Rudin Theorem 3.27

Solution 1:

Suppose $\{s_n\}$ is bounded, so that $-M \le s_n \le M$ for some $M > 0$. Let $k \in \mathbb{N}$. Choose $n_1$ such that $n_1 < 2^k$. By $(8)$,

$$-M \le s_{n_1} \le t_k$$

Now choose $n_2$ so that $n_2 > 2^k$. Then

$$2M \ge 2s_{n_2} \ge t_k$$

Hence, the sequence $\{t_n\}$ is bounded. Use similar reasoning to conclude that if $\{t_n\}$ is bounded, then $\{s_n\}$ is bounded.

Suppose $\{s_n\}$ is unbounded. Let $M > 0$. There exists $n \in \mathbb{N}$ such that $s_n > M$. Choose $k$ such that $2^k > n$. Then,

$$t_k \ge s_n > M$$

so $\{t_n\}$ is unbounded. The converse ($\{t_n\}$ unbounded $\implies$ $\{s_n\}$ unbounded) is similar.

Solution 2:

[I do not understand your confusion, so I'd just write out what appears obvious, using the word bounded/unbounded instead of converge (which we don't care as yet)]

If $s_n$ is unbounded, then by (8), $t_k$ is unbounded. If $t_k$ is unbounded, then by (9), $s_n$ is unbounded.

If $s_n$ is bounded, then by (9), $t_k$ is bounded. If $t_k$ is bounded, then by (8), $s_n$ is bounded.