Answer to Question #161764 in Calculus for Vishal

Question #161764

Let {an}∞n=1 be a convergent sequence. Prove that {an}∞n=1 satisfies Cauchy’s criterion


1
Expert's answer
2021-02-24T06:39:38-0500

Given:  {an}n=1  is  a  convergent  sequence.\mathrm{Given:\;\{a_n\}_{n=1}^{\infty}\;is\;a\;convergent\;sequence.}


Prove:{an}n=1  satisfies  Cauchys  criterion.\mathrm{Prove:\{a_n\}_{n=1}^{\infty}\;satisfies\;Cauchy's\;criterion.}


Proof:  Let  ϵ>0  be  given.Since  {an}n=1  is  a  convergent  sequence.  So,\mathrm{Proof:\;Let\;\epsilon >0\;be \;given.Since\;\{a_n\}_{n=1}^{\infty}\;is\;a\;convergent\;sequence.\;So,}

choose  N  such  that  if  n>N,  we  have  anα<ϵ2  .\mathrm{choose\;N\;such\;that\;if\;n>N,\;we\;have\;|a_n-\alpha |<\dfrac{\epsilon}{2}\;.}


Then  if  m,n>N,we  have  anam=anα+αam\mathrm{Then\;if\;m,n>N,we\;have\;|a_n-a_m|=|a_n-\alpha+\alpha-a_m|}

anα+amα\mathrm{\leq |a_n-\alpha|+|a_m-\alpha|}

<ϵ2+ϵ2\mathrm{< \dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}}

=ϵ\mathrm{=\epsilon}


{an}n=1  is  a  cauchy  sequence.\mathrm{\therefore \{a_n\}_{n=1}^{\infty}\;is\;a\;cauchy\;sequence.}



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