Limit of $\frac{n + (-1)^n n}{n+2}$

Question

I'm taking a university real analysis course and I was tasked with demonstarting a proof on regaurding the limit (or lack there of) of $\frac{n + (-1)^n n}{n+2}$. Here's my attempt at a proof that no limit exists.

First use algerbra of limits to seperate

$\frac{n + (-1)^n n}{n+2}$ $\implies$ $\frac{1}{1+\frac{2}{n}} + \frac{(-1)^n}{1 + \frac{2}{n}}$

Limit of $\frac{1}{1+\frac{2}{n}}$as $n \rightarrow \infty$ is simply 1.

Using the squeeze prinicible. Assuming $\frac{(-1)^n}{1 + \frac{2}{n}}$ does have a limit $y$ then sequences $x_n,y_n,z_n$ such that $x_n \leq y_n \leq z_n$ all sequences must converge on $y$.

Take...

$x_n = \frac{-1}{1+\frac{2}{n}}$

$ y_n = \frac{(-1)^n}{1 + \frac{2}{n}}$

$z_n = \frac{1}{1+\frac{2}{n}}$

There is no possible $x_n$ or $z_n$ that converge to similar limits and obeys $x_n \leq y_n \leq z_n$ . Therefore there is no limit $y$. Since $x_n = \frac{(-1)^n}{1+\frac{2}{n}}$ has no limit, by the algebra of limits neither does $\frac{n + (-1)^n n}{n+2}$.

Any feedback would be much appreciated!

Answer 1

What you did looks correct. However, a somewhat simpler method would be to use that for a sequence to converge to a limit, all subsequences must also converge to the same limit. For details about this, there are various posts about it on this site, such as Prove: If a sequence converges, then every subsequence converges to the same limit..

With

$$f(n) = \frac{n + (-1)^n n}{n+2} \tag{1}\label{eq1A}$$

you have for odd $n$ that $f(n) = 0$, while for even $n$, you have $f(n) = \frac{2n}{n + 2}$, with $\lim_{n \to \infty}\frac{2n}{n + 2} = \lim_{n \to \infty}\frac{2}{1 + \frac{2}{n}} = 2$. Since these $2$ subsequence limits are different, $f(n)$ doesn't converge as $n \to \infty$.