Let $a_n>0$; $\sum a_n$ diverges; find $b_n$ s.t. $b_n>0$; $b_n/a_n\to0$; $\sum b_n$ diverges

Note that the sequence of partial sums $S_n = \sum_{k=1}^n a_k$ is increasing and divergent. Hence,

$$\left|\sum_{k=n+1}^m b_k \right|= \sum_{k=n+1}^m \frac{a_k}{S_k} > \frac{1}{S_m}\sum_{k=n+1}^m a_k = \frac{S_m- S_n}{S_m} = 1 - \frac{S_n}{S_m}$$

Since $S_m \to \infty$ as $m \to \infty$, for any fixed $n$ there exists $m > n$ such that $S_n/S_m < 1/2$ and

$$\left|\sum_{k=n+1}^m b_k \right| > \frac{1}{2}$$

Therfore, $\sum b_n$ diverges as the Cauchy criterion is violated.

Let $N_1$ be the smallest natural such that $\sum_{n=1}^{N_1}a_n\geqslant1$. For each $n\leqslant N_1$, put $b_n=1$. Then $\sum_{n=1}^{N_1}b_n\geqslant1$.

Now, let $N_2$ be the smallest natural greater than $N_1$ such that $\sum_{n=N_1+1}^{N_2}a_n\geqslant2$. For each $n\in\{N_1+1,N_1+2,\ldots,N_2\}$, put $b_n=\frac12a_n$. Then $\sum_{n=N_1+1}^{N_2}b_n\geqslant1$.

Now, let $N_3$ be the smallest natural greater than $N_2$ such that $\sum_{n=N_2+1}^{N_3}a_n\geqslant3$. For each $n\in\{N_2+1,N_2+2,\ldots,N_3\}$, put $b_n=\frac13a_n$. Then $\sum_{n=N_2+1}^{N_3}b_n\geqslant1$.

And so on…