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 4 of 6: Start the alternating chain

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a_{m+2}>a_m \implies a_{m+2}\le a_{m+1},\quad a_{m+3}=a_{m+1}

Detailed analysis

Suppose some index $m$ has $a_{m+2}>a_m$ . By the contrapositive of the first lemma (with $n=m$ ), we cannot have $a_{m+1}\le a_m$ , so $a_m\le a_{m+1}$ ; the second lemma then gives $a_{m+2}\le a_{m+1}$ . Applying the first lemma at $n=m+1$ (since $a_{m+2}\le a_{m+1}$ ) yields $a_{m+3}=a_{m+1}$ .