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Given a sequence $\gamma_k\in\ell^2$ but not in $\ell^1$ and a convergent nonnegative sqeuence $r_k$, if $\sum\limits_{k=1}^\infty \gamma_k r_k<\infty$ then $\lim\inf r_k=0$.

My attempt: Assume for the sake of contradiction that $\lim\inf r_k=\tilde r>0$. Then we have,

$$\infty > \sum\limits_{k=1}^\infty \gamma_k r_k \geq \tilde r\sum\limits_{k=1}^\infty \gamma_k=\tilde r\cdot\infty$$

Is this correct?

Edit: Okay obviously it's not correct as the $\lim\inf$ is not a lower bound for the sequence. Here is a second attempt: Since $\tilde r>0$ is the $\lim\inf$ of the sequence $r_k$ we have that $\exists c>0$ and $\exists n_0$ such that $r_k \geq c$ for all $k\geq n_0$. Then we have,

$$\infty > \sum\limits_{k=n_0}^\infty \gamma_kr_k \geq c\sum\limits_{k=n_0}^\infty \gamma_k = c\cdot \infty$$

which contradicts our assumption about the product $\gamma_k r_k$ belonging to $\ell^1$.

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You have not mentioned that $\gamma_k$'s are non-negative. If $\gamma_k=\frac {(-1)^{n}} n$ and $r_n=\sqrt n$ then the hypothesis is satisfied but $\liminf r_n =\infty$. If $\gamma_n \geq 0$ then the conclusion is true, but your argument has some errors. $\liminf$ is not the same as infimum. If $\liminf r_n>0$ You can only say that there exists some $c>0$ and some integer $n_0$ such that $r_n >c$ for $n\geq n_0$. You can still complete your argument using this fact.

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  • $\begingroup$ You are correct in your suspicion, $\gamma_k\geq 0$ for all $k$. $\endgroup$ – Tony S.F. Mar 21 '18 at 10:09
  • $\begingroup$ Can you check my edit? Is it correct now? $\endgroup$ – Tony S.F. Mar 21 '18 at 10:28
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    $\begingroup$ Your argument is now correct. $\endgroup$ – Kavi Rama Murthy Mar 22 '18 at 5:04

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