Find sup$\{x_n\}$.

Question

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.

Answer 1

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 $x_{n+1}$.