If $(a_n)_n,\ (b_n)_n,$ are positive convex decreasing sequences, $\sum a_n$ converges and $\sum b_n$ diverges, then $\ \frac{a_n}{b_n}\to 0.$ Defintition: A real sequence $\ (x_n)_n\ $ is convex if $\ x_n - x_{n+1} \geq x_{n+1} - x_{n+2}\quad \forall\ n\in\mathbb{N}. $
Continuing on from this question here,

Proposition $\ 3:\ $ If $\ (a_n)_n,\ (b_n)_n,\ $ are positive convex decreasing sequences, $\ \displaystyle\sum  a_n \ $ converges and $\ \displaystyle\sum  b_n \ $ diverges, then $\ \frac{a_n}{b_n}\to 0.\ $

In the previous question, counter-examples were found if either $\ (a_n)_n,\ $ or $\ (b_n)_n,\ $ were not required to be convex (but were required to be decreasing), so requiring them both to be convex is a follow-up question I cannot resist investigating.

*

*If the proposition is false, then $\ \frac{a_n}{b_n} = c>0\ $ for infinitely many $\ n.\ $ (We may assume WLOG that $\ c=1,\ $ since $\ \displaystyle\sum  a_n \ $  converges $\ \iff \displaystyle\sum \lambda a_n \ $ converges).


*But in order for $\ \displaystyle\sum  a_n \ $ to converge and $\ \displaystyle\sum  b_n \ $ diverge, we need $\ a_n \ll b_n\ $ for most $\ n,\ $ meaning, I think, that for all $\varepsilon > 0$, $$\lim_{n\to\infty} \left( \frac{ \text{ The number of integers } \leq n \text{ with } \frac{a_n}{b_n} < \varepsilon }{n} \right) = 1.$$
I know as the question asker, I get to decide what is meant by "$\ll$". But I'm not sure what I want this to mean rigorously, but maybe the definition above is appropriate?
I suspect these two facts are at odds with one another, although I don't know how to make this rigorous.
 A: This is not true in general. In fact, given any strictly positive, convex, decreasing, summable $a_n$, I can construct a convex, decreasing, non-summable $b_n$ so that $\frac{a_n}{b_n} \not\to 0$.
Defining $(b_n)$
The method will be to select certain points in the graph of the sequence $(a_n)$, and form a sequence of points $(b_n)$ that linearly interpolate these points. By choosing these points carefully, we can ensure that $b_n$ is not summable, but it should retain the convexity requirement. At these points, obviously $\frac{a_n}{b_n} = 1$, which precludes the limit of the ratio being $0$.
First, choose $n_0 = 0$, and take $b_{n_0} = b_0 = a_0$.
Now, suppose $k \ge 0$, and assume we have defined already $n_0, \ldots, n_k$ such that all the following properties hold:

