An inequality for two positive series Let $\{ a_n \}_{n \in \mathbb{N} }$ and $\{ b_n \}_{n \in \mathbb{N} }$ be two non-increasing sequences of positive numbers,
$$
\sum _{n=1} ^\infty a_n  < \infty,  \ \ \  \   \ \   \  \  \  \  \  \ 
\sum _{n=1} ^\infty b_n  = \infty.
$$
Let $\{ u_n \}_{n \in \mathbb{N} }$ be an increasing sequence of positive numbers, $u _n \to \infty$ as $n \to \infty$.  Is it true that the inequality
$$
\sum _{n=1} ^m a_n u _n  \leq \sum _{n=1} ^m b_n u _n  \   \   \  \   \  \  \  \  \ \ \ 
 \   \   \  \   \  \  \  \  \ \ \ 
(1)
$$
holds for all sufficiently large $m \in \mathbb{N}$? If not, is it true that (1) holds for infinitely many $m \in \mathbb{N}$?
 A: The question that asks if the proposition must hold for all sufficiently large $m$ must be answered in the negative. I have at the moment no answer for the second question.
Set for  $n\ge 1$
$$a_n=\frac1{2^n},$$
and define $i_n \in \mathbb N$ recursively as
$$i_1=1, i_{n+1}=i_n+\frac2{a_{i_n}}, n \ge 1$$ and then
$$b_n=\frac{a_{i_k}}2, \text{ if } i_k \le n < i_{k+1}.$$
The $i_n$ sequence is easily seen to tend to $\infty$, so the definition of $b_n$ makes sense.
What happens is that $b_1$ starts out as half of $a_1=\frac12$, but then $b_n$ stays constant for as long as necessary to reach sum 1, which is 4 for summands for $b_1=\frac14$. Then $b_5$ is again half of $a_5=\frac1{32}$ and $b_6, b_7,\ldots$ are all the same, until again the sum of all $b_n$'s, starting from $b_5$ becomes one (64 summands, so $i_2=5+64=69$).
That means both $(a_n)$ and $(b_n)$ fullfill the conditions of the problem, but still $b_{i_k} = \frac{a_{i_k}}2 < a_{i_k}$ for each $k$.
The condition (1) is equivalent to
$$\sum_{n=1}^m (b_n-a_n)u_n \ge 0$$,
and that means whatever $\sum_{n=1}^{i_k-1} (b_n-a_n)u_n$ for any $k$ may be, you can choose $u_{i_k}$ big enough such that $\sum_{n=1}^{i_k} (b_n-a_n)u_n < 0$, because the last summand has a negative coefficient $(b_{i_k}-a_{i_k})$.
