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?

  • 3
    $\begingroup$ Yes, that is correct. $\endgroup$ – 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.

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