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Math Help - Cauchy Sequence Proof

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    Cauchy Sequence Proof

    I need to prove that {(2n+1/n} is Cauchy. I know that the theorem states that if |s_m-s_n| < epsilon for n>=N, then the sequence is Cauchy, but I am not sure how to start.
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    Quote Originally Posted by tarheelborn View Post
    I need to prove that {(2n+1/n} is Cauchy. I know that the theorem states that if |s_m-s_n| < epsilon for n>=N, then the sequence is Cauchy, but I am not sure how to start.
    If the sequence is s_n=\left(2n+\frac{1}{n}\right) then it is not a Cauchy sequence.

    Please review your post for correctness.
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    I am so sorry; I missed a closing parentheses. It should be {(2n+1)/n}.
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    Quote Originally Posted by tarheelborn View Post
    I am so sorry; I missed a closing parentheses. It should be {(2n+1)/n}.
    Well s_n=\frac{2n+1}{n}=2+\frac{1}{n}.
    So s_n-s_m=\left(2+\frac{1}{n}\right)-\left(2+\frac{1}{m}\right)=\frac{1}{n}-\frac{1}{m}.

    Is \frac{1}{n} a Cauchy sequence?
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  5. #5
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    Quote Originally Posted by tarheelborn View Post
    I am so sorry; I missed a closing parentheses. It should be {(2n+1)/n}.
    Note that \frac{2n+1}{n}=2+\frac{1}{n}. I claim this sequence converges to 2.

    \left|2+\frac{1}{n}-2\right|=\left|\frac{1}{n}\right|

    If we let N=\frac{1}{\epsilon}, n>N implies that \frac{1}{n}<\epsilon, so this sequence converges. If a sequence converges, then it is Cauchy* (but not necessarily the other way around).

    *If this is not obvious, I can provide a proof.
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    Yes, it is absolutely obvious, and thank you. We don't get much in class except the proofs of the theorems, no examples with actual numbers in them. I am sometimes slow to relate one to the other! I will use these calculations in formulating a proof and, if need be, as for further help, if that is ok. Thank you very much!
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    I got stuck quickly. Do I need to do the epsilon proof or will I be able to simply show convergence? My professor is all about the epsilon proofs, but she didn't say it had to be that way for this problem.
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    Super Member redsoxfan325's Avatar
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    Showing convergence rigorously makes use of \epsilon-proofs. A sequence a_n converges to a point a iff for all \epsilon>0, \exists~N such that n>N implies |a_n-a|<\epsilon, and a sequence a_n is Cauchy iff for all \epsilon>0, \exists~N such that m,n>N implies |a_m-a_n|<\epsilon.
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