Answer to Question #177816 in Real Analysis for Nikhil

Question #177816

Show that the sequence (an), where an= n/(n^2+4) is monotonic. Is (an ) a Cauchy sequence? Justify your answer


1
Expert's answer
2021-04-13T13:57:00-0400

Let us show that the sequence "(a_n)", where "a_n= \\frac{n}{n^2+4}" is monotonic. For this let us find the difference


"a_{n+1}-a_n=\\frac{n+1}{(n+1)^2+4}-\\frac{n}{n^2+4}=\\frac{(n+1)(n^2+4)-n((n+1)^2+4)}{n^2+4}=\n\\frac{n^3+n^2+4n+4-n(n^2+2n+5)}{n^2+4}=\\frac{-n^2-n+4}{n^2+4}<0" for "n\\ge2."


It folllows that "a_{n+1}<a_n" for "n\\ge 2", and hence the sequence is decreasing for "n\\ge 2."


Let us show that "(a_n)" is a Cauchy sequence. For any "\\varepsilon>0" let "n=\\lceil\\frac{2}{\\varepsilon}\\rceil". Then for any "m\\ge n,\\ k\\ge n" we have that "|a_m-a_k|\\le |a_m|+|a_k|=\\frac{m}{m^2+4}+\\frac{k}{k^2+4}<\\frac{m}{m^2}+\\frac{k}{k^2}=\\frac{1}{m}+\\frac{1}{k}\\le\\frac{2}{n}\\le\\varepsilon."

Therefore, for any "\\varepsilon>0" there exists "n\\in\\mathbb N" such that for any "m\\ge n,\\ k\\ge n" we have that "|a_m-a_k|<\\varepsilon", and we conclude that "(a_n)" is a Cauchy sequence.


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