Consider the $n$ sums:
$$s_k = \sum_{j=1}^k a_j$$
and their remainder after a division by $n$:
$$n|s_k-r_k$$
There are $n$ of them.
If one of them is 0: take $s_k$ and you are done.
Otherwise, there are $n-1$ possibilities for the possible values of $r_k$, so (pigeonhole principle) there are $p<q$ such as $r_p = r_q$, and so you take
$\sum_{j=p+1}^q a_j$ and you are done.
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.
Best Answer
The affirmative answer to a previous question of mine means that for every $k$ no matter how large, $A$ will contain a quasi-arithmetic progression of length $k.$
Now note that every quasi-A.P. of length $k$ has a concave subsequence of length at least $\ \log_2 k - 1.\ $ For example, using the above theorem, we can find a quasi-arithmetic progression of $A$ of length $k=64.$ That is, there exists intervals $\ I_0,\ I_1,\ I_2,\ I_3,\ \ldots,\ I_{63}$ of equal width, such that $\ a_0\in I_0,\ a_1\in I_1,\ a_2\in I_2,\ a_3\in I_3,\ \ldots,\ a_{63} \in I_{63}.$ Then $a_0,\ a_{32},\ a_{32+16},\ a_{32+16+8},\ a_{32+16+8+4},\ a_{32+16+8+4+2}\ $ is a concave subsequence of $A,\ $ and we are done, since we can do all this for arbitrarily large $k,$ and $\ \log_2 k - 1\to\infty\ $ as $\ k\to\infty.$