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Suppose is given a decreasing vanishing sequence $(b_n)_n \subset \mathbb{R}$. I want to find a sequence $(y_n)_n \subset \mathbb{R}$ s.t. $$ \sum_n y_n^2 < + \infty, \quad \quad \sum_n \left ( \frac{y_n}{b_n} \right )^2 = + \infty.$$

I tried to consider the sequence $y_n:= \sqrt{b_n-b_{n+1}}$ but I am not sure that the second condition is satisfied.

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  • $\begingroup$ It means that $(b_n)$ is decreasing and that $b_n \to 0$. There is a typo in your comment and I can't see it... $\endgroup$
    – Bremen000
    Jul 14 at 19:14
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Perhaps you should consider a subsequence of $(b_n)$. Since $b_ n \rightarrow 0$, consider a subsequence $b_{n_k}$ of $b_n$ such that $b_{n_k}<\dfrac{1}{k}$. Denote $I$ the ordered set of indices $n=n_k$ for which you have $b_{n_k}<\dfrac{1}{k}$. Set $y_n = 0$, if $n\in \mathbb{N}\setminus I$ and $y_n=\dfrac{1}{k}$, if $n=n_k \in I$. Then $\sum_{n}y_n^2 = \sum_{k} \dfrac{1}{k^2}< \infty$ and $\sum_n \left ( \frac{y_n}{b_n} \right )^2 >\sum_k \left ( \frac{y_{n_k}}{b_{n_k}} \right )^2 > \sum_k 1 = \infty.$

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Rather a long comment than an answer.

It is in fact quite hard to discard your attempt $y_n=\sqrt{b_n-b_{n+1}}$, it may actually work but I don't see an easy way to prove it.

The quicker $b_n\to 0$ the larger is $u_n=\left(\dfrac{y_n}{b_n}\right)^2$, here are some examples:

  • $b_n=\frac 1{n^2}\implies u_n\sim 2n$
  • $b_n=\frac 1n\implies u_n\sim 1$
  • $b_n=\frac 1{\sqrt{n}}\implies u_n\sim \frac 1{2\sqrt{n}}$
  • $b_n=\frac 1{\sqrt[4]{n}}\implies u_n\sim \frac 1{4\sqrt[4]{n^3}}$

All series are divergent, so to find a possible counter example we have to go for a very slow growing function but even for these we still end up with a divergent Bertrand series (since none of the logarithm exponent is bigger than $1$):

  • $b_n=\frac 1{\ln{n}}\implies u_n\sim \frac 1{n}$
  • $b_n=\frac 1{\ln(\ln(n))}\implies u_n\sim \frac 1{n\ln(n)}$
  • $b_n=\frac 1{\ln(\ln(\ln(n)))}\implies u_n\sim \frac 1{n\ln(n)\ln(\ln(n))}$
  • and so on...

I'm not even sure what would be the result with the iterated logarithm $\log^*(n)$.

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