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


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.

  • $\begingroup$ I understand your answer, but why are we allowed to take the absolute value of the sequence? Does this not change anything? $\endgroup$ – user66807 Mar 16 '13 at 17:43
  • $\begingroup$ @user66807 This is because $\lim |a_n|=0$ if and only if $\lim a_n=0$. If this bothers you, prove $-1/n\leq a_n\leq 1/n$ by using $-1\leq \cos y\leq 1$ and squeeze. But this is slightly less "elegant". $\endgroup$ – Julien Mar 16 '13 at 17:45
  • $\begingroup$ So to do that I would say: Observe that $-1\leq\cos(n)\leq1$ So $\frac{-1}{n}=\frac{-n}{n^2}\leq \frac {-n}{n^2+1}\leq \frac{n\cos n}{n^2+1}\leq \frac {n}{n^2+1}\leq \frac{n}{n^2}=\frac{1}{n}$ Since $\lim \frac{1}{n}=0$ and $\lim \frac{-1}{n}=-\lim \frac{1}{n}=0$ By the squeezing theorem, $\lim \frac{n\cos n}{n^2+1}=0$? $\endgroup$ – user66807 Mar 16 '13 at 18:08
  • $\begingroup$ @user66807 Yes, that's perfect. But you should quickly get used to the absolute value argument which yields half of the writing. $\endgroup$ – Julien Mar 16 '13 at 18:09
  • $\begingroup$ Okay thanks a bunch! $\endgroup$ – user66807 Mar 16 '13 at 18:10

You find the limit using a very similar approach:

$$ \left| \frac{n \cos n}{n^2 +1} \right| \le \left| \frac{n}{n^2 + 1} \right| \le \left| \frac{n}{n^2} \right| = \frac{1}{n} $$

And since $1/n \to 0$ as $n \to \infty$, we find that the limit of $a_n$ is indeed $0$.

This is sufficient to show that $a_n$ is Cauchy too, since every convergent sequence is Cauchy.


$$\left|\frac{n\cos(n)}{n^2+1}\right| =\frac{n}{n^2+1}|\cos(n)| \leq \frac{n}{n^2+1} \rightarrow 0$$


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