Let $a$ be a given positive integer. We consider the sequence $(x_n)_{n \ge 1}$ defined by $x_n=\frac{1}{1+na},$ for every positive integer $n.$ Prove that for any integer $k \ge 3,$ there exist positive integers $n_1<n_2<\ldots<n_k$ such that the numbers $x_{n_1},x_{n_2},\ldots,x_{n_k}$ are consecutive terms in an arithmetic progression.
Problem
Source: Romanian National Olympiad 2024 - Grade 9 - Problem 4
Tags: algebra, Arithmetic Progression, Sequence
dgkim
04.06.2024 18:00
Nice problem. I wonder why nobody has solved it...
Claim) for each $\epsilon >0$ and $k\geq 3$, there exists a sequence satisfying the given condition such that
$$(1+\epsilon )(1+an_{k-1})>(1+an_k)$$proof)
$\mathit{i)}$ $k=3$:
$$\frac{1}{(1+an)(1+a(n+1))} ,\frac{1}{(1+an)(1+a(n+2))} ,\frac{1}{(1+a(n+1))(1+a(n+2))}$$form an arithmetic sequence, and if we set $n$ very large, we can make it satisfy $(1+\epsilon )(1+an)>(1+a(n+1))$.
Let's assume that the statement holds for $k=m-1$.
$\mathit{i)}$ $k=m$:
Set some $\epsilon >0$. Then there exists a sequence $n_1,n_2,\cdots n_{m-1}$ such that
$$(1+\frac{\epsilon}{1+\epsilon} )(1+an_{m-2})>(1+an_{m-1})$$Let $t=2n_{m-2}-n_{m-1}$. Also let $n_i'=an_it+n_i+t$ for $i\in \{1,2,\cdots ,m-1\}$.
Then clearly $n_1', n_2',\cdots n_{m-1}'$ satisfies the condition, since $1+an_i'=(1+an_i)(1+at)$.
Set $n_m'=an_{m-1}n_{m-2}+n_{m-1}+n_{m-2}$. Then by some calculation, it is clear that $x_{n_1'},x_{n_2'},\cdots ,x_{n_m'}$ form an arithmetic sequence.
Also,
$$(1+\epsilon )(1+n_{m-1}'a)<(1+an_m')$$$$\Leftrightarrow (1+\epsilon )(1+an_{m-2})>(1+an_{m-1})$$Since $\frac{\epsilon}{1+\epsilon} <\epsilon$, This condition also holds.
Now it is left to prove that $t>0$.
This is easy by the fact that $$(1+\frac{\epsilon}{1+\epsilon} )(1+an_{m-2})>(1+an_{m-1})$$so the problem is solved. $\blacksquare$