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Algebra Difficulty 6.5 National olympiad Prove it China

The sequence $\{a_n\}$ is defined as follows: $a_1 = 0$ , and for integer $n \ge 2$ ,
$a_n = \frac{1}{n} + \frac{1}{\lceil \frac{n}{2} \rceil} \sum_{k=1}^{\lceil \frac{n}{2} \rceil} a_k,$
where $\lceil \frac{n}{2} \rceil$ denotes the smallest integer not less than $\frac{n}{2}$ . Find the maximum term of the sequence $\{a_n\}$ .

Solution

Proof. From the given definition, we have $a_2 = \frac{a_1}{2} + \frac{1}{2} = \frac{a_1+a_2}{2} + \frac{1}{3} = \frac{7}{12}$ .
We now prove by induction that $a_n \le \frac{7}{12}$ , with equality if and only if $n = 3$ . The cases $n = 1, 2, 3$ have been verified. Assume the statement holds for all $1, 2, \dots, n-1$ ( $n \ge 4$ ), then
$\begin{align*} a_n &= \frac{a_1 + \dots + a_{\lceil \frac{n}{2} \rceil}}{\lceil \frac{n}{2} \rceil} + \frac{1}{n} \\ &\le \frac{0 + \frac{1}{2} + \left(\lceil \frac{n}{2} \rceil - 2\right) \cdot \frac{7}{12}}{\lceil \frac{n}{2} \rceil} + \frac{1}{n} \\ &= \frac{7}{12} - \frac{2}{3\lceil \frac{n}{2} \rceil} + \frac{1}{n} \\ &= \frac{7}{12} - \frac{2n - 3\lceil \frac{n}{2} \rceil}{3n \cdot \lceil \frac{n}{2} \rceil} \\ &\le \frac{7}{12} - \frac{2n - 3\left(\frac{n+1}{2}\right)}{3n \cdot \lceil \frac{n}{2} \rceil} \\ &= \frac{7}{12} - \frac{\frac{n}{2} - \frac{3}{2}}{3n \cdot \lceil \frac{n}{2} \rceil} \\ &< \frac{7}{12}. \end{align*}$
Therefore, the maximum term of the sequence $\{a_n\}$ is $a_3 = \frac{7}{12}$ . $\square$

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