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If a sequence $(x_n)^{\infty}_{n=1}$ in $\mathbb{R}^n$ satisfies $\displaystyle \sum_{n\geq{1}} ||x_n -x_{n+1}|| < \infty$, Show that it is a Cauchy sequence.


By definition, a sequence $x_k$ in $\mathbb{R}^n$ is Cauchy if for every $\varepsilon >0$, there is an integer $N$ such that $||x_k-x_l||<\varepsilon$ for all $k,l\geq{N}$

A set $S \subset {\mathbb{R}^n}$ is complete if every Cauchy sequence of points in $S$ converges to a point in $S$

So, in this question, every two consecutive elements will create a new $\varepsilon$, say $(\varepsilon_1, ...., \varepsilon_n)$

$||x_n-x_{n+1}||<\varepsilon_1$, $||x_{n+1}-x_{n+2}||<\varepsilon_2$, $||x_{n+2}-x_{n+3}||<\varepsilon_3$, .......

we have $\varepsilon_1 > \varepsilon_2 > \varepsilon_3 > ....$

Sum up all the $\varepsilon$, we must get an exact number because the sum wont get infinitely large as its getting smaller and smaller with always $>0$


marked as duplicate by Mhenni Benghorbal, Amzoti, Davide Giraudo, Emily, mdp Mar 12 '13 at 15:08

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By triangle inequality we get (WLOG $k\geq l$) $$ \lVert x_k - x_l \rVert \leq \sum\limits_{i=k}^{l-1} \lVert x_i-x_{i+1}\rVert \leq \sum\limits_{i=k}^\infty \lVert x_i - x_{i+1}\rVert $$

Choose $\varepsilon > 0$. As $\sum\limits_{i=0}^\infty \lVert x_i - x_{i+1}\rVert < \infty$, $\sum\limits_{i=N}^\infty \lVert x_i - x_{i+1}\rVert \to 0$ (as $N\to\infty$) . Choose $N$ big enough, so that $\sum\limits_{i=N}^\infty \lVert x_i - x_{i+1}\rVert < \varepsilon$. Now for all $k,l \geq N$ (again WLOG $k \leq l$) $$\lVert x_k - x_l \rVert \leq \sum\limits_{i=N}^\infty \lVert x_i - x_{i+1}\rVert<\varepsilon$$


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