Question #159452

Prove that if a sequence {an}∞n=1 satisfies Cauchy’s criterion, then it is bounded


1
Expert's answer
2021-02-03T03:53:01-0500

Let (an)n=1n=(a_n)_ {n = 1}^{n = \infty} be a Cauchy sequence. This implies that given ϵ>0\epsilon > 0 \exist NNN \in \mathbb{N} \ni anam<ϵ\mid a_n - a_m \mid < \epsilon whenever n,mNn,m \geq N . When m=Nm = N , we have that anaN<ϵ\mid a_n - a_N \mid < \epsilon

But anaNanaNanaN<ϵ\mid a_n \mid - \mid a_N \mid \leq \mid \mid a_n \mid - \mid a_N \mid \mid \leq \mid a_n - a_N \mid < \epsilon


an<ϵ+aN\mid a_n\mid < \epsilon + \mid a_N \mid

Let K:=max(a0,a1,a2,,aN1,aN,ϵ+aN)K := \max (\mid a_0 \mid, \mid a_1 \mid, \mid a_2 \mid, \cdot \cdot \cdot , \mid a_{N-1} \mid, \mid a_N \mid, \epsilon + \mid a_N \mid )

So we have that anK\mid a_n \mid \leq K as desired.


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