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Question Number 148519 by learner001 last updated on 28/Jul/21
  a_n =(n/(n+1))  let ε>0 be given, ∣a_m −a_n ∣=∣(m/(m+1))−(n/(n+1))∣=∣((m−n)/((m+1)(n+1)))∣=((m−n)/((m+1)(n+1))) provided  m>n, ((m−n)/((m+1)(n+1)))<((m+1)/((m+1)(n+1)))=(1/(n+1))<ε. if N>((1−ε)/ε) then ∣a_m −a_n ∣<ε ∀ n,m≥N
$$ \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}} \\ $$$$\mathrm{let}\:\epsilon>\mathrm{0}\:\mathrm{be}\:\mathrm{given},\:\mid\mathrm{a}_{\mathrm{m}} −\mathrm{a}_{\mathrm{n}} \mid=\mid\frac{\mathrm{m}}{\mathrm{m}+\mathrm{1}}−\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}}\mid=\mid\frac{\mathrm{m}−\mathrm{n}}{\left(\mathrm{m}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}\mid=\frac{\mathrm{m}−\mathrm{n}}{\left(\mathrm{m}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}\:\mathrm{provided} \\ $$$$\mathrm{m}>\mathrm{n},\:\frac{\mathrm{m}−\mathrm{n}}{\left(\mathrm{m}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}<\frac{\mathrm{m}+\mathrm{1}}{\left(\mathrm{m}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}<\epsilon.\:\mathrm{if}\:\mathrm{N}>\frac{\mathrm{1}−\epsilon}{\epsilon}\:\mathrm{then}\:\mid\mathrm{a}_{\mathrm{m}} −\mathrm{a}_{\mathrm{n}} \mid<\epsilon\:\forall\:\mathrm{n},\mathrm{m}\geqslant\mathrm{N} \\ $$
Commented by learner001 last updated on 28/Jul/21
can someone check if the proof I made is correct? I was trying to show that the sequence is Cauchy.

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