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Question Number 148609 by learner001 last updated on 29/Jul/21

$$\mathrm{prove}\mathrm{that}{\left({\mathrm{a}}_{\mathrm{n}}\right)}_{\mathrm{n}⩾1}\mathrm{defined}\mathrm{by}{\mathrm{a}}_{\mathrm{n}}=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{\mathrm{n}(\mathrm{n}+1)}\mathrm{is}$$$$\mathrm{cauchy}\mathrm{sequence}.$$

Commented by learner001 last updated on 29/Jul/21

$$\mathrm{This}\mathrm{is}\mathrm{what}\mathrm{i}\mathrm{tried}.$$$$\mathrm{let}ϵ>0\mathrm{be}\mathrm{given}\mathrm{i}\mathrm{need}\mathrm{an}{\mathrm{n}}^{\ast }\mathrm{such}\mathrm{that}\mathrm{p}\in \mathbb{N}\mathrm{if}\mathrm{\forall }\mathrm{n}⩾{\mathrm{n}}^{\ast }\mathrm{then}$$$$\mid {\mathrm{a}}_{\mathrm{n}+\mathrm{p}}-{\mathrm{a}}_{\mathrm{n}}\mid <ϵ.$$$$\mid {\mathrm{a}}_{\mathrm{n}+\mathrm{p}}-{\mathrm{a}}_{\mathrm{n}}\mid =\mid (\frac{1}{\mathrm{n}+1}-\frac{1}{\mathrm{n}+2})+(\frac{1}{\mathrm{n}+2}-\frac{1}{\mathrm{n}+3})+...+(\frac{1}{\mathrm{n}+\mathrm{p}}-\frac{1}{\mathrm{n}+\mathrm{p}+1})\mid$$$$=\mid \frac{1}{\mathrm{n}+1}-\frac{1}{\mathrm{n}+\mathrm{p}+1}\mid ⩽\mid \frac{1}{\mathrm{n}+1}\mid +\mid \frac{1}{\mathrm{n}+\mathrm{p}+1}\mid <\frac{1}{\mathrm{n}}+\frac{1}{\mathrm{n}+\mathrm{p}}<\frac{1}{\mathrm{n}}<ϵ$$$$\mathrm{if}{\mathrm{n}}^{\ast }⩾\frac{1}{ϵ}\mathrm{then}\mid {\mathrm{a}}_{\mathrm{n}+\mathrm{p}}-{\mathrm{a}}_{\mathrm{n}}\mid <ϵ\mathrm{\forall }\mathrm{n}⩾{\mathrm{n}}^{\ast }.$$

Commented by learner001 last updated on 29/Jul/21

is this correct?

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