Let $(a_j)$ and $(b_j)$ be two monotonically decreasing sequences both converging to zero, such that both $\sum_{j=1}^\infty a_j$ and $\sum_{j=1}^\infty b_j$ diverge with \begin{align*} \limsup_{n\to\infty}\gamma_n=\infty, \end{align*} where \begin{align*} \gamma_n:=\frac{\sum_{j=1}^na_j}{\sum_{j=1}^nb_j}. \end{align*}
QUESTION: Is it always possible to find a sequence $(y_j)$ of non-negative terms such that $\sum_{j=1}^\infty a_jy_j$ stays divergent while $\sum_{j=1}^\infty b_jy_j$ becomes convergent?
I suppose not, but I couldn't come up with a counterexample so far. In order to find such $y_j$, I tried something like $y_j:=1/A_j$, where $A_n:=\sum_{j=1}^n a_j$. Then one has on the one hand that $\sum_{j=1}^\infty a_jy_j=\sum_{j=1}^\infty\frac{a_j}{A_j}$ diverges by Abel's Theorem. On the other hand, setting $B_n:=\sum_{j=1}^nb_j$ we get by partial summation \begin{align*} \sum_{j=1}^n b_jy_j&=B_n/A_n+\sum_{j=1}^{n-1} B_j\frac{a_{j+1}}{A_jA_{j+1}} \\ &=\frac{1}{\gamma_n}+\sum_{j=1}^{n-1} \frac{a_{j+1}}{\gamma_jA_{j+1}}. \end{align*} Here, I know that $\sum_{j=1}^\infty \frac{a_{j+1}}{A_jA_{j+1}}$ converges, whereas $\sum_{j=1}^\infty \frac{a_{j+1}}{A_{j+1}}$ diverges, but I think that in general $B_j$ doesn't grow sufficiently "slowly" to keep $\sum_{j=1}^n B_j \frac{a_{j+1}}{A_jA_{j+1}}$ converging.
EDIT: Another indication that those $(y_j)$ don't have to exist is that from $A_n=\gamma_n B_n$ we get $\gamma_n/n=\frac{A_n}{n B_n}$, and since $A_n/n\to0$ and $B_n\to\infty$ we obtain $\gamma_n/n\to0$, i.e. $\gamma_n$ grows slower than linear, which also means that $A_n$ grows almost as fast as $B_n$...
Any help is highly appreciated, thanks in advance!
Such a sequence $y_n$ always exists, and we don't need monotonicity or the $\to 0$ hypothesis. Here's the idea: First prove that for any $M\in \mathbb N,$
$$\limsup_{N\to \infty} \frac{ \sum_{n=M}^N a_n}{\sum_{n=M}^N b_n }= \infty.$$
This allows us to choose positve integers $1=N_1 < N_2 < \cdots $ such that
$$\tag 1\frac{ \sum_{n=N_k}^{N_{k+1}-1} a_n }{\sum_{n=N_k}^{N_{k+1}-1} b_n} > 2^k\,\text { for } k=1,2,\dots$$
Let $A_k,B_k$ denote the numerator, denominator on the left of $(1).$ We define the sequence $y_n$ in blocks. For $k=1,2,\dots$ and $N_k\le n \le N_{k+1}-1$ define
$$y_n = \frac{1}{k^2B_k}.$$
Then
$$\sum_{n=1}^{\infty}y_na_n =\sum_{k=1}^{\infty}\sum_{n=N_k}^{N_{k+1}-1} y_na_n = \sum_{k=1}^{\infty}\frac{A_k}{k^2B_k} \ge \sum_{k=1}^{\infty}\frac{2^k}{k^2} = \infty.$$
In the same way we see
$$\sum_{n=1}^{\infty}y_nb_n=\sum_{k=1}^{\infty}\frac{1}{k^2} < \infty.$$