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Let $(V,\|\|)$be a normed space. Let $(x_n) \subset V^{\Bbb{N}}$. We say that $\sum x_n$ converges if, $\lim_{n\to \infty} \sum_{i=1}^{n}x_i$ exists.

Show that if $\sum x_n$ converges then $x_n \to 0$

Consider $s_n = \sum_{i=1}^{n}x_i$. Then since $\sum x_n$ converges we have that $s_n \to l$ for some $l \in V$. So I wan't to see that given $\epsilon >0 $ there exist $n_0$ such that $\|x_n\| < \epsilon$ if $n \geq n_0$. So I'm trying to use the fact that $(s_n)$ is a Cauchy sequence, but im only managing to get an expression like

$$\|s_{n+p}-s_n\|=\|x_{n+p}+...+x_{n+1}\|< \epsilon$$

If $n$ is sufficiently big. But I don't know how to bound my $\|x_n\|$ from that expression. Any hints?

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  • $\begingroup$ Whatever proof you know for real series applies to this situation easily. $\endgroup$ – zhw. Jun 28 '17 at 2:47
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Hint

Try letting $p=1$.

...........

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  • $\begingroup$ I don't know why I was thinking taking $p=1$ didn't imply $\|x_n\|<\epsilon$ for all $n \geq n_0$. Actually it does. $\endgroup$ – Joaquin Liniado Jun 16 '15 at 23:24
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Let the sum be $s$; then, for every $\varepsilon>0$, there exists $m$ such that, for every $n>m$, $$ \biggl\|\sum_{i=1}^n x_i-s\biggr\|<\varepsilon $$ By the triangle inequality, for $n>m$, $$ \|x_{n+1}\|= \biggl\| \biggl(\sum_{i=1}^{n+1} x_i-s\biggr)+ \biggl(s-\sum_{i=1}^{n} x_i\biggr) \biggr\| \le \biggl\|\sum_{i=1}^{n+1} x_i-s\biggr\|+ \biggl\|s-\sum_{i=1}^{n} x_i\biggr\| <2\varepsilon $$

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Just use the definition:

If $S_{n}=\sum _{k=0}^{n}x^{n}$ converges, then it is Cauchy so for all $\epsilon >0$ there is an integer $N$ such that whenever $n,m\geq N$, $\vert S_{n}-S_{m}\vert <\epsilon$.

For this $N$, take $n=m+1$ and then $\vert S_{m+1}-S_{m}\vert=\vert x_{m+1}\vert $ and this is $<\epsilon $.

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Following your notation, let $s_n = \sum_{i=1}^n x_i$. Assuming the series converges, $s_n$ is Cauchy. Rewriting this, for any $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $m \geq n > N$, we have $$\sum_{i=n}^m ||x_n|| < \varepsilon.$$ In particular, setting $m = n$ shows that $x_n < \varepsilon$ for all $n \geq N$, as desired. As another user pointed out, setting $p = 1$ gives you this result, but seeing it spelled out with this notation could help you see the solution better.

A shorter solution: both $s_{n+1}$ and $s_n$ converge, so $x_n = s_{n+1} - s_n \to 0$.

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