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I am in desperate need for hints to get me in the right direction for this proof.

Let $(x_n)_{n\in \mathbb {N}}$ be a monotone decreasing null sequence. Prove that:

$$\sum_{n=1}^\infty x_n \text{ converges } \ \iff \sum_{n=0}^\infty 2^nx_{2^n} \text{ converges}$$

Thanks.

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Hint:This is standard cauchy condensation test. Please try to group terms in powers of 2. Thus $$\sum x_n = x_1 + (x_2 + x_3) + (x_4 + x_5 + x_6 + x_7) + \cdots$$ now use he fact that $x_n$ is decreasing and try to bound the above sum in terms of $\sum 2^{n}x_{2^{n}}$.

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The direction ($\implies$) is trivial, so assume that $$\sum_{n\ge 0}2^nx_{2^n}$$ converges. For $m\in\Bbb N$ let $$s_m=\sum_{n=0}^mx_n\;.$$ Suppose that $2^k\le m<2^{k+1}$; then

$$\begin{align*} s_m&=\sum_{n=0}^{2^k-1}x_n+\sum_{n=2^k}^mx_n\\\\ &=x_0+\sum_{\ell=0}^{k-1}\sum_{n=2^\ell}^{2^{\ell+1}-1}x_n+\sum_{n=2^k}^mx_n\\\\ &\le x_0+\sum_{\ell=0}^{k-1}2^\ell x_{2^\ell}+\sum_{n=2^k}^mx_n\\\\ &\le x_0+\sum_{\ell=0}^k2^\ell x_{2^\ell}\;. \end{align*}$$

That is, for each $m\in\Bbb N$ we have

$$s_m\le\sum_{\ell=0}^{\lfloor\log_2m\rfloor}2^\ell x_{2^\ell}\;.$$

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This is the Cauchy condensation test.

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