# $\sum_{n=1}^\infty n\beta_n<\infty$ implies $\sum_{n=1}^\infty n\alpha_n\beta_n<\infty$, for $\alpha_n\to\infty$?

Suppose that $\beta_n$ is a sequence of positive real numbers satisfying $$\sum_{n=1}^\infty n\beta_n<\infty$$

Is it possible to find a sequence $\alpha_n$ of positive numbers such that $\alpha_n\to\infty$ and $$\sum_{n=1}^\infty n\alpha_n\beta_n<\infty$$

Thank you.

• This was already asked (and answered) on the site. – Did Mar 30 '13 at 18:14
• Please @Did, give me the link. – Tomás Mar 30 '13 at 18:44
• It seems I am unable to find it. Maybe I was remembering this, which is, in a way, the complement result: for every divergent series $\sum\limits_na_n$, there exists another divergent series $\sum\limits_nb_n$ such that $b_n\ll a_n$. – Did Mar 30 '13 at 19:45

The "$n$" part in $n\beta_n$ is not important. The question really is whether for positive $\gamma_n$ such that $\sum \gamma_n\lt \infty$, we can find $\alpha_n$ that goes to $\infty$ such that $\sum\gamma_n\alpha_n$ converges.

Suppose that $\sum \gamma_n$ converges to $c$, and let $\delta_n=c-\sum_{i=1}^n \gamma_i$. Then $\delta_n$ has limit $0$. Let $\alpha_n=\sqrt{1/\delta_n}$. We show that this $\alpha_n$ works.

It is enough to show that the sequence $t_n=\sum_{i=1}^n \gamma_i\alpha_i$ is Cauchy. Let $\epsilon \gt 0$ be given.

So look at $|t_n-t_m|$, where $n\gt m$. Take $N$ large enough so that $\delta_N\lt \epsilon^2$. If $m$ and $n$ are $\gt N$, then $|t_n-t_m|\lt \epsilon^2/\sqrt{\epsilon^2}=\epsilon$.