a_k+a_n/1+a_ka_n is constant when k+n is constant

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a_k+a_n/1+a_ka_n is constant when k+n is constant

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Source: SMO senior 2024 Q5

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#1 h Aug 3, 2024, 2:45 AM • 2 Y

Y by Resolut1on07, GeoKing

Let $a_1,a_2,\dots$ be a sequence of positive numbers satisfying, for any positive integers $k,l,m,n$ such that $k+n=m+l$ , $\frac{a_k+a_n}{1+a_ka_n}=\frac{a_m+a_l}{1+a_ma_l}.$ Show that there exist positive numbers $b,c$ so that $b\le a_n\le c$ for any positive integer $n$ .

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#2 h Aug 3, 2024, 3:14 AM • 4 Y

Y by Resolut1on07, GeoKing, gghx, centslordm

We show that the sequence is bounded above. In fact, this finishes because if $a_1,a_2,\ldots$ satisfies the given constraint then so does $1/a_1,1/a_2,\ldots$ , and we can apply the same argument to that sequence.

First note that $\frac{a+b}{1+ab} \geq 1$ if and only if $a-1,b-1$ have different sign, so for any fixed value of $i+j$ , $(a_i-1)(a_j-1)$ should have fixed sign (positive, negative, or zero). Clearly if $a_i=1$ then $a_j=1$ for all $j>i$ as well, else take $(i,2j-i)$ and $(j,j)$ , and then the desired claim is true. Thus assume that no terms in the sequence are $1$ . Then if $a_i<1$ then $a_{i+2}<1$ is forced by comparing $(i,i+2)$ with $(i+1,i+1)$ , and by induction eventually either every term is at most $1$ or exactly other term is at most $1$ . We only care about the latter scenario, in which case we care only about the subsequence $a_k,a_{k+2},\ldots$ of terms greater than $1$ , which "behaves" the same way as a complete sequence in itself would. Thus assume that every term is greater than $1$ .

Let $a_i=\coth x_i$ for each $i$ , so each $x_i$ is a positive, and it suffices to show $x_i$ is bounded below by a positive real. The condition (after reciprocating both sides) then becomes $\coth(x_k+x_n)=\coth(x_m+x_l) \implies x_k+x_n=x_m+x_l \implies 2x_n=x_{n-1}+x_{n+1}$ . This is then a linear recurrence whose solutions (over $\mathbb{R}$ ) are $x_n=an+b$ . But if $a<0$ then we eventually have $x_n<0$ which is forbidden, so $a \geq 0$ and the $x_i$ are bounded below by $x_1>0$ . $\blacksquare$

This post has been edited 4 times. Last edited by IAmTheHazard, Aug 3, 2024, 8:25 PM

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#3 h Apr 20, 2025, 7:26 AM

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Note that $\frac{a_k+a_n}{1+a_ka_n}=\frac{a_m+a_l}{1+a_ma_l}$ can be expressed as $\frac{a_k + a_n}{(1+a_k)(1+a_n)} = \frac{a_m + a_l}{(1+a_m)(1+a_l)}$
Let $b_i = 1+a_i$ , where $b_i \in (1, \infty)$ . This implies that $a_i = b_i-1$ , so after substitution, we get
$\frac{1}{b_n} + \frac{1}{b_k} - \frac{2}{b_kb_n} = \frac{1}{b_m} + \frac{1}{b_l} - \frac{2}{b_mb_l}$
Let $c_i = \frac{1}{b_i}$ , where $c_i \in (0, 1)$ . We then get
$c_n+c_k-2c_nc_k=c_l+c_m-2c_lc_m$ which we then multiply both sides by $-1$ and factor to get
$(2c_n-1)(2c_k-1)=(2c_m-1)(2c_l-1)$
Let $d_i = 2c_i-1$ , where $d_i \in (-1, 1)$ . We then get
$\frac{d_n}{d_m} = \frac{d_m}{d_k}$ Note that $n-m=l-k$ , so when $n-m=l=k=1$ , it forms a geometric progression where $d_1 > d_2 > \dots > d_n$ as $|$ common ratio $|$ $\leq 1$

So let $d_1 = x$ , where $x \in \mathbb{N}$
$d_i \in [-x, x]$ $c_i \in \left[ \frac{1 - x}{2}, \frac{1 + x}{2} \right]$ $b_i \in \left[ \frac{2}{1 + x}, \frac{2}{1 - x} \right]$ $a_i \in \left[ \frac{1 - x}{1 + x}, \frac{1 + x}{1 - x} \right]$
So obviously $\frac{1-x}{1+x} \leq a_n \leq \frac{1+x}{1-x}$

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