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Suppose that $a_n \geq 0$ for all $n$ and $a_n \downarrow 0$ as $n \rightarrow \infty.$ If $\sum_{k=1}^{\infty}2^ka_{2^k}$ diverges then $\sum_{n=1}^{\infty}a_n$ diverges.

Not sure if this is true because it is given as a conjecture. Would it be easier to prove the contrapositive of the statement? Please give me a hint on how to do this problem!

2 Answers 2

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That's called Cauchy condensation test. If $a_n\geq 0$ and $a_{n+1}\leq a_n$ for all $n$, and $\lim a_n =0$, then $\sum a_n$ converges if and only if $\sum 2^n a_{2^n}$ converges.

Note that \begin{equation*} \begin{aligned} \sum_{1}^{\infty} a_n= &a_1+a_2+(\underbrace{a_3}_{\geq a_4}+a_4)+(\underbrace{a_5}_{\geq a_8}+\underbrace{a_5}_{\geq a_8}+\underbrace{a_7}_{\geq a_8}+a_8)+\cdots \\ &\geq a_1+a_2+2a_4+4a_8+\cdots \\ &=a_1+\frac{1}{2}\sum_{1}^{\infty}2^na_{2^n}.\\ \end{aligned} \end{equation*} Thus, if $\sum a_n$ converges then $\sum 2^n a_{2^n}$ converges, that is, if $\sum 2^n a_{2^n}$ diverges, then $\sum a_n$ diverges.

We can also show that $$\sum_{1}^{\infty} a_n \leq a_1 + \sum_{1}^{\infty} 2^na_{2^n},$$ so if $\sum 2^n a_{2^n}$ converges then $\sum a_n$ converges.

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Write $$\sum_{n=1}^\infty a_n = \sum_{n=1}^{\infty}\sum_{i=2^{n-1}}^{2^n-1}a_{i} \geq \sum_{n=1}^\infty\sum_{i=2^{n-1}}^{2^n-1}a_{2^{n}} = \sum_{n=1}^\infty (2^n-1-2^{n-1} + 1) a_{2^n} $$ $$ = \sum_{n=1}^\infty 2^{n-1} a_{2^n} = \frac{1}{2}\sum_{n=1}^\infty 2^{n} a_{2^n} = \infty $$

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    How can you prove that $\sum_{n=1}^{\infty}2^na_{2^n}$ diverges? – 2017-01-27
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    @user3000482 As per your question "If $\sum_{k=1}^{\infty}2^ka_{2^k}$ diverges then ..." we are *assuming* this sum diverges, that is the point of the question. – 2017-01-27
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    But does it really diverge? Or is it just for the sake of proving the theorem? I mean, we made an assumption that it diverges. But given that $a \downarrow 0$ and $a_n$ are non-negative, is it possible to prove it diverges? – 2017-01-27
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    "Does it really diverge?" That depends on what $a_n$ is. With $a_n \geq 0$ decreasing to $0$, this alone is not enough information to determine whether $\sum_n 2^n a_{2^n}$ diverges. For example $\sum_n 2^n a_{2^n} < \infty$ if $a_{n} = 1/n^2$, then $a_{2^n}=1/4^n$ and $\sum_n 2^n a_{2^n}= \sum_n 2^{-n} < \infty$. – 2017-01-27