Q:

Find sup$\{x_n\}$.

I don't understand the part yellow highlighted. The sequence $x_n$ is defined by 0 when it is even, but how can we justify $\frac {n+2}{n+1}$ when $n$ is even? 
Thank you in advance. 

A:

The odd subsequence is decreasing and positive. 
The even subsequence is $0$.
Hence for even $n$, the supremum  of  $\{ x_k: k \ge n \}$ is attained at the first odd index that is at least as big as $n$. Hence the supremum is $$x_{n+1}=\frac{(n+1)+1}{(n+1)}=\frac{n+2}{n+1}.$$
Edit:
When $n$ is even,
\begin{align}\{ x_k: k \ge n \}&=\{ x_n , x_{n+1}, x_{n+2}, x_{n+3}, \ldots \} \\
&=\{ 0, x_{n+1}, 0, x_{n+3}, \ldots\}\end{align}
We know that $x_{n+1} > x_{n+3} > x_{n+5} > \ldots$ and they are positive.
Hence the supremumis $x_{n+1}$.

