Problem 2

Show that

r=2

is the largest real number

r

that satisfies the following condition: if a sequence

a_1,a_2,\ldots

of positive integers fulfills the inequalities

a_n\le a_{n+2}\le\sqrt{a_n^2+r\,a_{n+1}}

for every positive integer

n

, then there exists a positive integer

M

such that

a_{n+2}=a_n

for every

n\ge M

Step 5 of 6: Induct to freeze the odd terms and bound the even terms

Previous step Next step

a_{m+2k+1}=a_{m+1}\ (k\ge1),\qquad a_{m+2k}\uparrow,\ a_{m+2k}\le a_{m+1}

Detailed analysis

Repeating the previous step (comparing $a_{m+2k}$ with $a_{m+2k+1}=a_{m+1}$ via the two lemmas) shows by induction that $a_{m+2k+1}=a_{m+1}$ for every $k\ge1$ and $a_{m+2k}\le a_{m+1}$ for every $k\ge0$ . The hypothesis of the problem already gives $a_m\le a_{m+2}\le a_{m+4}\le\cdots$ , so the even-offset subsequence $(a_{m+2k})_{k\ge0}$ is a non-decreasing sequence of positive integers bounded above by $a_{m+1}$ , hence it is eventually constant: there is $K$ with $a_{m+2k}=a_{m+2K}$ for all $k\ge K$ .