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 6 of 6: Conclude

Previous step Next step

M=\max(m+1,\,m+2K) \implies a_{n+2}=a_n\ (n\ge M)

Detailed analysis

If no index $m$ has $a_{m+2}>a_m$ , then $a_{n+2}=a_n$ for all $n$ (equality throughout, by the given $a_n\le a_{n+2}$ together with the previous lemmas applied everywhere), so $M=1$ works. Otherwise, with $m,K$ as above, take $M=\max(m+1,m+2K)$ : for $n\ge M$ of the same parity as $m+1$ , $a_{n+2}=a_n=a_{m+1}$ ; for $n\ge M$ of the same parity as $m$ , $a_{n+2}=a_n=a_{m+2K}$ . In both cases $a_{n+2}=a_n$ , so $r=2$ satisfies the required condition, and by the first step it is the largest such real number.