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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.

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

2 Answers 2

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

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  • $\begingroup$ 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. $\endgroup$ Commented Nov 22, 2022 at 10:54
  • $\begingroup$ I haven't actually read your answer properly, but at a glance I suspect it is similar to mine, at least in parts... $\endgroup$ Commented Nov 22, 2022 at 23:58
  • $\begingroup$ @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. $\endgroup$ Commented Nov 23, 2022 at 0:21
  • $\begingroup$ (+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. $\endgroup$
    – Mittens
    Commented Jan 30 at 16:56
  • $\begingroup$ @Mittens Thanks! I'm glad you found it useful. $\endgroup$ Commented Jan 31 at 0:13
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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 $\ +\infty.$

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