[Math] Absolute convergence of series of norm of elements implies that sequence of partial sums is Cauchy.

Question

Consider a sequence $(x_n)$ in some inner product space $X$ such that the series $$\sum_{i=1}^{\infty}\|x_i\|$$
is convergent in $\mathbb{R}$. I am required to prove that the sequence of partial sums of $(x_n)$, say $(s_n)$ is Cauchy in $X$. Obviously then the sequence $(\|x_n\|)$ converges to $0$. After some fairly easy calculations we get that $\forall n,n \in \mathbb{N}$ such that $n>m$
$$\|s_n-s_m\|\leq\sum_{k=m+1}^{n}\|x_k\|$$
From the above 2 facts it is thus quite easy for any fixed $\epsilon>0$ and fixed difference $n-m=d$ to find a natural $N_{\epsilon,d}$ such that
$$\|s_n-s_m\|<\epsilon \ \ \ \ \ \ \ \forall n\forall m\big((n>m>N_{\epsilon,d})\land (n-m=d)\big)$$
This gets me close to the proof I need but it does not quite show that $(s_n)$ is Cauchy as I need to be able to find a natural number independent of the distance. I can't quite see how to reach that point though, maybe I need to use some properties of inner product spaces. This question is in a section about orthonormal sequences but I don't see how I could apply that at all. Any tips would be appreciated.

Answer 1

If $\sum_k \|x_k\|$ is convergent, then $\lim_n \sum_{k \ge n} \|x_k\| = 0$.

Pick $\epsilon>0$, then there is some $N$ such that $\sum_{k \ge N} \|x_k\| < \epsilon$.

Then if $n,m \ge N$ we have (assuming $n \ge m$) that $\|s_n-s_m\| \le \sum_{k=m+1}^n \|x_k\| \le \sum_{k=N}^\infty \|x_k\|< \epsilon$.