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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 $s_n \le t_k$.

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    $\begingroup$ See here for a more general statement. $\endgroup$
    – JavaMan
    Jan 8, 2013 at 21:30
  • $\begingroup$ So, it is not like that either $n<2^k$ or $n>2^k$? Even without comparing $n$ and $2^k$, one should be able to get $s_n\le t_k\le 2s_n$? The wiki link of "Cauchy condensation test" is helpful. Thank you! $\endgroup$ Jan 8, 2013 at 21:48

2 Answers 2

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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.

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[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.

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  • $\begingroup$ So, one should use both cases of $n<2^k$ and $n>2^k$? I thought that each case should be considered separately. Thank you! $\endgroup$ Jan 8, 2013 at 21:50
  • $\begingroup$ @SangcheolChoi, if $\{t_k\}$ is bounded, then $\{t_k\} \leq M$, $\forall k \in \mathbb{N}$. So $n \lt 2^k$ really means $\forall n \in \mathbb{N}$. $\endgroup$
    – Sun
    Aug 19, 2020 at 4:14

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