IMC 2024 · Day 1 · P2

Hint: $S_n$ is (close to) a Riemann sum of a certain integral.

Transform $S_n$ as $$\begin{align*} S_n &= \frac{1}{n^2} \sum_{k=1}^{n} k \log k - \frac12 \log n \\ &= \frac{1}{n} \sum_{k=1}^{n} \frac{k}{n} \left( \log\frac{k}{n} + \log n \right) - \frac12 \log n \\ &= \frac{1}{n} \sum_{k=1}^{n} \frac{k}{n} \log\frac{k}{n} + \frac{\log n}{n^2} \sum_{k=1}^{n} k - \frac12 \log n \\ &= \frac{1}{n} \sum_{k=1}^{n} \frac{k}{n} \log\frac{k}{n} + \frac{\log n}{2n}. \end{align*}$$ Here the last term $\dfrac{\log n}{2n}$ converges to 0. The sum $\dfrac{1}{n} \sum\limits_{k=1}^{n} \dfrac{k}{n} \log\dfrac{k}{n}$ is a Riemann sum for the integrable function $f(x) = x \log x$ on the segment $[0, 1]$ with the uniform grid $\left\{ \dfrac1n, \dfrac2n, \dots, \dfrac{n-1}{n}, 1 \right\}$. Therefore $$\lim \frac{1}{n} \sum_{k=1}^{n} \frac{k}{n} \log\frac{k}{n} = \lim \frac{1}{n} \sum_{k=1}^{n} f\left( \frac{k}{n} \right) = \int_0^1 x \log x\,dx = \left[ \frac{x^2}{2} \log x - \frac{x^2}{4} \right]_0^1 = -\frac14.$$ Hence, $\lim S_n$ exists, and $\lim S_n = -\dfrac14$.