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

Falak Kinney 2021-08-18 Answered
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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Expert Answer

Szeteib
Answered 2021-08-19 Author has 14740 answers

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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Answered 2021-10-23 Author has 10829 answers

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