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Can anyone prove from first principles, that $$a_n = \left(\frac{2n-1}{n}\right)_{n=1}^\infty$$ is a Cauchy sequence.

Thank you all.

Noa Even
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LoveMath
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1 Answers1

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Large HINT: Let $$a_n=\frac{2n-1}n=2-\frac1n\;.$$ Then

$$|a_n-a_m|=\left|\left(2-\frac1n\right)-\left(2-\frac1m\right)\right|=\left|\frac1m-\frac1n\right|=\frac{|m-n|}{mn}\le\frac{m+n}{mn}\;.$$

Brian M. Scott
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