MA.7.4 级数的应用

一、用级数方法计算积分

Example

例1 计算积分 $\int_{0}^{1} \frac{\ln (1 + x)}{x} d x$
$0$ 处未定义，但为可去间断点

Analysis

原函数非初等函数，无法求解

Solution

\begin{aligned} \ln (1 + x) & = \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1}}{n} x^{n} (- 1 < x \leq 1) \\ ⟹ \frac{\ln (1 + x)}{x} & = \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1}}{n} x^{n - 1} (x = 1 为收敛点 - Abel lemma) \\ ∴ \int_{0}^{1} \frac{\ln (1 + x)}{x} d x & = \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1}}{n} \int_{0}^{1} x^{n - 1} d x = \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1}}{n^{2}} = \dots \end{aligned}

Jan.11:续上

\begin{aligned} \int_{0}^{1} \frac{\ln (1 + x)}{x} d x & = \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1}}{n^{2}} \\ = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} - 2 \sum_{n = 1}^{\infty} \frac{1}{(2 n)^{2}} \\ = \frac{1}{2} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} \\ = \frac{π^{2}}{12} \end{aligned}

Example

例2 计算椭圆积分 $\int_{0}^{1} \frac{d u}{\sqrt{(1 - u^{2}) (1 - k^{2} u^{2})}}, (k^{2} < 1)$

Tips

变量替换 #Wallis公式

Solution

\begin{aligned} 原积分 & \overset{u = \sin x}{=} \int_{0}^{π / 2} \frac{\cos x}{\sqrt{1 - \sin^{2} x} \sqrt{1 - k^{2} \sin^{2} x}} d x = \int_{0}^{π / 2} \frac{1}{\sqrt{1 - k^{2} \sin^{2} x}} d x \\ Recall: \frac{1}{\sqrt{1 - u}} = 1 + \sum_{n = 1}^{\infty} \frac{(2 n - 1)!!}{(2 n)!!} u^{n} \\ = \int_{0}^{π / 2} [1 + \sum_{n = 1}^{\infty} \frac{(2 n - 1)!!}{(2 n)!!} (k^{2} \sin^{2} x)^{n}] d x \\ [0, k^{2}] 收敛，积分求和 \\ = \frac{π}{2} + \sum_{n = 1}^{\infty} \frac{(2 n - 1)!!}{(2 n)!!} k^{2 n} \cdot \int_{0}^{π / 2} \sin^{2 n} x d x \\ \overset{Wallis 公式}{=} \frac{π}{2} (1 + \sum_{n = 1}^{\infty} {[\frac{(2 n - 1)!!}{(2 n)!!}]}^{2} k^{2 n}) \end{aligned}

二、近似计算

Example

例3 计算 $\ln 2$ 的近似值(误差不超过 $10^{- 4}$ )

Analysis

$\ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots + \frac{(- 1)^{n - 1}}{n} | + \underset{(- 1)^{n} [\frac{1}{n + 1} - \frac{1}{n + 2} + \dots]}{\underset{⏟}{\dots}}$

Solution

\begin{aligned} \ln (1 + x) & = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \dots \\ \ln (\frac{1}{1 - x}) & = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots \\ ⟹ \ln \frac{1 + x}{1 - x} & = 2 (x + \frac{x^{3}}{3} + \frac{x^{5}}{5} + \dots) \\ x := \frac{1}{3} : \\ \ln 2 & = \frac{2}{3} (1 + \frac{1}{3} \cdot \frac{1}{3^{2}} + \frac{1}{5} \cdot \frac{1}{3^{4}} + \dots) \\ 大约第五项即可 \end{aligned}