*

*$n_{i+1} > n_i$,

*$(n_{i+1} - n_i)(a_{n_{i+1}} + a_{n_i}) \ge 2$,

for all $i = 0, \ldots, k - 1$.
Choose:
$$n_{k+1} \ge n_k + \frac{2}{a_{n_k}}.$$
Clearly, $n_{k+1} > n_k$, satisfying property 1, and
$$(n_{k+1} - n_k)(a_{n_{k+1}} + a_{n_k}) \ge (n_{k+1} - n_k)a_{n_k} \ge 2.$$
Thus, property 2 is satisfied, and we can recursively choose an entire sequence $(n_k)_k$ satisfying these properties.
We then define $(b_n)$ as a linear interpolation of these points. Specifically, given fixed $n \in \Bbb{N}$ let $k$ be the unique natural number such that $n_k \le n < n_{k+1}$, and define
$$b_n = \frac{n - n_k}{n_{k+1} - n_k}a_{n_{k+1}} + \frac{n_{k+1} - n}{n_{k+1} - n_k}a_{n_k}. \tag{1}$$
Proving $(b_n)$ works
Clearly $(b_n)$ is a positive sequence; at every point it is a convex combination of a sequence of positive numbers: $(a_{n_k})$. We need to show $(b_n)$ is decreasing, convex, and not summable.
To show $(b_n)$ is decreasing, fix $n \in \Bbb{N}$, and let $k \in \Bbb{N}$ such that $n_k \le n < n_{k+1}$. If $n+1 < n_{k+1}$, then using $(1)$,
$$b_{n+1} - b_n = \frac{a_{n_k} - a_{n_{k+1}}}{n_{k+1} - n_k} > 0.$$
This also holds true when $n + 1 = n_{k+1}$. Either way, the sequence is decreasing.
Now we show convexity. Clearly, from the above calculation, if we choose $n$ so that $n_k \le n < n + 2 \le n_{k+1}$, then
$$b_n - b_{n+1} \ge b_{n+1} - b_{n+2}, \tag{2}$$
and in fact, the two sides are equal.
Otherwise, $n + 1 = n_{k+1}$ for some $k$, and so
\begin{align*}
b_n &= \frac{n - n_k}{n_{k+1} - n_k}a_{n_{k+1}} + \frac{n_{k+1} - n}{n_{k+1} - n_k}a_{n_k} \\
b_{n+1} &= a_{n_{k+1}} \\
b_{n+2} &= \frac{n + 2 - n_{k+1}}{n_{k+2} - n_{k+1}}a_{n_{k+2}} + \frac{n_{k+2} - n - 2}{n_{k+2} - n_{k+1}}a_{n_{k+1}}.
\end{align*}
Note: this still holds even if $n + 2 = n_{k+2}$. In this case, we get
\begin{align*}
b_{n+1} - b_n &= \frac{a_{n_k} - a_{n_{k+1}}}{n_{k+1} - n_k} \\
b_{n+2} - b_{n+1} &= \frac{a_{n_{k+1}} - a_{n_{k+2}}}{n_{k+2} - n_{k+1}}.
\end{align*}
Now, using the convexity of $a$,
\begin{align*}
a_{n_k} - a_{n_{k+1}} &= \sum_{i = n_k}^{n_{k+1} - 1} (a_i - a_{i+1}) \\
&\ge (n_{k+1} - n_k)(a_{n_{k+1}-1} - a_{n_{k+1}}) \\
&= (n_{k+1} - n_k)(a_{n-1} - a_n).
\end{align*}
This is due to the fact that minimum term in the sum is the last term. That is,
$$b_{n+1} - b_n \ge a_{n-1} - a_n.$$
Similarly, still using the convexity of $a$, but now bounding with the largest term of the corresponding sum,
$$b_{n+2} - b_{n+1} \le a_n - a_{n+1}.$$
Thus $(2)$ holds, once again, by the convexity of $(a_n)$. That is, in any case, $(b_n)$ is convex.
Finally, we just need to show $(b_n)$ is not summable. We have,
\begin{align*}
\sum_{n=0}^\infty b_n &= \sum_{k=0}^\infty \sum_{i=n_k}^{n_{k+1}-1} b_i \\
&= \sum_{k=0}^\infty \sum_{i=n_k}^{n_{k+1}-1} \left(\frac{i - n_k}{n_{k+1} - n_k}a_{n_{k+1}} + \frac{n_{k+1} - i}{n_{k+1} - n_k}a_{n_k}\right) \\
&= \sum_{k=0}^\infty \left(-a_{n_{k+1}} + \sum_{i=n_k}^{n_{k+1}} \left(\frac{i - n_k}{n_{k+1} - n_k}a_{n_{k+1}} + \frac{n_{k+1} - i}{n_{k+1} - n_k}a_{n_k}\right) \right) \\
