Rudin Theorem 3.27 Theorem 3.27 of Rudin's book Principles of mathematical analysis at pages 61-62 states that,
Suppose $a_1\ge a_2\ge a_3\ge \cdots \ge 0.$ Then the series $\sum_{n=1}^{\infty}a_{n}$ converges if and only if the series
$\sum_{k=0}^{\infty}2^{k}a_{2^{k}}=a_{1}+2a_{2}+4a_{4}+8a_{8}+\cdots$ converges.
I could follow all of the arguments except the the last sentence, which is
By (8) and (9), the sequences $\left\{ s_{n}\right\}$ and $\left\{ t_{k}\right\}$ are either both bounded or both unbounded.
Here (8) and (9) are 
(8) For $n<2^k$, $s_n \le t_k$.
(9) For $n>2^k$, $2s_n \ge t_k$.
where 
$s_{n}=\sum_{i=1}^{n}a_{i}$, $t_{k}=\sum_{i=0}^{k}2^{i}a_{2^{i}}.$
Why are the two sequences either both bounded or both unbounded? The (8) seems to imply that if $t_k$ converges, then $s_n$ converges. The (9) seems to imply that if $s_n$ converges, then $t_k$ converges. I could not further more arguments to see how the last sentence works.
Thank you for any help. 
BTW, for $n=2^k$, it seems like $s_n \le t_k$.
 A: 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.
A: [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.
