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Let $X$ be the space of the real sequences $\{(x_n)_{n\in\mathbb{N}}|x_n\in\mathbb{R}\}$ with the metric $$ d((x_n),(y_n)) = \sum_{k=1}^{\infty}\frac{1}{k^2}\min\{|x_n-y_n|, 1\} $$ Show that the sequence $$ ((x_n^i)_{n\in\mathbb{N}})_{i\in\mathbb{N}} $$ in $X$ converges to $(x_n)_{n\in\mathbb{N}}$ if for all $n\in\mathbb{N}$, $$ \lim_{i\to\infty} x_n^i = x_n $$

Given $\varepsilon>0$, I need to find a $I>0$ such that $i>I$ implies $d((x_n^i),(x_n)) < \varepsilon$. But $$ \|(x_k^i) - (x_k)\| = \|(x_1^i-x_1,x_2^i-x_2,...)\|= \sum_{k=1}^{\infty}\frac{1}{k^2}\min\{|x_k^i - x_k|, 1\} $$

This is what I tried so far, and still I don't know how to use the limit in the hypothesis. Any clarification will be appreciated.

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Let $\epsilon >0$. Choose $N$ such that $\sum_{k=N}^{\infty} \frac 1 {k^{2}} <\epsilon$. Then $\sum_{k=N}^{\infty} \frac 1 {k^{2}} \min \{|x^{i}_k-x_k|,1\} <\epsilon$. Now consider $\sum_{k=1}^{N-1} \frac 1 {k^{2}} \min \{|x^{i}_k-x_k|,1\}$. Make this small by choosing $i$ large enough.

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