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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?)

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  • $\begingroup$ 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} $. $\endgroup$ – copper.hat Jun 18 at 17:49
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Hint: $\sum_{j\geq 1} j^{-2}<\infty$.

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  • $\begingroup$ 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? $\endgroup$ – user3766553 Jun 18 at 16:59

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