Let $(a_n)$ be a sequence of positive real numbers such that $\sum a_n$ diverges. Prove that there exists a sequence $(b_n)$ of positive real numbers such that $b_n/a_n \to 0$, but $\sum b_n$ is divergent.
I found a possible candidate solution for the sequence $(b_n)$ where
$$
b_n = \frac{a_n}{\sum_{k=1}^n a_k}.$$
In fact, since $\sum a_k$ diverges, $b_n/a_n$ tends to 0. However I don't know how to show wether $\sum b_n$ actually diverges.
Best Answer
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.