Show that the sequence $a_n=\frac{cos(n^2+n)}{n^2}$ converges to $0$.

Question

Question: Show that the sequence $a_n=\frac{cos(n^2+n)}{n^2}$ converges to $0$.

My attempt: Need to show that $\forall\epsilon>0$, $\exists N\in\mathbb{N}$ such that $n>N \implies|a_n-0|<\varepsilon.$

Fix $\epsilon>0$.

We have $\left|a_n-0\right|=\left|\frac{cos(n^2+n)}{n^2}\right|$.

Since $cos(n^2+n)$ is bounded by $-1$ and $1$, we then have

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

Then $\frac{1}{n^2}<\epsilon \implies n^2>\frac{1}{\epsilon} \implies n>\sqrt\frac{1}{\epsilon}$

So choose $N=\lfloor\sqrt\frac{1}{\epsilon}\rfloor+1$ and the definition is satisfied. Q.E.D.

Not sure about this one so I would appreciate some input. Thanks

Answer 1

Simply note that

$$-\frac1{n^2}\le\frac{\cos(n^2+n)}{n^2}\le\frac1{n^2}$$

thus by squeeze theorem since $\frac1{n^2}\to 0$ we have that

$$\frac{\cos(n^2+n)}{n^2}\to 0$$

Answer 2

Your proof is correct, but if you want to make it easier, please observe that $$ \frac1{n^2}<\frac1n, $$ so it is enough to take $N=\lfloor 1/\epsilon\rfloor+1$.