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

timer Asked: Nov 24th, 2015
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Question Description

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

Tutor Answer

School: UCLA


The central part is to consider |a_m1 - a_m2|  (a_m = m/(2m+1)) and to estimate this.

|a_m1 - a_m2| = |m1/(2m1+1) - m2/(2m2+1)| = |[2m1m2 + m1 - 2m1m2 - m2]/[(2m1+1)*(2m2+1)]| =

= |[m1 - m2]/[(2m1+1)*(2m2+1)]|.

This <= (|m1| + |m2|)/[2|m1|*2|m2|] = (1/4)*(1/|m2| + 1/|m1|) <= (1/2)/N

if m1, m2 >= N. And this should be <e. No problem.

For given e>0 choose N(e) > 1/(2e)  (entire part of 1/(2e) + 1).
Then for m1, m2 >= n(e) we know |a_m1 - a_m2| < e.

Please ask if something is unclear.

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