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Problem 839

AIME late

Number theory Difficulty 5.6 Prove it

Example 4 Let the positive divisors of 8128 be denoted as $a_{1}, a_{2}, \cdots$ , $a_{n+1}$ , where $a_{1}=1, a_{n+1}=8128$ . Prove:
$\sum_{k=2}^{n} \frac{a_{k}}{k\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}\right)}<\frac{1}{2} .$

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Prove: Since $8128=2^{6}\left(2^{7}-1\right)$ is a perfect number, we have
$\begin{array}{l} \frac{1}{a_{2}}+\frac{1}{a_{3}}+\cdots+\frac{1}{a_{n+1}}=1 . \\ \text { Then } \sum_{k=2}^{n} \frac{a_{k}}{k\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{k}^{2}\right)} \\ <\sum_{k=2}^{n} \frac{a_{k}}{k a_{k}^{2}} \\ =\sum_{k=2}^{n} \frac{1}{k a_{k}} \\ \leqslant \frac{1}{2} \sum_{k=2}^{n} \frac{1}{a_{k}} \\ <\frac{1}{2} \sum_{k=2}^{n+1} \frac{1}{a_{k}} \\ =\frac{1}{2} . \end{array}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.