Answer to Question #177816 in Real Analysis for Nikhil

Question

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

Answer 1

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

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