Question

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

Answer 1

$\displaystyle a_n = \frac{n}{n^2+4}$

The sequence is monotonic if $a_{n+1} < a_n \Rightarrow (a_{n+1}-a_n) <0$.

$\displaystyle a_{n+1}=\frac{n+1}{(n+1)^2+4}=\frac{n+1}{n^2 +2n+5}$

$\displaystyle \frac{n+1}{n^2+2n+5} - \frac{n}{n^2+4} = \frac{(n+1)(n^2+4)-(n^2+2n+5)n}{(n^2+2n+5)(n^2+4)} = \frac{n^3+4n+n^2+4 -n^3-2n^2-5n}{(n^2+2n+5)(n^2+4)} = \frac{-n^2-n+4}{(n^2+2n+5)(n^2+4)}$

$-(n^2+n-4) <0$ is true for any $n\geq 2$.

Nevertheless, this sequence is not monotonic because $a_2>a_1$.

This sequence is Cauchy sequence because there exists such a $n=n_0$ that for all $n>n_0, p>n_0$ and any $\epsilon>0:$ $| a_{n+p} -a_n| < \epsilon$ . Starting from n=2 this indeed works for any $\epsilon>0$.