= Solution
Suppose $\sum_na_n$ diverges. Positivity implies $s_n\to\infty$. Put $t_n=a_n/s_n\in(0,1)$ for $n\geq2$. Since $s_{n-1}=s_n-a_n$,
$$
\frac{s_n}{s_{n-1}}=\frac1{1-t_n}.
$$
If $\sum_nt_n$ converged, then eventually $t_n\leq1/2$ and
$$
\log\frac{s_n}{s_{n-1}}=-\log(1-t_n)\leq2t_n.
$$
Summing would bound the telescoping quantity $\log(s_n/s_1)$, contradicting $s_n\to\infty$. Therefore
$$
\boxed{\sum_{n=1}^\infty\frac{a_n}{s_n}=\infty}.
$$
The converse is true. If $\sum_na_n<\infty$, then $s_n\geq s_1>0$ and
$$
\sum_n\frac{a_n}{s_n}\leq\frac1{s_1}\sum_na_n<\infty.
$$
Taking the contrapositive shows that divergence of the ratio series forces divergence of $\sum_na_n$.
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