Does every positive, decreasing, real sequence whose series converges have a corresponding convex sequence greater than it whose series converges? 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}. $
Suppose $\ (x_n)_n\ $ is a positive, decreasing sequence of real numbers such that $\ \displaystyle\sum x_n\ $ converges.
Proposition: There exists a convex sequence $\ (y_n)_n,\ $ such that:

*

*$\ y_n\geq x_n\quad \forall\ n\in\mathbb{N},\ $ and

*$\ \displaystyle \sum_n y_n\ $ converges.

I suspect there is some counter-example, and I think my best attempt to find one is that maybe there is an increasing subsequence $\ (A_n)_n\ $ of $\ \mathbb{N}\ $ such that $\ x_n:= \frac{1}{A_k}\ $ for all $\ n\ $ with $\ A_{k-1} < n \leq A_k,\ $ and then maybe the convexity of $\ (y_n)_n\ $ sort of forces $\ y_n \approx \frac{1}{n}\ ?$ But I'm not sure if this is true or how to prove this.
Edit: I have spent more time thinking about this problem without further progress towards a solution. Therefore I have added a bounty of $+50$.
 A: I think the answer is yes. The idea is to consider a finitary version of the problem, which can be handled via LP.
Let $M = 2\sum_{i = 1}^\infty x_i$.
Lemma: for every $n \geq 0$, there exists a sequence $Y_n = (y_1, \cdots, y_n)$ that satisfies

*

*$y_i \geq x_i$.

*$y_i$ is convex, meaning $y_i + y_{i + 2} \geq 2y_{i + 1}$ for all $1 \leq i \leq n -2$.

*$\sum_{i = 1}^{n - 2} y_i \leq M$.

Proof: We can pose this as a standard LP problem. Let $y_i' = y_i - x_i$. Then the LP is

*

*Constraint 1: $y_i' \geq 0$ for all $1 \leq i \leq n$.

*Constraint 2: $y_i' - 2y'_{i + 1} + y'_{i + 2} \geq - x_{i} + 2x_{i + 1} - x_{i + 2}$ for all $1 \leq i \leq n - 2$.

*Objective: minimize $\sum_{i = 1}^n y_i'.$
By the fundamental lemma for LPs, this has the same solution as its dual LP, which is posed as follows.

*

*Constraint 1: $z_i \geq 0$ for all $1 \leq i \leq n - 2$.

*Constraint 2: $z_i - 2z_{i + 1} + z_{i + 2} \leq 1$ for all $-1 \leq i \leq n - 2$, where we assume $z_{-1} = z_0 = z_{n - 1} = z_n = 0$.

*Objective: maximize $\sum_{i = 1}^{n - 2} (-x_{i} + 2x_{i + 1} - x_{i + 2})z_i.$
We can arrange the objective as
$$\sum_{i = 1}^{n - 1} (x_i - x_{i + 1})(z_{i - 1} - z_{i}).$$
Let $d_i = z_{i - 1} - z_i$. Then we have $d_0 = d_n = 0$ and $d_{i - 1} \leq d_i + 1$. The important observation is that $\boxed{d_i \leq i}$. To justify this claim, we divide it into two cases.
Case 1: $i \leq n / 2$. Then we have
$$z_{2i - 1} = -\sum_{j = 1}^{2i - 1} d_j.$$
We can split the sum into two parts. For $j < i$, we have $d_j \geq -j$ since $d_i \geq d_{i - 1} - 1$ and $d_0 = 0$. For $j \geq i$ we have $d_j \geq d_i - (j - i)$ for the same reason. So we conclude that
$$z_{2i - 1} \leq - \sum_{j = 0}^{i - 1} (-j + d_i - j).$$
However, $z_{2i - 1} \geq 0$, so we conclude that $d_i \leq i$.
Case 2: $i > n / 2$. In this case, we have $d_i \leq d_n + (n - i) =n - i \leq i$.
So we have shown our observation. The objective of the dual LP is at most (btw, this is where the assumption that $\{x_n\}$ is decreasing is used)
$$\sum_{i = 1}^{n - 1} (x_i - x_{i + 1}) i \leq \sum_{i = 1}^\infty x_i.$$
So the objective of the original LP is also at most this value, and we conclude the existence of $Y_n$ by finding the optimal solution to the original LP $\{{y_i'}^*\}$ and setting $y_i = {y_i'}^* + x_i$.
Pad the sequence $Y_n$ with zeros to make it an infinite sequence $Y_n = (y_1, \cdots, y_n, 0, 0,\cdots) \subset \mathbb{R}^{\mathbb{N}}$. The rest of the argument is compactness.
Note that the entries of $Y_n$ are bounded by $M$. By diagonalization or Tychonoff's theorem, we can find a subcollection $\{Y_{n_i}\}_i$ of $\{Y_n\}_n$ that converges pointwise(i.e. in $\mathbb{R}^{\mathbb{N}}$) to a sequence $Y_\infty$. Now $Y_{\infty}$ is convex and upper bounds $x_i$ by pointwise convergence(in more fancy terms, these properties are ``open" in the topology on $\mathbb{R}^{\mathbb{N}}$), and lies in $L^1(\mathbb{N})$ by Fatou's lemma.
