# Sequence of series is Cauchy sequence

If $$(x_j)$$ is a sequence in an inner product space $$X$$ such that the series $$\|x_1\|+\|x_2\|+\cdots$$ converges, show that $$(s_n)$$ is a Cauchy sequence, where $$s_n=x_1+\cdots+x_n.$$

My proof is as follows. Given a sequence $$(x_j)$$ in the inner product space $$X$$ such that $$\sum_{j=1}^{\infty}\|x_j\|<\infty$$, this implies there exists $$N_1\in\mathbb{N}$$ such that whenever $$k>N_1$$, $$\sum_{j=k+1}^{\infty}\|x_j\|<\epsilon$$.

WLOG, assume $$n>m$$. Then, for any $$\epsilon>0$$, choose $$N=N_1\in\mathbb{N}$$. Then, whenever $$m,n>N$$ we have $$\|s_n-s_m\|=\|\sum_{j=m+1}^{n}x_j\|\leq\sum_{j=m+1}^{n}\|x_j\|\leq\sum_{j=m+1}^{\infty}\|x_j\|<\epsilon$$ as required.

Am I correct?

• Yes, that is correct. – José Carlos Santos Nov 29 '19 at 5:24

Yes, your proof is the standard (probably only?) way to prove this. I would point out that you are not using that $$X$$ is an inner product space; you are only using that $$X$$ is a normed space, which makes the proof more general.
I would also like to mention that, because it is not assumed that $$X$$ is complete, you cannot talk about the series being convergent, even if it is Cauchy.