Question #158007

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


Expert's answer

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

∣an−L∣<ϵ|a_n-L|<\epsilon .

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

⇒1<ϵ⋅n⇒1ϵ<n⇒n>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>N, ∣1n−0∣=∣1n∣|\dfrac{1}{ n}-0|=|\dfrac{1}{ n}|

Also, n>N>1ϵ⇒ϵ>1n⇒1n<ϵ⇒1n<ϵn>N>\dfrac{1}{\epsilon} \Rightarrow \epsilon>\dfrac1 n \Rightarrow {\dfrac1n}<\epsilon \Rightarrow {\dfrac1{ n}}<\epsilon

Thus, ∣1n−0∣=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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