IMC 1997 · Day 1 · P1

It is well known that $$-1 = \int_0^1 \ln x\,dx = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left( \frac{k}{n} \right)$$ (Riemman's sums). Then $$\frac{1}{n} \sum_{k=1}^{n} \ln \left( \frac{k}{n} + \varepsilon_n \right) \ge \frac{1}{n} \sum_{k=1}^{n} \ln \left( \frac{k}{n} \right) \xrightarrow[n \to \infty]{} -1.$$ Given $\varepsilon > 0$ there exist $n_0$ such that $0 < \varepsilon_n \le \varepsilon$ for all $n \ge n_0$. Then $$\frac{1}{n} \sum_{k=1}^{n} \ln \left( \frac{k}{n} + \varepsilon_n \right) \le \frac{1}{n} \sum_{k=1}^{n} \ln \left( \frac{k}{n} + \varepsilon \right).$$ Since $$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left( \frac{k}{n} + \varepsilon \right) = \int_0^1 \ln (x + \varepsilon)\,dx = \int_{\varepsilon}^{1 + \varepsilon} \ln x\,dx$$ we obtain the result when $\varepsilon$ goes to 0 and so $$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left( \frac{k}{n} + \varepsilon_n \right) = -1.$$