I need to study the limit behavior of $a_n=\frac{n\cos(n)}{n^2+1}$, which can be written as $\frac{n}{n^2+1}\cos(n).$ I knew that it wasn't going to be monotone because $cos(n)$ oscillates between -1 and 1, so I attempted to prove it was Cauchy. I think I did it the right way: $a_n$ is a Cauchy sequence if for every $\epsilon>0$ there exists an $N$ such that $|a_m-a_n|<\epsilon$ for $m,n>N$
$$|a_m-a_n|= \bigg|\frac{m\cos(m)}{m^2+1}-\frac{ncos(n)}{n^2+1}\bigg|\leq\bigg|\frac{m\cos(m)}{m^2+1}\bigg|+\bigg|\frac{n\cos(n)}{n^2+1}\bigg|=\frac{|m\cos(m)|}{m^2+1}+\frac{|n\cos(n)|}{n^2+1}\leq\frac{m}{m^2+1}+\frac{n}{n^2+1}\leq \frac{m+n}{mn}=\frac{1}{n}+\frac{1}{m}<\frac{1}{N}+\frac{1}{N}=\frac{2}{N}<\epsilon$$ Thus there exists an $N$, namely $N>\frac{2}{\epsilon}$ such that $m,n>N$ implies $|a_m-a_n|<\epsilon$.
That is how I proved that the sequence is a Cauchy sequence. Was my work correct? Also, I have a feeling that the limit of this sequence is $0$ because $\frac{n}{n^2+1}$ tends to $0$. This assumption makes me want to show that $\lim \bigg|\frac{a_{n+1}}{a_n}\bigg|=L<1$ which would imply that $\lim|a_n|=0$. This doesn't get me very far because I cant find a way to simplify it and find the limit L, and I end up with a big mess.
Any hints on how I should approach finding the limit?
Thanks in advance
Best Answer
A sequence in $\mathbb{R}$ converges if and only if it is Cauchy. That's the completeness of $\mathbb{R}$. In this case, it is easier and much more natural to show that it converges directly, than to show it is Cauchy. And anyway, completeness or not, convergence implies Cauchy. So if you can prove convergence directly, do it first.
Using $|\cos x|\leq 1$ for all $x\in\mathbb{R}$, you get $$ 0\leq \lvert\frac{n\cos n}{n^2+1}\rvert\leq \frac {n}{n^2+1}\leq \frac{n}{n^2}=\frac{1}{n}\qquad \forall n\geq 1. $$ By squeezing, the limit of your sequence is $0$. In particular, it is Cauchy.