# 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

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.

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…