Question #158007

Let be any positive integer. Use sandwich theorem for sequences to prove that the sequence { 1 n} converges to 0.


1
Expert's answer
2021-01-27T04:06:51-0500

We know that {an}\{a_n\} converges to L, or, limn=L\lim_{n \rightarrow \infty}=L if \forall ϵ\epsilon >0>0 \exist N\in Z+Z^+ such that \forall n>Nn>N,

anL<ϵ|a_n-L|<\epsilon .

Consider 1n0=1n=1n<ϵ|\dfrac{1}{ n}-0|=|\dfrac{1}{ n}|=\dfrac{1}{ n}<\epsilon

1<ϵn1ϵ<nn>1ϵ\Rightarrow 1<\epsilon\cdot n \Rightarrow \dfrac{1}{\epsilon}< n \Rightarrow n>\dfrac{1}{\epsilon}

Proof: Let ϵ\epsilon >0>0 . Choose N>1ϵ>\dfrac{1}{\epsilon} . Then, \forall n>Nn>N1n0=1n|\dfrac{1}{ n}-0|=|\dfrac{1}{ n}|

Also, n>N>1ϵϵ>1n1n<ϵ1n<ϵn>N>\dfrac{1}{\epsilon} \Rightarrow \epsilon>\dfrac1 n \Rightarrow {\dfrac1n}<\epsilon \Rightarrow {\dfrac1{ n}}<\epsilon

Thus, 1n0=1n<ϵ|\dfrac{1}{ n}-0|=\dfrac{1}{ n}<\epsilon

Hence, by above statement, {1n}\{\dfrac{1}{n}\} converges to 0.


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