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This question already has an answer here:

Let $(a_n)_{n\in\mathbb{N}}\in[0,\infty[^{\mathbb{N}}$ be a monotonically decreasing sequence, so that $\sum_{k=0}^{\infty}2^ka_{2^k}$ converges. Show that $\sum_{n=1}^{\infty}a_n$ converges and test $\sum_{n=1}^{\infty}\frac{1}{n\sqrt{n}}$ for convergence.


Can anybody…explain to me what this exercise even asks of me? I have trouble understanding it in the first place. Is $a_n=2^ka_{2^k}$ here or something else? If so am I simply supposed to show that $\sum_{n=1}^{\infty}2^{n}a_{2^{n}}$ converges? Seems unlikely. I don't get it.

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marked as duplicate by Robert Z calculus Jan 5 '18 at 22:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Observe $$ \sum_{n=1}^\infty a_n = \sum_{k=0}^\infty \sum_{n=2^{k}}^{2^{k+1}-1} a_n \le \sum_{k=0}^\infty 2^{k} a_{2^{k}} < \infty. $$ So you group the elements of the original sequence into exponentially larger blocks, and use monotonicity in each block.

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This is Cauchy's condensation test, see https://en.wikipedia.org/wiki/Cauchy_condensation_test

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