Given a vanishing sequence find a series in $\ell^2$

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.

• 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... Jul 14 at 19:14

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 $$\sum_n \left ( \frac{y_n}{b_n} \right )^2 >\sum_k \left ( \frac{y_{n_k}}{b_{n_k}} \right )^2 > \sum_k 1 = \infty.$$

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 $$\log^*(n)$$.