International Mathematics Competition
for University Students
2024

IMC2024: Day 1, Problem 2

Problem 2. For \(\displaystyle n=1,2,\ldots\) let

\(\displaystyle S_n = \log\left( \sqrt[n^2]{1^1\cdot 2^2\cdot\ldots\cdot n^n}\right) - \log (\sqrt{n}), \)

where \(\displaystyle \log\) denotes the natural logarithm. Find \(\displaystyle \mathop{\lim}\limits_{n\to\infty} S_n\).

Sergey Chernov, Belarusian State University, Minsk

Solution. Transform \(\displaystyle S_n\) as

$$\begin{align*} S_n &= \dfrac1{n^2}\sum_{k=1}^n k\log k -\dfrac12\log n\\ &= \dfrac1{n}\sum_{k=1}^n \Bigg( \dfrac{k}{n}\bigg(\log\dfrac{k}{n}+\log n\bigg)\Bigg) - \dfrac12\log n\\ &= \dfrac1{n}\sum_{k=1}^n \dfrac{k}{n}\log\dfrac{k}{n} + \dfrac{\log n}{n^2}\sum_{k=1}^nk -\dfrac12\log n\\ &= \dfrac1{n}\sum_{k=1}^n \dfrac{k}{n}\log\dfrac{k}{n} + \dfrac{\log n}{2n}.\\ \end{align*}$$

Here the last term \(\displaystyle \dfrac{\log n}{2n}\) converges to \(\displaystyle 0\). The sum \(\displaystyle \dfrac1{n}\sum\limits_{k=1}^n \dfrac{k}{n}\log\dfrac{k}{n}\) is an integral sum for the Riemann-integrable function \(\displaystyle f(x) = x\log x\) on the segment \(\displaystyle [0,1]\) on the uniform grid \(\displaystyle \left\{\dfrac1n, \dfrac2n, \ldots, \dfrac{n-1}n, 1\right\}\). Therefore

\(\displaystyle \lim \dfrac1{n}\sum_{k=1}^n \dfrac{k}{n}\log\dfrac{k}{n} = \lim \dfrac1{n}\sum_{k=1}^n f\bigg(\dfrac{k}{n}\bigg) = \int_0^1 x\log x \mathrm{d}x = \bigg[ \dfrac{x^2}{2}\log x-\dfrac{x^2}{4}\bigg]_0^1 = -\dfrac14. \)

Hence, \(\displaystyle \lim S_n\) exists, and \(\displaystyle \lim S_n=-\dfrac14\).