&= \sum_{k=0}^\infty \left(-a_{n_{k+1}} + \frac{a_{n_{k+1}} + a_{n_k}}{n_{k+1} - n_k}\sum_{i=0}^{n_{k+1} - n_k} i \right) \\
&= \sum_{k=0}^\infty \left(-a_{n_{k+1}} + \frac{a_{n_{k+1}} + a_{n_k}}{n_{k+1} - n_k} \cdot \frac{1}{2}(n_{k+1} - n_k)(n_{k+1} - n_k + 1) \right) \\
&= \sum_{k=0}^\infty \frac{1}{2}(a_{n_k}-a_{n_{k+1}} + (a_{n_{k+1}} + a_{n_k})(n_{k+1} - n_k)).
\end{align*}
Using the second defining property of $(b_n)$, we therefore have
$$\sum_{n=0}^\infty b_n \ge \sum_{k=0}^\infty \frac{1}{2}(a_{n_k}-a_{n_{k+1}} + 2) > \sum_{n=0}^\infty 1 = \infty.$$
A: Given any strictly positive, convex, decreasing, summable $\ (a_n).$
Let: $\ k_1 = 1,\quad k_{j+1} = 2 \left\lceil \frac{1}{a_{k_j}} \right \rceil + k_j,\qquad \forall\ j\in\mathbb{N}$
Then for each $\ j\in\mathbb{N},\ $let:
$b_n = \left( \frac{ n - k_j}{ k_{j+1} - k_j }\right)\ a_{k_{j+1}} + \left( 1 - \frac{ n - k_j}{ k_{j+1} - k_j } \right)\ a_{k_j} \ $ for each $\ n\ $ with $\ k_j \leq n < k_{j+1},\ $ that is, if
$\ k_j \leq n < k_{j+1},\ $ then $\ (n,b_n)\ $ lies on the straight line joining $\ (k_j, a_{k_j})\ $ to $\ (k_{j+1}, a_{k_{j+1}}),\ $ and so, since $\ (a_n)_n\ $ is convex, $\ (b_n)_n\ $ is also convex.
The idea behind the choice of $\ (k_j)_{j\in\mathbb{N}}\ $ is so that the area of the trapezium under the straight line joining $\ (k_j, a_{k_j}) = (k_j, b_{k_j})\ $ to $\ (k_{j+1}, a_{k_{j+1}}) = (k_{j+1}, b_{k_{j+1}})\ $ is, due to positivity of all $\ a_k,\ $ greater than the area of the triangle bounded by the $\ x-$axis ( $\ n-$axis ) and$\ (k_j, a_{k_j})\ $ to $\ (0, a_{k_{j+1}}).$ Formally, for each $\ j\in\mathbb{N}:$
$$ \sum_{n=k_j}^{n=k_{j+1} - 1} b_n = \sum_{n=k_j}^{n=k_{j+1} - 1} \frac{ n - k_j}{ k_{j+1} - k_j }\ a_{k_{j+1}} + \sum_{n=k_j}^{n=k_{j+1} - 1} \left( \frac{ k_{j+1} - n }{ k_{j+1} - k_j } \right)\ a_{k_j} $$
$$= \frac{ 1 }{ k_{j+1} - k_j } \left( \sum_{n=k_j}^{n=k_{j+1} - 1} \left( n - k_j \right)\ a_{k_{j+1}} + \sum_{n=k_j}^{n=k_{j+1} - 1} \left( k_{j+1} - n \right) a_{k_j} \right) $$
$$= \frac{ 1 }{ k_{j+1} - k_j } \left( \frac{ \left( k_{j+1} - k_j - 1 \right) \left( k_{j+1} - k_j \right) }{ 2 } a_{k_{j+1}} + \frac{ \left( k_{j+1} - k_j \right) \left( k_{j+1} - k_j + 1 \right) }{ 2 } a_{k_j} \right) $$
$$ = \frac{1}{2} \left( \left( k_{j+1} - k_j - 1 \right) a_{k_{j+1} } + \left( k_{j+1} - k_j + 1 \right) a_{k_j} \right) $$
$$ = \frac{1}{2} \left( \left( k_{j+1} - k_j \right) \left( a_{k_{j+1} } + a_{k_J} \right) + \underbrace{a_{k_j} - a_{k_{j+1}}}_{ \geq 0,\ \text{ since } (a_{k_j})_j \text{ is decreasing} } \right) $$
$$ \geq \frac{1}{2} \left( k_{j+1} - k_j \right) \left( a_{k_{j+1} } + a_{k_J} \right) = \left\lceil \frac{1}{a_{k_j}} \right \rceil  \left( a_{k_{j+1} } + a_{k_J} \right) \geq \left\lceil \frac{1}{a_{k_j}} \right \rceil a_{k_j} \geq 1. $$
This shows formally that $\ \displaystyle\sum b_n\ $ diverges to $\ +\infty.$
