Two real positive sequences are given $a_n , b_n \ge 0 ,n\ge0$ and I have $$ \sum_{n=1}^{\infty} a_n = \infty $$ after some work I get $$ M \ge \sum_{n=1}^{\infty} a_n b_n $$ where $M$ is some positive number (it can be $0$ in some cases but in general it doesn't have to) and conclusion is $$ \liminf_{k \to \infty }\ b_k =0$$ I get it in general but can someone give some theorem or something from where we get this conclusion.

My toughts are something like splitting $a_n$ in two subsequences, one has infinite sum and the other finite. then $b_n$ corresponding to first subsequence need to tend to 0, and other subsequnce just need to not make infinite sum. But on the other hand since sum is bounded from top $a_n b_n \to0 $. And then subsequence of $b_n$ (correspondiing to subsequence of $a_n$ with infinite sum ) tend to $0$ and we got that $\liminf$. Or do we ?


Suppose that $\liminf_{n\to\infty} b_n>0$. Therefore, there must exist $h>0$ and $N\in \Bbb N$ such that $b_n>h$ for all $n>N$.

But then, by positivity and the fact that $\sum_{n=1}^\infty a_n=+\infty$, it must hold $$\sum_{n\ge N}a_nb_n\ge h\sum_{n\ge N}a_n=+\infty$$ Now you only need notice that $\sum\limits^\infty_{n=1}a_nb_n\ge \sum\limits_{n\ge N}a_nb_n$.

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    $\begingroup$ Thanks looks good, if i get it right. is it also true for regular limit instead of $\liminf$ ( I mean your proof )? $\endgroup$ – jack Feb 18 '16 at 13:41
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    $\begingroup$ Well, not really. In fact, pick $$a_n=\begin{cases}1&\text{if }n\text{ is even}\\0&\text{if }n\text{ is odd}\end{cases}$$ then, if $\sum_{n\in\Bbb N} a_nb_n\le M$, it must hold $$\liminf_{n\to\infty}\ b_{2n}=0$$ but $b_{2n+1}$ can be anything. For instance, $b_n=a_{n+1}$ is ok. $\endgroup$ – user228113 Feb 18 '16 at 13:44
  • $\begingroup$ yes ! $b_n$ does not have to converge , forgot about that $\endgroup$ – jack Feb 18 '16 at 13:58
  • $\begingroup$ (of course, for convergent sequences $\lim=\liminf$ ) $\endgroup$ – user228113 Feb 18 '16 at 13:59

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