# Proving that a sequence in $\ell^2$ is a Cauchy sequence

Let $$(x^{(n)})_{n\in\mathbb{N}}, x^{(n)}:= \sum\limits_{i=1}^n \frac{1}{i} e_i$$, where $$e_i$$ is the sequence that is $$0$$ everywhere but $$1$$ in the $$i^{th}$$ element.

I would like to prove that this sequence is a Cauchy-sequence in a subspace of $$\ell^2$$.

I have $$||x^{(n)}-x^{(m)}||^2 = … = \sum\limits_{i=m+1}^n \frac{1}{i^2}$$. This can be of course estimated by $$||x^{(n)}-x^{(m)}||^2 \leq \frac{n-m-1}{(m+1)^2}$$ or $$||x^{(n)}-x^{(m)}||^2 \leq\frac{n-m}{m^2}$$ or $$||x^{(n)}-x^{(m)}||^2 \leq\frac{n}{m^2}$$.

Now I'm wondering if (and how) this helps.

Can someone please give me hint on that?

The Axiom of Archimedes doesn't seem to work here (right?)

• Since $\sum_k {1 \over k^2}$ is summable, you know that $\lim_n \sum_{k \ge n} {1 \over k^2} = 0$. You know that $\|x^{(n)}-x^{(m)}\| \le \sum_{k \ge \min(m,n)} {1 \over k^2}$. – copper.hat Jun 18 at 17:49

Hint: $$\sum_{j\geq 1} j^{-2}<\infty$$.
• I know, but how does this help here? I want to show that $\sum\limits_{i=m+1}^n 1/i^2\to 0$ for $n, m\to\infty$. Or that $\frac{n}{m^2}\to 0$ for $n, m\to\infty$. It is clear that $m^2$ is increasing faster than $n$, but this is no proof for this, right? – user3766553 Jun 18 at 16:59