Need math help to prove that {m/(2m+1)} is a cauchy sequence

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Prove {m/(2m+1)} is a cauchy sequence using the cauchy sequence definition.

Cauchy sequence definition: For all epsilon > 0, there is some N in natural numbers such that if n is a natural number, then n >= N implies |a_m - a_n| < epsilon for any natural numbers n and m.

Sp the proof would start with, let epsilon > 0, let N = some number, and suppose n >= N. Then

|\frac{n}{2n+1} - \frac{m}{2m+1}| ... < \epsilon
Nov 24th, 2015

Thank you for the opportunity to help you with your question!

we just need to prove that for any e>0, there exist N that  for n,m> N

|n/(2n+1) - m/(2m+1)|<e 

|n/(2n+1)-m/(2m+1)| = |(n-m)/(2m+1)(2n+1)|< |N/(2N+1)(2N+1)|<1/(2(2N+1))<1/(4N) 

thus for any e>0, we just need let  N > 1/(4e)

then we have     

|an-am| = |n/(2n+1)-m/(2m+1)|<1/(4N) <1/(4*(1/(4e)) = e

 


Please let me know if you need any clarification. I'm always happy to answer your questions.
Nov 24th, 2015

Thank you for answering!

How do you know |n-m| < |N|?

Nov 24th, 2015

One of the conditions of a cauchy sequence is n >= N

Nov 24th, 2015

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