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.
 A: 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.
A: 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…