三、微分方程的幂级数解法

例4 求Airy方程 $y^{''} - x y = 0$ 的幂级数解

$y = \sum a_{n}$

Solution

\begin{aligned} 设 y = \sum_{n = 0}^{\infty} a_{n} x^{n} 是解，则 \\ y^{'} = \sum_{n = 1}^{\infty} n a_{n} x^{n}, \begin{aligned} y^{″} & = \sum_{n = 2}^{\infty} n (n - 1) x^{n - 2} \\ \overset{n - 2 := n}{=} \sum_{n = 0}^{\infty} (n + 2) (n + 1) x^{n} \end{aligned} \\ 又 x y = \sum_{n = 0}^{\infty} a_{n} x^{n + 1} \overset{n + 1 := n}{=} \sum_{n = 1}^{\infty} a_{n - 1} x^{n} 代入: \\ a_{2} + \sum_{n = 1}^{\infty} [(n + 2) (n + 1) a_{n + 2} - a_{n - 1}] x^{n} = 0 \\ 故 a_{2} = 0 及 a_{n + 2} = \frac{a_{n - 1}}{(n + 2) (n + 1)} (n = 1, 2, \dots) \\ a_{0, 1} 可以任取 \\ 从而 a_{3 k + 2} = 0; \\ a_{3 k} = \frac{a_{3 k - 3}}{3 k (3 k - 1)} = \frac{1 \cdot (3 k - 2)}{3 k (3 k - 1) (3 k - 2) (3 k - 3) (3 k - 4) \dots 3 \cdot 2} \cdot a_{0} \\ = \frac{1 \cdot 4 \cdot 7 \dots (3 k - 2)}{(3 k)!} c_{1} (a_{0} = c_{1}) \\ a_{3 k + 1} = \frac{a_{3 k - 2}}{(3 k + 1) 3 k} = \dots = \frac{1}{(3 k + 1) \cdot 3 k \cdot (3 k - 2) \dots 4 \cdot 3} \cdot a_{1} \\ = \frac{2 \cdot 5 \cdot 8 \dots (3 k - 1)}{(3 k + 1)!} \cdot c_{2} \\ ⟹ 解 y = \sum_{k = 0}^{\infty} a_{3 k} x^{3 k} + \sum_{k = 0}^{\infty} a_{3 k + 1} x^{3 k + 1} = \\ = c_{1} \sum_{k = 0}^{\infty} \frac{1 \cdot 4 \dots (3 k - 2)}{(3 k)!} x^{3 k} + c_{2} \sum_{k = 0}^{\infty} \frac{2 \cdot 5 \dots (3 k - 1)}{(3 k + 1)!} x^{3 k + 1} \\ \overset{def}{=} c_{1} y_{1} + c_{2} y_{2} \end{aligned}

Example

例5 设 $y (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$ 适合 $y (0) = 0, y^{'} (0) = 1$ 和

y^{''} - 2 x y^{'} - 4 y = 0

证明 $a_{n + 2} = \frac{2}{n + 1} a_{n}, n \in N$
求 $y (x)$ 的表达式

Solution

\begin{aligned} (1) \\ \begin{aligned} y^{'} = \sum_{n = 1}^{\infty} n a_{n} x^{n} \\ y^{″} = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n} \\ = \sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n} \\ 2 x y^{'} = \sum_{n = 1}^{\infty} 2 n a_{n} x^{n} + 2 n a_{n} x^{n} |_{n = 0} = \sum_{n = 0}^{\infty} 2 n a_{n} x^{n} \\ \overset{代入}{\Rightarrow} \sum_{n = 0}^{\infty} [(n + 2) (n + 1) a_{n + 2} - 2 n a_{n} - 4 a_{n}] x^{n} = 0 \\ ⟹ (n + 2) (n + 1) a_{n + 2} = 2 a_{n} (n + 2) \\ 即 a_{n + 2} = \frac{2 a_{n}}{n + 1} (n = 0, 1, \dots) \\ 题目已知 a_{0} = 0, a_{1} = y^{'} (0) = 1 \end{aligned} \\ (2) \\ \begin{aligned} 故 a_{2 k} = 0, a_{2 k + 1} = \frac{2 a_{2 k - 1}}{2 k} = \dots = \frac{2 \cdot 2 \dots 2 \cdot 2}{(2 k) (2 k - 2) \dots 4 \cdot 2} \cdot a_{1} = \frac{1}{k!} \\ 从而 y = \sum_{k = 1}^{\infty} a_{2 k + 1} x^{k + 1} = \sum_{k = 0}^{\infty} \frac{x^{2 k + 1}}{k!} = x \sum_{k = 0}^{\infty} \frac{(x^{2})^{k}}{k!} = x \cdot e^{x^{2}} \end{aligned} \end{aligned}

四、Stirling公式: 有效处理 $n!$

Wallis 引理

#级数/Wallis引理

$\frac{π}{2} = lim_{n \to \infty} \frac{1}{2 n + 1} {(\frac{(2 n)!!}{(2 n - 1)!!})}^{2}$

