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Question Number 148519 by learner001 last updated on 28/Jul/21

$$$$ $${\mathrm{a}}_{\mathrm{n}}=\frac{\mathrm{n}}{\mathrm{n}+1}$$ $$\mathrm{let}ϵ>0\mathrm{be}\mathrm{given},\mid {\mathrm{a}}_{\mathrm{m}}-{\mathrm{a}}_{\mathrm{n}}\mid =\mid \frac{\mathrm{m}}{\mathrm{m}+1}-\frac{\mathrm{n}}{\mathrm{n}+1}\mid =\mid \frac{\mathrm{m}-\mathrm{n}}{(\mathrm{m}+1)(\mathrm{n}+1)}\mid =\frac{\mathrm{m}-\mathrm{n}}{(\mathrm{m}+1)(\mathrm{n}+1)}\mathrm{provided}$$ $$\mathrm{m}>\mathrm{n},\frac{\mathrm{m}-\mathrm{n}}{(\mathrm{m}+1)(\mathrm{n}+1)}<\frac{\mathrm{m}+1}{(\mathrm{m}+1)(\mathrm{n}+1)}=\frac{1}{\mathrm{n}+1}<ϵ.\mathrm{if}\mathrm{N}>\frac{1-ϵ}{ϵ}\mathrm{then}\mid {\mathrm{a}}_{\mathrm{m}}-{\mathrm{a}}_{\mathrm{n}}\mid <ϵ\mathrm{\forall }\mathrm{n},\mathrm{m}⩾\mathrm{N}$$

Commented bylearner001 last updated on 28/Jul/21

can someone check if the proof I made is correct? I was trying to show that the sequence is Cauchy.

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