# Using the definition of a convergent sequence, prove \lim_{n\to\infty}\frac{1}{n^2}=0 Don't use any

Using the definition of a convergent sequence, prove
$$\displaystyle\lim_{{{n}\to\infty}}{\frac{{{1}}}{{{n}^{{2}}}}}={0}$$
Don't use any theorems about convergent sequences

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Szeteib

Let $$\displaystyle\epsilon$$ be any positive number
We need to show that $$\bigg|\frac{1}{n^2}-0\bigg|<\epsilon$$
Now take $$\displaystyle{N}\epsilon{N},\ {\frac{{{1}}}{{\epsilon}}}{<}{N}.\ \because{\frac{{{1}}}{{{N}}}}{<}\epsilon$$
$$\displaystyle\forall{n}{>}{N}{\left(\text{i.e.}{\frac{{{1}}}{{{n}}}}{<}{\frac{{{1}}}{{{N}}}}\right)}.\ {\frac{{{1}}}{{\epsilon}}}{<}{N}$$
and $$\bigg|\frac{1}{n^2}-0\bigg|=\bigg|\frac{1}{n^2}\bigg|=\frac{1}{n^2}<\frac{1}{N}<\epsilon; \ \bigg(\because\frac{1}{n^2} \ \text{is always positive}\bigg)$$
Hence $$\bigg|\frac{1}{n^2}-0\bigg|<\epsilon$$
$$\displaystyle\lim_{{{n}\to\infty}}{\frac{{{1}}}{{{n}^{{2}}}}}={0}$$

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