# If $\lim_{i\to\infty} x_n^i = x_n$ then $((x_n^i)_{n\in\mathbb{N}})_{i\in\mathbb{N}}\to(x_n)_{n\in\mathbb{N}}$

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.

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.