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Let $\langle a_n \rangle$ be a sequence of positive numbers. If $p>1/2$ and $\sum_{n=1}^{\infty} a_n$ converges, then how can I prove that $\sum_{n=1}^{\infty} \sqrt{a_n a_{n+1}}$ and $\sum_{n=1}^{\infty} \sqrt{a_n}/n^p$ also converges? It seems that any kind of convergence test does not work.

Also, does the converse hold? I think it doesn't, but I have trouble finding a counterexample.

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  • $\begingroup$ hint for the first: you have a geometric mean there. $\endgroup$ – Max Freiburghaus Jan 26 '17 at 21:11
  • $\begingroup$ @MaxFreiburghaus Thank you! I solved for the 1st case. Also, I found all of the counterexamples. $\endgroup$ – bellcircle Jan 26 '17 at 21:21
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Hints: (substantial)

  • For the first, use the AM–GM inequality: $$ \sqrt{a_na_{n+1}} \leq \frac{a_n+a_{n+1}}{2} $$ and comparison theorems.

  • For the second, use the Cauchy—Schwarz inequality* on the partial sums: $$ \sum_{n=1}^N \frac{\sqrt{a_n}}{n^p} \leq \sqrt{\sum_{n=1}^N a_n}\sqrt{\sum_{n=1}^N \frac{1}{n^{2p}}} $$ then the fact that $p>\frac{1}{2}$ and and comparison theorems.

For the counterexamples, you may want to find sequences such that

  • first: $a_{2n}$ is always "big," yet $a_{2n+1}$ is really "small."
  • second: for $p=1$, $\frac{\sqrt{a_n}}{n} = \frac{1}{n \ln^2 n}$ (recall that $\sum_{n=1}^\infty \frac{1}{n \ln^2 n}$ is a convergent series)

$\ast$ As mentioned in a comment below by Martin R, the AM–GM inequality would do as well for the second (and Cauchy–Schwarz for the first), but it's good to know several tools.

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  • $\begingroup$ AM-GM should work in the second case as well. $\endgroup$ – Martin R Jan 26 '17 at 21:22
  • $\begingroup$ Thanks a lot! I solved it! Also, I already found counterexamples of the converse. $\endgroup$ – bellcircle Jan 26 '17 at 21:22
  • $\begingroup$ @bellcircle I edited to give clues for counterexamples as well. $\endgroup$ – Clement C. Jan 26 '17 at 21:25
  • $\begingroup$ @MartinR Indeed, but knowing how to use both the AM-GM and the Cauchy-Schwarz inequalities is better than having just one feather to one's cap. $\endgroup$ – Clement C. Jan 26 '17 at 21:26

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