Does $\sum\limits_{n=1}^\infty\frac{a_{n+1}-a_n}{b_{n+1}-b_n}<\infty$ imply $\sum\limits_{n=1}^\infty\frac{a_n}{b_n}<\infty?$ Let $(a_n)_{n\geq 1}$ and $(b_n)_{n\geq 1}$ be sequnces of positive numbers such that $a_n\to 0$ and $b_n\to 0$ as $n\to\infty$. Then it is known that if $(b_n)_{n\geq 1}$ is strictly monotone
and
$$\lim\limits_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=0$$
then
$$\lim\limits_{n\to\infty}\frac{a_n}{b_n}=0.$$
If we additionally have that
$$\sum\limits_{n=1}^\infty\frac{a_{n+1}-a_n}{b_{n+1}-b_n}<\infty$$
does it then follow that
$$\sum\limits_{n=1}^\infty\frac{a_n}{b_n}<\infty?$$
 A: It does not follow that $\sum \frac{a_n}{b_n}$ converges. To see why it doesn't follow, and how to construct examples, I think it's best to take a different point of view, concentrating on the differences $\beta_n := b_n - b_{n+1}$ and the quotients
$$q_n := \frac{a_n - a_{n+1}}{b_n - b_{n+1}}\,.$$
Let's also suppose that $(a_n)$ is strictly monotonic. Then we have two positive sequences, $(q_n)$ and $(\beta_n)$, and by assumption the series $\sum q_n$ and $\sum \beta_n$ both converge. And by construction we have
\begin{align}
b_n &= \sum_{m = n}^{\infty} \beta_m, \\
a_n &= \sum_{m = n}^{\infty} q_m\beta_m, \\
\frac{a_n}{b_n} &= \frac{\sum_{m = n}^{\infty} q_m\beta_m}{\sum_{m = n}^{\infty} \beta_m}\,.
\end{align}
From the last we can immediately read off that $\sum \frac{a_n}{b_n}$ converges if $(q_n)$ is monotonic, since then $\frac{a_n}{b_n} \leqslant q_n$. And little more thought shows that $\sum \frac{a_n}{b_n}$ converges if there is any bound $C \in [1,+\infty)$ and $n_0$ such that $n_0 \leqslant n \leqslant m$ implies $q_m \leqslant C\cdot q_n$. For then we have $\frac{a_n}{b_n} \leqslant C\cdot q_n$ for $n \geqslant n_0$.
Thus to construct an example where $\sum \frac{a_n}{b_n}$ diverges, the sequence $(q_n)$ must contain a subsequence $(q_{n_k})$ such that for each $k$ the value $q_{n_k}$ is huge in comparison to many of the values $q_n$ for $n < n_k$. But of course $\sum q_n$ must still converge.
This type of construction often occurs in counterexamples, one such sequence is
$$q_n = \begin{cases} k^{-2} &\text{if } n = n_k \\ 2^{-n} &\text{otherwise} \end{cases}$$
where of course $k$ is not fixed, but ranges over all positive integers, and $(n_k)$ is a suitably fast growing sequence of natural numbers.
Then for every $k > 1$ we have
$$\sum_{n = n_{k-1}+1}^{n_k} \frac{a_n}{b_n} > \sum_{n = n_{k-1}+1}^{n_k} \frac{q_{n_k}\beta_{n_k}}{b_n} > \frac{n_k - n_{k-1}}{k^2} \frac{\beta_{n_k}}{b_{n_{k-1}+1}}$$
and we can achieve our goal if we choose $(\beta_n)$ such that $b_{n_{k-1}+1} \leqslant 3\beta_{n_k}$ (of course any constant would do in place of $3$), and $(n_k)$ in such a way that
$$\sum_{k = 2}^{\infty} \frac{n_k - n_{k-1}}{k^2}$$
diverges. For the latter, we can take $n_k = k^2$ for example (or anything growing faster, say $k^3$, $2^k$, or $k!$). For the former, we can choose
$$\beta_n = \begin{cases} (k!)^{-1} &\text{if } n = n_k \\ 2^{-n}(k!)^{-1} &\text{if } n_{k-1} < n < n_k\end{cases}$$
setting $n_0 = 0$ to have $\beta_n$ defined for all $n$.
With that choice we indeed have
$$\sum_{n = n_{k-1} + 1}^{n_k} \beta_n = \frac{1}{k!}\sum_{n = n_{k-1} + 1}^{n_k-1} 2^{-n} + \frac{1}{k!} < \frac{2}{k!}$$
and hence
$$b_{n_{k-1}+1} = \sum_{m = k}^{\infty} \sum_{n = n_{m-1}+1}^{n_m} \beta_n < \sum_{m = k}^{\infty} \frac{2}{m!} < \frac{2}{k!}\cdot \frac{1}{1 - \frac{1}{k+1}} < \frac{3}{k!} = 3\beta_{n_k}\,.$$
