Find a subsequence of $\{(-1)^nn\}$ which is monotone. Is there a convergent subsequence?
I know that the sequence is basically the sequence of $n$ where the elements alternate between negative and positive. eg. $-1,2,-3,4-5,...,\pm n$. I understand that I can take a subsequence of all the positive or all the negative elements and I would have a monotone subsequence. However, I am not sure how to denote that subsequence in terms of $\{a_{n_k}\}$. Could someone clarify? Thanks.
I also assume that there wouldn't be a convergent subsequence because $\{(-1)^nn\}$ is not bounded.
$(a_{2n})$ is monotone. As for your argument about a convergent subsequence, it is not valid, take $u_{2n}=0$ and $u_{2n+1}=n$, then $(u_n)$ is not bounded but $(u_{2n})$ converges. There is indeed no convergent subsequence, because if $(a_{\sigma(n)})$ is convergent, there is an infinitely number of $n$ such that $\sigma(n)$ is even, or an infinite number of $n$ such that $\sigma(n)$ is odd. In the first case, let $\varphi$ such that $\sigma\circ\varphi(n)$ is even for all $n$, then $(a_{\sigma\circ\varphi(n)})$ is a subsequence of $(a_{\sigma(n)})$ and thus converges. But for all $n$, $a_{\sigma\circ\varphi(n)}=\sigma\circ\varphi(n)\underset{n\rightarrow +\infty}{\longrightarrow}+\infty$. In the second case, by the same argument there exists $\varphi$ such that $\sigma\circ\varphi(n)$ is odd for all $n$ and $a_{\sigma\circ\varphi(n)}=-\sigma\circ\varphi(n)\underset{n\rightarrow +\infty}{\longrightarrow}-\infty$ which is a contradiction.