If $p_n = \frac {2}{n}+ (-1)^n$, compute sequence $P_n = sup\{p_m: m\ge n\}$ and find $\lim_{n\to 0}$sup $p_n$. If $p_n = \frac {2}{n}+ (-1)^n$, compute sequence $P_n = sup\{p_m: m\ge n\}$ and find $\lim_{n\to \infty}$sup $p_n$.

If n is even, $p_n = \frac{2}{n} + 1$ and $p_n \to 1$as $n\to \infty$

If n is odd, $p_n = \frac{2}{n} - 1$ and $p_n \to -1$as $n\to \infty$

Then $\lim_{n\to \infty}$sup $p_n = 1$

But $P_n =sup\{p_m: m\ge n\}=sup\{p_n, p_{n+1}, ...\} $

If $n=1$, then $P_1 = sup\{1,2,-\frac{1}{3}, \frac{3}{2}\,-\frac{3}{5}, \frac{4}{3}....\}$ ($P_1=2$, I think)

This is where I am having problem.
Note that the subsequence $\langle p_{2n}:n\in\Bbb Z^+\rangle$ is strictly decreasing, and that $p_{2n}&gt;p_{2m-1}$ for all $m,n\in\Bbb Z^+$: every term with an even index is greater than $1$, and no term with an odd index is greater than $1$. Thus, $P_n$ is simply going to be $p_m$, where $m$ is the smallest even integer greater than or equal to $n$. That is,

$$P_n=\begin{cases}
p_n,&amp;\text{if }n\text{ is even}\\
p_{n+1},&amp;\text{if }n\text{ is odd}\;.
\end{cases}$$

From here I expect that you can easily find $\limsup\limits_{n\to\infty}p_n=\lim\limits_{n\to\infty}P_n$.
if $n = 2k-1$ is odd, $P_n = (k+1)/k$ 

if $n = 2k$ is even, $P_n = (k+1)/k$

and $\lim_{n \rightarrow \infty} \sup(p_n) = \infty$