Series $\sum^{\infty}_{k=1}a_k$ and $\sum^{\infty}_{k=1}b_k$ converge but $\lim_{n \rightarrow \infty}\frac{b_k}{a_k}$ diverges EDIT: The are many good, simple examples to try here, but I want to understand this "hint," given below, opaque as it is.
I'm working on an existence proof that shows two series $\sum^{\infty}_{k=1}a_k$ and $\sum^{\infty}_{k=1}b_k$ converge but $\lim_{n \rightarrow \infty}\frac{b_k}{a_k}$ diverges. 
So I need $\forall M \in \mathbb{R}$ with $M >0$ $\exists$ $N \in \mathbb{N}$ such that $\forall k \gt N$ we have $\frac{b_k}{a_k} \gt M$. 
My hint is to construct an increasing sequence $(N_j)$ such that $\sum^{\infty}_{k=N_j}a_k \leq \frac{1}{j^3}$ and to let  $b_k = ja_k$ for $Nj \leq k \leq N_{j+1}$. 
Good. Then the sequence (a_k) can be broken down into blocks, with $\{a_1 + a_2 + ... + a_{N_1-1}\}$ less than some M, and the subsequent blocks less than $\frac{1}{j^3}$. So we have a way to estimate the sum: $\sum^{\infty}_{k=1}a_k \leq  M + \sum^{\infty}_{j+1}\frac{1}{j^3}$.
I'm befuddled because I can't figure out how this information about the sums relates to the convergence of $\frac{b_k}{a_k}$ unless there's some test that I'm not thinking of. I thought to construct sequences out of the partial sums, but that seems counter-indicated by the phrasing of the prompt.  
Guidance appreciated.
 A: One example where both $\lim\limits_{n\to\infty} \frac{a_n}{b_n}$ and $\lim\limits_{n\to\infty} \frac{b_n}{a_n}$ diverge would be
$$a_n = \frac{1}{n^2}\sin \left(\frac{n}{2}\right),\ b_n = \frac{1}{n^2}\sin\left(n\right)$$
A: I don't find the given hint particularly helpful. Instead, think of it in the following way: if it were true, then that would imply that whenever $\sum a_n$ and $\sum b_n$ converge, the quotient $a_n/b_n$ has a nonzero limit.
This is too strong to be true: As in Martin R's comment, Take any two examples of convergent series, say
$$\sum _{n = 1}^\infty \frac{1}{2^n}, \quad \sum _{n = 1}^\infty \frac{1}{3^n}$$
or
$$\sum _{n = 1}^\infty \frac{1}{n^2}, \quad \sum _{n = 1}^\infty \frac{1}{n^3}$$
There are plenty of counterexamples.

The hint allows you to show that for all convergent $\sum a_k$ with $a_k > 0$ there exists a convergent series $\sum b_k$ such that $a_k/b_k$ diverges. Indeed:
$$\begin{align*}
\sum_{k = 1}^\infty b_k
& = \sum_{j = 1}^\infty \sum_{k=N_j}^{N_{j+1}-1} b_k \\
& = \sum_{j = 1}^\infty j \sum_{k=N_j}^{N_{j+1}-1} a_k \\
& \leq \sum_{j = 1}^\infty j \cdot \frac{1}{j^3} \\
& < \infty
\end{align*}$$
while we have
$$\frac{b_k}{a_k}\geq j$$
for $k \geq N_j$ so that $\frac{b_k}{a_k} \to \infty$.
A: Let $\sum a_i$ be a convergent series and let $a_i\not=0$. Let $b_i=a_i$. Then $\sum b_i$ is also convergent series and $\frac{a_i}{b_i}=1$. So -- counterexample.
