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 1 of 6: Rule out r>2 with a growing example

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a=\left\lceil\tfrac{1}{r-2}\right\rceil,\quad a_n=a+\left\lfloor\tfrac{n}{2}\right\rfloor

Detailed analysis

Suppose $r>2$ and fix an integer $a\ge 1/(r-2)$ . Let $a_n=a+\lfloor n/2\rfloor$ . Then $a_n\le a_{n+2}=a_n+1$ , and since $a_{n+1}\ge a_n\ge a\ge 1/(r-2)$ we get $(r-2)a_n\ge 1$ , i.e. $2a_n+1\le r\,a_n\le r\,a_{n+1}$ , so $a_{n+2}^2=(a_n+1)^2=a_n^2+2a_n+1\le a_n^2+r\,a_{n+1}$ . Thus the sequence satisfies the hypothesis for this $r$ , yet $a_{n+2}=a_n+1\ne a_n$ for every $n$ , so no such $M$ exists. Hence $r$ cannot exceed $2$ .