Proof

\begin{aligned} 考虑 \int_{0}^{π / 2} \sin^{2 n} x d x \\ 由于 \int_{0}^{π / 2} \sin^{2 n - 1} d x < \int_{0}^{π / 2} \sin^{2 n} x d x < \int_{0}^{π / 2} \sin^{2 n - 1} d x \\ 即 \frac{(2 n)!!}{(2 n + 1)!!} < \frac{(2 n - 1)!!}{(2 n)!!} \frac{π}{2} < \frac{(2 n - 2)!!}{(2 n - 1)!!} \\ 交叉相乘+补项 ⟹ \\ \frac{2 n}{2 n + 1} \frac{π}{2} < \frac{1}{(2 n + 1)} {[\frac{(2 n)!!}{(2 n - 1)!!}]}^{2} < \frac{π}{2} \\ 令 n \to \infty 即证 \end{aligned}

Tips
类似地，取 $n \to \infty$ ，两边开根号可得

lim_{n \to 0} \frac{1}{\sqrt{n}} \cdot \frac{(2 n)!!}{(2 n - 1)!!} = \sqrt{π}

Stirling公式

$n! = \sqrt{2 n π} {(\frac{n}{e})}^{n} e^{\frac{θ_{n}}{12 n}} (0 < θ_{n} < 1)$

估计阶乘：数量级 $\approx$ ${(\frac{n}{e})}^{n}$

Analysis

令 $a_{n} = \frac{n! e^{n}}{n^{n + 1 / 2}}$ 要证： $a_{n} = \sqrt{2 π} \cdot e^{\frac{θ_{n}}{12 n}}$
则 $\frac{a_{n}}{a_{n + 1}} = \frac{n! e^{n}}{n^{n + 1 / 2}} \cdot \frac{((n + 1)^{n + 1 + 1 / 2})}{e}$

下面考察 $(n + \frac{1}{2}) \ln (1 + \frac{1}{n})$ 与 $1$ 的大小关系

Proof

\begin{aligned} \ln (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \dots (- 1 < x \leq 1) \\ \ln \frac{1}{1 - x} = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots (- 1 \leq x < 1) \\ \ln \frac{1 + x}{1 - x} = 2 x (1 + \frac{x^{2}}{3} + \frac{x^{4}}{5} + \dots) (- 1 < x < 1) \\ \overset{x = \frac{1}{2 n + 1}}{\Rightarrow} \\ \begin{aligned} (n + \frac{1}{2}) \ln (1 + \frac{1}{n}) & = 1 + \frac{1}{3} \cdot \frac{1}{(2 n + 1)^{2}} + \frac{1}{5} \cdot \frac{1}{(2 n + 1)^{4}} + \dots \\ < 1 + \frac{1}{3} [\frac{1}{(2 n + 1)^{2}} + \frac{1}{(2 n + 1)^{4}} + \dots] \\ = 1 + \frac{1}{3} \cdot \frac{\frac{1}{(2 n + 1)^{2}}}{1 - \frac{1}{(2 n + 1)^{2}}} = 1 + \frac{1}{12 n (n + 1)} \end{aligned} \\ ⟹ \\ e < {(1 + \frac{1}{n})}^{n + 1 / 2} < e^{1 + 1 / (12 n (n + 1))} \\ 故 1 < \frac{a_{n}}{a_{n + 1}} < e^{1 / (12 n (n + 1))} 即 {a_{n}} 有界 \\ 设 lim_{n \to \infty} a_{n} = a 又 \\ a_{n} e^{- 1 / 12 n} < a_{n + 1} < a_{n + 1} e^{- 1 / 12 (n + 1)} \\ 即 {a_{n} e^{- 1 / 12 n}} 单增 \to a 从而有 \\ a < a_{n}; a_{n} e^{- 1 / 12 n} < a \\ ⟹ a \cdot e^{0 / 12 n} < a_{n} < a \cdot e^{1 / 12 n} \\ \overset{介值}{\Rightarrow} \exists Q_{n} \in (0, 1) s.t. a_{n} = a \cdot e^{Q_{n} / 12 n} \\ 下证： a = \sqrt{2 π} 即可 \\ 由 n! = \sqrt{n} {(\frac{n}{e})}^{n} \cdot a_{n} 有 \\ (Squared) (n!)^{2} = n \cdot ({\frac{n}{e}}^{2 n}) \cdot a_{n}^{2} \\ (n := 2 n) (2 n)! = \sqrt{2 n} {(\frac{2 n}{e})}^{2 n} \cdot a_{2 n} \\ ⟹ \frac{\sqrt{n}}{\sqrt{2}} \frac{1}{2^{2 n}} \cdot \frac{a_{n}^{2}}{a_{2 n}} = \frac{(n!)^{2}}{(2 n)!} \\ ⟹ \frac{a_{n}^{2}}{a_{2 n}} = \frac{\sqrt{2}}{\sqrt{n}} \cdot \frac{2^{2 n} \cdot (n!)^{2}}{(2 n)!} = \frac{\sqrt{2}}{\sqrt{n}} \cdot \frac{(2^{n} \cdot n!)^{2}}{(2 n)!} = \frac{[(2 n)!!]^{2}}{(2 n)!} \cdot \frac{\sqrt{2}}{\sqrt{n}} \\ = \frac{\sqrt{2}}{\sqrt{n}} \cdot \frac{(2 n)!!}{(2 n - 1)!!} \\ \overset{Wallis}{\Rightarrow} a = lim_{n \to \infty} \frac{a_{n}^{2}}{a_{2 n}} \sqrt{2 π} \end{aligned}

