# 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.

1. 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).

2. 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.

• One still has that $\liminf_n\frac{a_n}{b_n}=0$. Commented Jan 21 at 21:00

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:

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

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 $$\sum_{n=0}^\infty b_n \ge \sum_{k=0}^\infty \frac{1}{2}(a_{n_k}-a_{n_{k+1}} + 2) > \sum_{n=0}^\infty 1 = \infty.$$

• Oh yeah, if we build $b_n$ given $a_n$ then it's easy. I was trying the opposite: building $a_n$ given $b_n$... This makes me think of a simple counter-example, which I'll post as an answer shortly. Commented Nov 22, 2022 at 10:54
• I haven't actually read your answer properly, but at a glance I suspect it is similar to mine, at least in parts... Commented Nov 22, 2022 at 23:58
• @AdamRubinson Yes, our constructions are essentially the same. There is a superficial difference is in how we pick our subsequence, in that I instruct the reader to pick $n_{k+1}$ greater than $n_k + \frac{2}{n_k}$, whereas you select one such number with the ceiling function. Commented Nov 23, 2022 at 0:21
• (+1) That was smart construction. I used your answer to provide a solution to a recent posting here which is the equivalent question of Adam's but for integrals. Commented Jan 30 at 16:56
• @Mittens Thanks! I'm glad you found it useful. Commented Jan 31 at 0:13

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.$$