International Mathematics Competition
for University Students
2016

IMC2016: Day 2, Problem 6

6. Let $(x_1,x_2,\ldots)$ be a sequence of positive real numbers satisfying ${\displaystyle \sum_{n=1}^{\infty}\frac{x_n}{2n-1}=1}$. Prove that $$ \displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{k} \frac{x_n}{k^2} \le2. $$

Proposed by Gerhard J. Woeginger, The Netherlands

Solution. By interchanging the sums, $$ \sum_{k=1}^{\infty} \sum_{n=1}^{k} \frac{x_n}{k^2} = \sum_{1\le n\le k} \frac{x_n}{k^2} = \sum_{n=1}^{\infty} \left( x_n \sum_{k=n}^{\infty} \frac{1}{k^2} \right). $$ Then we use the upper bound $$ \sum_{k=n}^{\infty} \frac{1}{k^2} \le \sum_{k=n}^{\infty} \frac{1}{k^2-\frac14} = \sum_{k=n}^{\infty} \left(\frac{1}{k-\frac12} - \frac{1}{k+\frac12}\right) = \frac{1}{n-\frac12} $$ and get $$ \sum_{k=1}^{\infty} \sum_{n=1}^{k} \frac{x_n}{k^2} = \sum_{n=1}^{\infty} \left( x_n \sum_{k=n}^{\infty} \frac{1}{k^2} \right) \lt \sum_{n=1}^{\infty} \left( x_n \cdot \frac1{n-\frac12} \right) = 2\sum_{n=1}^{\infty}\frac{x_n}{2n-1} = 2. $$