Example

例6 讨论级数 $\sum_{n =}^{\infty} \frac{n! e^{n}}{n^{n + p}} (p \in R)$ 的敛散性.

Solution

\begin{aligned} \begin{aligned} 通项 u_{n} & = \frac{\sqrt{2 n π} {(\frac{n}{e})}^{n} e^{θ_{n} / 12 n} e^{n}}{n^{n + p}} \\ \sim \frac{\sqrt{2 π}}{n^{p - 1 / 2}} \end{aligned} \\ 故当且仅当 p > \frac{3}{2} 时 收 敛 \end{aligned}

五、 $\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{6}$ 的幂级数证明

Example

例7 证明

1. 函数项级数

\sin x + \sum_{n = 1}^{\infty} \frac{(2 n - 1)!!}{(2 n)!!} \cdot \frac{\sin^{2 n + 1} x}{2 n + 1}

在 $[0, π / 2]$ 上一致收敛于 $x$ .

2. 求级数 $\sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ 的和.

Tips

回忆公式: $\arcsin t = t + \sum_{n = 1}^{\infty} \frac{(2 n - 1)!!}{(2 n)!!} \cdot \frac{t^{2 n + 1}}{2 n + 1}, t \in [- 1, 1]$

Laabe: $x = 1$ 收敛=>Abel.II $[0, 1]$ 收敛

Proof

由于 $t + \sum_{n = 1}^{\infty} \frac{(2 n - 1)!!}{(2 n)!!} \cdot \frac{t^{2 n + 1}}{2 n + 1} \overset{[0, 1]}{⇉} \arcsin t$

故 $\sin x + \sum_{n = 1}^{\infty} \frac{(2 n - 1)!!}{(2 n)!!} \frac{\sin^{2 n + 1} x}{2 n + 1} \overset{[0, \frac{π}{2}]}{⇉} x$

通项积分：

\int_{0}^{π / 2} x d x = 1 + \sum_{n = 1}^{\infty} \frac{(2 n - 1)!!}{(2 n)!!} \frac{1}{2 n + 1} \int_{0}^{π / 2} \sin^{2 n + 1} x d x

即

\frac{π^{2}}{8} = 1 + \sum_{n = 1}^{\infty} \frac{(2 n - 1)!!}{(2 n + 1) (2 n + 1)!!} = 1 + \sum_{n = 1}^{\infty} \frac{1}{(2 n + 1)^{2}} = \sum_{n = 1}^{\infty} \frac{1}{(2 n - 1)^{2}}

故

\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \sum_{n = 1}^{\infty} \frac{1}{(2 n - 1)^{2}} + \sum_{n = 1}^{\infty} \frac{1}{(2 n)^{2}} = \frac{π^{2}}{8} + \frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}}

\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{6}

Example

例8 设 $f (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$ 在 $(- R, R)$ 内收敛, 且 $\sum_{n = 0}^{\infty} \frac{a_{n}}{n + 1} R^{n + 1}$ 收敛.

证明: 无论 $\sum_{n = 0}^{\infty} a_{n} x^{n}$ 在 $x = R$ 处是否收敛, 都有

\int_{0}^{R} f (x) d x = \sum_{n = 0}^{\infty} \frac{a_{n}}{n + 1} R^{n + 1}

并由此证明

\int_{0}^{1} \frac{1}{1 + x} d x = \ln 2 = \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1}}{n}

