单重求和转一重积分，双重求和转二重积分

一、前言

在本文中，「荒原之梦考研数学」将通过两道题目给同学们展示一下如何将“求和”转为“积分”，内容涵盖考研数学中常考的一重求和转一重积分，以及二重求和转二重积分.

二、正文

根据定积分的定义，定积分就是将一个区间分成无数份，并对每一份做求和操作. 所以，对于以 $n$ 为变量，且 $n \rightarrow \infty$ 形式的求和，我们要做的就是要让 $n$ 都去到分母上，产生 “$\frac{i}{n}$” 或者 “$\frac{j}{n}$” 这样的部分，从而使求和与积分产生关联.

§2.1 单重求和转一重积分

题目

解析

$$
\begin{aligned}
& \lim_{n \rightarrow + \infty} \sum_{i=1}^{n} \frac{(2i – 1)^{2} }{n^{3} } \\ \\
= \ & \lim_{n \rightarrow + \infty} \sum_{i=1}^{n} \frac{(2i – 1)^{2} }{n^{2} } \cdot \frac{1}{n} \\ \\
= \ & \lim_{n \rightarrow + \infty} \sum_{i=1}^{n} \left( \frac{2i – 1}{n} \right)^{2} \cdot \frac{1}{n} \\ \\
= \ & \lim_{n \rightarrow + \infty} \sum_{i=1}^{n} \left( 2 \cdot \frac{i}{n} – \textcolor{tan}{ \frac{1}{n} } \right)^{2} \cdot \frac{1}{n} \\ \\
\leadsto \ & \textcolor{gray}{ \text{当 } n \rightarrow \infty \text{ 时}, \ \textcolor{tan}{\frac{1}{n}} \rightarrow 0, \text{ 所以}, \ \left( 2 \cdot \frac{i}{n} – \textcolor{tan}{ \frac{1}{n} } \right)^{2} \text{ 中的 } \textcolor{tan}{\frac{1}{n}} \text{ 可以舍去} } \\ \\
= \ & \lim_{n \rightarrow + \infty} \sum_{i=1}^{n} \left( 2 \cdot \frac{i}{n} – \textcolor{red}{\cancel{ \textcolor{tan}{ \frac{1}{n} }} } \right)^{2} \cdot \frac{1}{n} \\ \\
= \ & \lim_{n \rightarrow + \infty} \sum_{i=1}^{n} \left( 2 \cdot \textcolor{lightgreen}{ \frac{i}{n} } \right)^{2} \cdot \textcolor{orangered}{ \frac{1}{n} } \\ \\
\leadsto \ & \textcolor{gray}{ \textcolor{lightgreen}{x_{i} = \frac{i}{n}}, \ \textcolor{orangered}{\Delta x = \frac{1}{n}}, \ \lim_{n \rightarrow \infty} \frac{i}{n} \Big|_{i=1} = 0, \ \lim_{n \rightarrow \infty} \frac{i}{n} \Big|_{i=n} = 1 } \\ \\
= \ & \int_{0}^{1} \left( 2x \right)^{2} \mathrm{~d} x \\ \\
= \ & \textcolor{springgreen}{\int_{0}^{1} 4x^{2} \mathrm{~d} x} \\ \\
\end{aligned}
$$

Tip

通过上面这道题，我们可以知道，在求和转积分的过程中，遇到纯粹用加减法连接的无穷小量，一般是可以直接舍去的，并且通常可以极大地降低计算的复杂度，特别适用于不要求呈现计算过程的选择题或者填空题.

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§2.2 双重求和转二重积分

题目

解析

$$
\begin{aligned}
& \lim_{n \rightarrow \infty} \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{n}{(n+i)(n^{2}+j^{2})} \\ \\
= \ & \lim_{n \rightarrow \infty} \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{n}{n \cdot n^{2} \cdot \left[1 + \left(\frac{j}{n}\right)^{2}\right] \cdot \left(1 + \frac{i}{n}\right)} \\ \\
= \ & \lim_{n \rightarrow \infty} \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{1}{n^{2}} \cdot \frac{1}{1 + \left(\frac{j}{n}\right)^{2}} \cdot \frac{1}{1 + \frac{i}{n}} \\ \\
= \ & \lim_{n \rightarrow \infty} \sum_{i=1}^{n} \sum_{j=1}^{n} \left( \textcolor{lightgreen}{\frac{1}{n} \cdot \frac{1}{1 + \left(\frac{j}{n}\right)^{2}}} \right) \cdot \textcolor{orangered}{ \frac{1}{n} \cdot \frac{1}{1 + \frac{i}{n}} } \\ \\
= \ & \lim_{n \rightarrow \infty} \sum_{i=1}^{n} \left[ \left( \textcolor{lightgreen}{\sum_{j=1}^{n} \frac{1}{n} \cdot \frac{1}{1 + \left( \frac{j}{n} \right)^{2}} } \right) \cdot \textcolor{orangered}{ \frac{1}{n} \cdot \frac{1}{1 + \frac{i}{n}} } \right] \\ \\
\leadsto \ & \textcolor{gray}{ \begin{cases}
\textcolor{lightgreen}{ \Delta y = \frac{1}{n} } \\
\textcolor{lightgreen}{ y_{i} = \frac{j}{n} }
\end{cases}, \begin{cases}
\textcolor{orangered}{\Delta x = \frac{1}{n}} \\
\textcolor{orangered}{ x_{i} = \frac{i}{n} }
\end{cases} } \\ \\
\leadsto \ & \textcolor{gray}{\begin{cases}
\lim_{n \rightarrow \infty} \frac{j}{n} = \textcolor{orange}{0}, j = 1 \\ \lim_{n \rightarrow \infty} \frac{i}{n} = \textcolor{orange}{0}, i = 1
\end{cases}}, \textcolor{gray}{\begin{cases}
\lim_{n \rightarrow \infty} \frac{j}{n} = \textcolor{orange}{1}, j = n \\ \lim_{n \rightarrow \infty} \frac{i}{n} = \textcolor{orange}{1}, i = n
\end{cases}} \\ \\
= \ & \int_{\textcolor{orange}{0}}^{\textcolor{orange}{1}} \left[ \left( \textcolor{lightgreen}{ \int_{\textcolor{orange}{0}}^{\textcolor{orange}{1}} \frac{1}{1 + y^{2}} \mathrm{~d} y } \right) \cdot \frac{1}{1 + x} \right] \mathrm{~d} x \\ \\
= \ & \int_{0}^{1} \left[ \left( \int_{0}^{1} \frac{1}{1 + y^{2}} \cdot \frac{1}{1 + x} \mathrm{~d} y \right) \right] \mathrm{~d} x \\ \\
= \ & \textcolor{springgreen}{ \int_{0}^{1} \mathrm{d} x \int_{0}^{1} \frac{1}{(1+x)(1+y^{2})} \mathrm{~d} y }
\end{aligned}
$$

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