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Rudin proves a statement in his classical book. Most of this proof is straightforward, while a subtle point confuses me.

Claim: Suppose $a_1 \geq a_2 \geq \dots \geq 0$. Then, the series $\sum_{n=1}^{\infty} a_n$ converges if and only if the series $\sum_{k=0}^{\infty} 2^k a_{2^k}$ converges.

Proof: Let $s_n=a_1+a_2+\dots +a_n$, and $t_k = a_1 + 2a_2 + \dots +2^k a_{2^k}$. Obviously, if $n<2^k$, then $s_n \leq t_k$; if $n>2^k$, then $2s_n\geq t_k$.

Then, Rudin concludes that $s_n$ and $t_k$ are both bounded or both unbounded. Well, what if $n<2^k$ and $t_k$ is unbounded? In this case, we have no idea whether $s_n$ is bounded or not. Thanks for any explanation.

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  • $\begingroup$ If $\sup \{t_k: k\in \Bbb N\}=\infty$ then $\sup \{s_n:n\in \Bbb N\}\geq$ $ \sup\{s_{(1+2^k)}:k\in \Bbb N\}\geq$ $ \sup \{\frac {1}{2}t_k: k\in \Bbb N\}=\infty.$ $\endgroup$ – DanielWainfleet Aug 5 '18 at 2:06
  • $\begingroup$ If you have not made a typo in copying the book then the book has a typo, as the last term in the definition of $t_k$ should be $2^{k-1}a_{(2^k)}.$ $\endgroup$ – DanielWainfleet Aug 5 '18 at 2:29
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Assume $s_n$ is bounded by some positive constant $A>0$. Fix $k$ and choose $n$ such that $n>2^k$. Then, $t_k\leq 2s_n \leq 2A$. This shows that $(t_k)$ is bounded.

Assume now that $s_n$ is unbounded. Let $A>0$ be any positive constant. Then, there exists some $N$ such that for any $n\geq N$, we have $s_n > A$. Let $K=\log_2(N)+1$. Take $k\geq K$. Then $N<2^k$ so that $t_k \geq s_N > A$. It follows that $(t_k)$ is unbounded.

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Suppose that $\sum_{k=0}^{\infty}2^ka_{2^k}<\infty$. Then observe that, for each n, we can choose $k$ large enough so that $n<2^k$, so

$s_n \leq t_k \leq \sum_{k=0}^{\infty}2^ka_{2^k}<\infty$ $\implies$ $\lim_{n\rightarrow \infty}s_n=\sum_{n=1}^{\infty}a_n<\infty$.

Suppose that $\sum_{n=1}^{\infty}a_n<\infty$. Then, for each k, we can choose $n$ large enough so that $n>2^k$, so

$t_k \leq 2s_n \leq 2\sum_{n=1}^{\infty}a_n<\infty$ $\implies $ $\lim_{k\rightarrow \infty } t_k=\sum_{k=0}^{\infty}2^ka_{2^k}<\infty$.

Likewise, you can do the unbounded case.

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  • $\begingroup$ When $\sum_{k=0}^{\infty} 2^k a_{2^k}<\infty$, why do you say, "for each $n$, we can choose $k$" instead of "for each $k$, we can choose $n$?" I am a little confused why $n$ is fixed here. $\endgroup$ – Q.L. Aug 6 '18 at 21:28
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    $\begingroup$ In that case, our goal is to show that $s_n\leq \sum_{k=0}^{\infty}2^k a_{2^k}<\infty$ for every $n\in \mathbb{N}$ so that $\lim_{n\rightarrow \infty} s_n\leq \sum_{k=0}^{\infty}2^k a_{2^k}<\infty$. $\endgroup$ – Lev Bahn Aug 6 '18 at 23:05
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    $\begingroup$ Once we can bound every $s_n$ by a finite number, we know the limit is bounded by the finite number. Moreover, once we bound $s_n$ by $t_k$ for some $t_k$, $s_n$ is bounded by the sum. $\endgroup$ – Lev Bahn Aug 6 '18 at 23:08
  • $\begingroup$ $s_n$ can be unbounded if each $s_n$ is bounded by a finite number. For example, $s_n=2^n$ and $p_n=2^n+1$. Then, $s_n<p_n$ for each $n$, however, both sequences go to infinity as $n\rightarrow \infty$. $\endgroup$ – Q.L. Aug 6 '18 at 23:34
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    $\begingroup$ In my case, ‘a’ finite number was $\sum_{k=0}^\infty 2^k a_{2^k}$. If you just bound each $s_n$ by varying function or sequence $b_n$, you cannot conclude that the limit is bounded $\endgroup$ – Lev Bahn Aug 6 '18 at 23:45
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This may help to illustrate the idea of the proof:

Let $B_n =\sum_{j=1}^n 2^ja_{(2^j)}.$

If $\{B_n:n\in \Bbb N\}$ is unbounded above then $\{s_n:n\in \Bbb N\}$ is unbounded above, because $s_{(2^n)}\geq a_1+\frac {1}{2} B_n.$

For example $$s_{16}=a_1+a_2+(a_3+a_4)+(a_5+...+a_8)+(a_9+...+a_{16})\geq$$ $$\geq a_1+a_2+2a_4+4a_8+8a_{16}=a_1+\frac {1}{2}B_4.$$

If $\{s_n: n\in \Bbb N\}$ is unbounded above then $\{s_{(2^{n-1}-1)}:n\in \Bbb N\}$ is unbounded above ( because the sequence $(s_n)_{n\in \Bbb N}$ is increasing) whereupon $\{B_n:n\in \Bbb N\}$ is unbounded above, because $s_{(2^{n-1}-1)}\leq a_1+B_n.$ For example $$s_{15}=a_1+(a_2+a_3)+(a_4+...+a_7)+(a_8+...a_{15})\leq$$ $$ \leq a_1+2a_2+4a_4+8a_8=a_1+B_3.$$

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