Solution

\begin{aligned} 由于 \sum_{n = 0}^{\infty} \frac{a_{n}}{n + 1} x^{n + 1} 在 x = R 收敛，故 \\ \forall x \in (- R, R) \\ 逐项积分 \int_{0}^{x} f (t) d t = \sum_{n = 0}^{\infty} a_{n} \int_{0}^{x} t^{n} d t = \sum_{n = 0}^{\infty} \frac{a_{n}}{n + 1} x^{n + 1} (已知收敛) \\ \overset{Abel.III}{\Rightarrow} \underset{←定积分：直接等于，瑕积分：记为该值}{\underset{⏟}{\int_{0}^{R} f (t) d t = lim_{x \to R^{-}} \int_{0}^{x} f (t) d t}} < \sum_{n = 0}^{\infty} \frac{a_{n}}{n + 1} R^{n + 1} \\ 令 a_{n} = (- 1)^{n} : f (x) = \sum_{n = 0}^{\infty} (- x)^{n} = \frac{1}{1 + x} 故 R = 1 \\ \overset{t = x, R = 1}{\Rightarrow} \int_{0}^{1} \frac{1}{x + 1} d x = \int_{0}^{1} \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{n + 1} d x = \int_{0}^{1} \sum_{n = 1}^{\infty} \frac{(- 1)^{n - 1}}{n} d x \end{aligned}

Addition

\begin{aligned} 又取 a_{n} = \frac{1}{n + 1} 则 R = 1, 且 \forall x \in [0, 1) \\ f (x) = \sum_{n = 0}^{\infty} \frac{x^{n}}{n + 1} = \frac{1}{x} \sum_{n = 0}^{\infty} \frac{x^{n + 1}}{n + 1} = \frac{1}{x} \cdot \ln \frac{1}{1 - x} \\ 故 \int_{0}^{1} \frac{1}{x} \ln \frac{1}{1 - x} d x = \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2}} = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{6} \end{aligned}

Example

例9 求级数 $\sum_{n = 2}^{n} \frac{1}{(n^{2} - 1) 2^{n}}$ 的和

Solution

\begin{aligned} 令 f (x) & = \sum_{n = 2}^{\infty} \frac{x^{n}}{n^{2} - 1} R = \pm 1 \\ = \sum_{n = 2}^{\infty} \frac{x^{n}}{(n - 1) (n + 1)} = \frac{1}{2} \sum_{n = 2}^{\infty} (\frac{1}{n - 1} - \frac{1}{n + 1}) x^{n} \\ = \frac{1}{2} \sum_{n = 2}^{\infty} \frac{x^{n}}{n - 1} - \sum_{n = 2}^{\infty} \frac{x^{n}}{n + 1} \\ = \frac{x}{2} \sum_{n = 1}^{\infty} \frac{x^{n}}{n} - \frac{1}{2 x} \underset{缺项补项即可}{\underset{⏟}{\sum_{n = 3}^{\infty} U \frac{x^{n}}{n}}} \\ = \frac{x}{2} \ln \frac{1}{1 - x} - \frac{1}{2 x} (\ln \frac{1}{1 - x} - x - \frac{x^{2}}{2}) \dots \frac{1}{1 - x} 在 1 处无意义不影响整体 \\ 故 \sum_{n = 2}^{\infty} \frac{1}{n^{2} - 1} \cdot \frac{1}{2^{n}} = f (\frac{1}{2}) = \frac{5}{8} - \frac{3 \ln 2}{4} \end{aligned}

Example

例10 $\sum_{n = 1}^{\infty} \frac{n^{2} + 1}{n} x^{n}$

Solution

\begin{aligned} 令 f (x) & = \sum_{n = 1}^{\infty} \frac{n^{2} + 1}{n} x^{n}, R = 1, x \in (- 1, 1) \\ = \underset{见上次笔记例3}{\underset{⏟}{\sum_{n = 1}^{\infty} n x^{n}}} + \sum_{n = 1}^{\infty} \frac{x^{n}}{n} \\ = \frac{x}{(1 - x)^{2}} + \ln \frac{1}{1 - x} \end{aligned}

MA.7.4 级数的应用

一、用级数方法计算积分

二、近似计算

三、微分方程的幂级数解法

四、Stirling公式: 有效处理 $n!$

Wallis 引理

Stirling公式

五、 $\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{6}$ 的幂级数证明

🎉 MA.I 完结撒花！

👎 MA.II 蓄势待发！

一、用级数方法计算积分

二、近似计算

三、微分方程的幂级数解法

四、Stirling公式: 有效处理n!

Wallis 引理

Stirling公式

五、 ∑n=1∞1n2=π26 的幂级数证明

🎉 MA.I 完结撒花！

👎 MA.II 蓄势待发！

四、Stirling公式: 有效处理 $n!$

五、 $\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{6}$ 的幂级数证明