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All sequences here are sequences of positive reals. If $\sum x_n^2$ diverges, does there always exist a sequence $y_n$ such that $\sum y_n^2$ converges but $\sum x_ny_n$ does not?

My attempt is as follows:

Suppose not. Then choose $x_n$ with $\sum x_n^2$ divergent such that $\sum y_n^2 < \infty \implies \sum s_ny_n < \infty$. Then there exists $K$ such that $Ky_n^2>x_ny_n$. So $K^2y_n^2>x_n^2$ (??). So $\sum y_n^2$ diverges. Contradiction.

Although I am not overly convinced by this. Is the statement even true?

Jakobian
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awsomeguy
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  • Such a sequence ${y_n}$ always exists and this can be proves using Uniform Boundedness Principle. But I don't have an elementary proof at the moment. – Kavi Rama Murthy Aug 04 '19 at 11:57
  • I can provide a proof if use of Functional Analysis is OK. – Kavi Rama Murthy Aug 04 '19 at 12:00
  • Or this one https://math.stackexchange.com/questions/37647/if-sum-a-n-b-n-infty-for-all-b-n-in-ell2-then-a-n-in-ell2, or this https://math.stackexchange.com/questions/58565/is-there-a-constructive-proof-of-this-characterization-of-ell2. – Martin R Aug 04 '19 at 12:03
  • @KaviRamaMurthy: Do you mean a functional analysis argument like this https://math.stackexchange.com/a/37662/42969? – Martin R Aug 04 '19 at 12:07
  • @MartinR Exactly. That is what I meant. – Kavi Rama Murthy Aug 04 '19 at 12:08

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