If $\sum_{j=1}^\infty \| x_{j+1} - x_j\| < \infty$ then is it Cauchy?

Question

Let $X$ be a normed space. Assume for $x = (x_1, x_2, ...)\in X$, $$ \sum_{j=1}^\infty \| x_{j+1} - x_j\| < \infty. $$ Then is $x = (x_1, x_2, ...)$ Cauchy sequence with respect to $\|. \|$? If so, how can I prove it?

Answer 1

If $$\sum_{j=1}^\infty\Vert x_{j+1}-x_j\Vert<\infty$$ then $$\lim_{n\to\infty}\sum_{j=n}^\infty\Vert x_{j+1}-x_j\Vert=0$$ Let $\varepsilon>0$, then there exists an integer $n$ such that $$\sum_{j=n}^\infty\Vert x_{j+1}-x_j\Vert<\varepsilon$$ Hence for all positive integer $p$, we have: $$\Vert x_{n+p}-x_n\Vert\leqslant\sum_{j=n}^{n+p-1}\Vert x_{j+1}-x_j\Vert\leqslant\sum_{j=n}^\infty\Vert x_{j+1}-x_j\Vert<\varepsilon$$ i.e. $x$ is a Cauchy sequence in $X$.

Answer 2

HINT: Can you give a bound on $\|x_{j+k}-x_j\|$ for $j$ sufficiently large, $k\ge 1$?

Answer 3

Let $\varepsilon > 0$. For some $N_\varepsilon$, $$\sum \limits_{i=N_\varepsilon}^\infty \left\| x_{i} - x_{i+1} \right\| < \varepsilon.$$ Let $N_\varepsilon < a < b$. Then, by triangle inequality (why), $$\left\| x_a - x_b \right\| \leq \sum \limits_{i=a}^{b-1} \left\| x_i - x_{i+1} \right\| \leq \sum \limits_{i=N_\varepsilon}^ \infty \left\| x_i - x_{i+1} \right\| < \varepsilon.$$