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### Let $$x_n := -\frac{2}{n^2}$$. Prove that $$(x_n) \rightarrow 0$$.

Proof: Given $$\varepsilon >0$$, let $$K > \sqrt{\tfrac{2}{\varepsilon}}$$. Then if $$n \ge K$$ we have $$|x_n - 0| =$$ $$\left|-\frac{2}{n^2}\right|=$$ $$\frac{2}{n^2}\le$$ $$\frac{2}{K^2}<$$ $$\frac{2}{(\sqrt{2/\varepsilon})^2}=$$ $$\varepsilon$$. Thus $$\left( -\frac{2}{n^2}\right)\rightarrow 0$$ by the Definition of Sequence Converence.

Move on to Example 3