Problem

Source: Romanian TST 2018 Problem 3 Day 4

Tags: floor function, algebra, inequalities



Given an integer $n \geq 2$ determine the integral part of the number $ \sum_{k=1}^{n-1} \frac {1} {({1+\frac{1} {n}}) \dots ({1+\frac {k} {n})}}$ - $\sum_{k=1}^{n-1} (1-\frac {1} {n}) \dots(1-\frac{k}{n})$