IMC 2016 · Day 2 · P6

By interchanging the sums, $$\sum_{k=1}^{\infty} \sum_{n=1}^{k} \frac{x_n}{k^2} = \sum_{1 \le n \le k} \frac{x_n}{k^2} = \sum_{n=1}^{\infty} x_n \left( \sum_{k=n}^{\infty} \frac{1}{k^2} \right).$$ 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 $$\sum_{k=1}^{\infty} \sum_{n=1}^{k} \frac{x_n}{k^2} = \sum_{n=1}^{\infty} x_n \left( \sum_{k=n}^{\infty} \frac{1}{k^2} \right) < \sum_{n=1}^{\infty} x_n \cdot \frac{1}{n - \frac12} = 2 \sum_{n=1}^{\infty} \frac{x_n}{2n-1} = 2.$$