x_n|$ converges so given $\epsilon>0$, there is $N$ such that for all $m\geq n\geq N$, we have $\sum_{j=n}^m|x_j|<\epsilon$. Then, for such $m,n$, we also have $$ \left|\sum_{j=n}^mx_j\right|\leq\sum_{j=n}^m|x_j|<\epsilon. $$ This proves that $\sum x_n$ is Cauchy and hence convergent since $X$ is a Banach space.

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  • $\begingroup$ I have edited my solution based on your input, and the other comment - how does it look now? @Kim $\endgroup$ – user3784030 Oct 31 '14 at 0:19
  • $\begingroup$ Strictly speaking, the objects to which we attribute properties such as "Cauchy" or "convergent" is the sequence of partial sums $\{S_n\}$ where $S_n=\sum_{j=1}^s x_j$. When people say $\sum_{n=1}^\infty x_n$ is Cauchy, they implicitly mean that $\{S_n\}$ is Cauchy. Be sure that you understand this. $\endgroup$ – Kim Jong Un Oct 31 '14 at 0:22
  • $\begingroup$ So would it be more proper to add in, prior to the last line of my proof, something along the lines of "As the sequence of partial sums $\{S_n\}$ are Cauchy, we know that $\sum^\infty x_n$ is Cauchy"? @Kim $\endgroup$ – user3784030 Oct 31 '14 at 0:27

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