2025年考研数二第13题解析：定积分的定义、数列求和

一、题目

二、解析

$$
\begin{aligned}
& \ \lim_{{n} \to \infty} \frac{1}{{n}^{{2}}} \left[ \ln \frac{1}{{n}} + 2 \ln \frac{2}{{n}} + \cdots + \left( {n} – 1 \right) \ln \frac{{n} – 1}{{n}} \right] \\ \\
= & \ \lim_{{n}\to{\infty}} \left(\frac{{1}}{{n}}\ln\frac{{1}}{{n}}+\frac{{2}}{{n}} \ln \frac{{2}}{{n}}+\cdots+\frac{{n-1}}{{n}} \ln \frac{{n-1}}{{n}}\right)\cdot\frac{{1}}{{n}} \\ \\
= & \ \lim_{{n}\to{\infty}} \left(\frac{{1}}{{n}}\ln\frac{{1}}{{n}}+\frac{{2}}{{n}} \ln \frac{{2}}{{n}}+\cdots+\frac{{n-1}}{{n}} \ln \frac{{n-1}}{{n}} + \textcolor{orangered}{0} \right) \cdot \frac{{1}}{{n}} \\ \\
= & \ \lim_{{n}\to{\infty}} \left(\frac{{1}}{{n}}\ln\frac{{1}}{{n}}+\frac{{2}}{{n}} \ln \frac{{2}}{{n}}+\cdots+\frac{{n-1}}{{n}} \ln \frac{{n-1}}{{n}}+ \textcolor{orangered}{\frac{{n}}{{n}} \ln \frac{{n}}{{n}}} \right) \cdot \frac{{1}}{{n}} \\ \\
= & \ \lim_{{n}\to{\infty}} \sum_{{i}={1}}^{{n}} \frac{{i}}{{n}}\ln\frac{{i}}{{n}} \cdot \frac{{1}}{{n}} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{x = \frac{i}{n}} \\ \\
= & \ \int_{{0}}^{{1}} x \ln x \mathrm{~d}x \\ \\
= & \ \frac{{1}}{{2}} \int_{{0}}^{{1}} \ln x \mathrm{~d}x^{{2}} \\ \\
= & \ \frac{{1}}{{2}} \left(\left.x^{{2}} \ln x \right|_{{0}}^{{1}}-\int_{{0}}^{{1}}x^{{2}} \cdot \frac{{1}}{{x}}\mathrm{~d}x\right) \\ \\
= & \ \frac{{-1}}{{2}} \int_{{0}}^{{1}} x \mathrm{~d}x \\ \\
= & \ \frac{-1}{2} \cdot \frac{1}{2} x^{2} \Bigg|_{0}^{1} \\ \\
= & \ \frac{{-1}}{{4}}
\end{aligned}
$$

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