So I've been trying to figure out how to prove the following.
Let $(a_n)$ be a sequence of positive numbers such that $\sum\limits_{n=1}^\infty a_n =\infty$, and define $s_n=\sum\limits_{i=1}^n a_i$. Then $\sum\limits_{n=1}^\infty\frac{a_n}{s_n} =\infty$ as well.
I can prove it by comparing it to $\int_1^\infty \frac{1}{x} \, dx $ if the sequence $a_n$ is bounded by some $M$, but that's as far as I've been able to get.
Since $s_n\to+\infty$ when $n\to\infty$, for each $n\geqslant1$ there exists some finite $m\gt n$ such that $s_m\geqslant2s_n$. In particular, $\sum\limits_{k=n+1}^m\frac{a_k}{s_k}\geqslant\sum\limits_{k=n+1}^m\frac{a_k}{s_m}=1-\frac{s_n}{s_m}\geqslant1/2$. Thus, the rests of the series $\sum\limits_k\frac{a_k}{s_k}$ do not converge to zero, QED.
Alternative proof 1, using the inequality $x\ge \log(x+1)$: $$ \frac{a_n}{s_n}\ge\log\left(1+\frac{a_n}{s_n}\right)=\log\left(\frac{s_n+a_n}{s_n}\right)=\log(s_{n+1})-\log(s_n)\tag1 $$ Summing (1) from $n=1$ to $n=N$, the RHS telescopes: $$ \sum_{n=1}^N\frac{a_n}{s_n}\ge\log(s_{N+1})-\log(s_1). $$ But $s_n\to\infty$, so the series $\sum\frac{a_n}{s_n}$diverges.
Alternative proof 2, using theory of integration (apologies for the heavy machinery): Define the sequence $(f_n)$ of functions on $[0,\infty)$ by $f_n:=\frac1{s_n} I_{[0,s_n]}$. (So $f_n$ is a rectangle of height $\frac1{s_n}$ placed over the interval $[0,s_n]$.) Then $\int f_n=1$ for each $n$. But $f_n\to0$ pointwise (since $s_n\to\infty$), so $\int\lim f_n=0$. By the dominated convergence theorem, any $g$ that satisfies $|f_n|\le g$ for all $n$ must have $\int g=\infty$. Apply this conclusion to $g:=\sum_{n=1}^\infty\frac1{s_n}I_{[s_{n-1},s_n]}$, which has integral $\sum\frac{a_n}{s_n}$.
