高等数学-级数习题

高等数学-级数习题

级数审敛

级数收敛定义与性质判敛

例1
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例2
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例3
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例4
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例5
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例6
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例7
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正项级数审敛

比较审敛与极限审敛

比较审敛法的一般形式,以及基本不等式的应用
例1
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(主要是根据不等式:\(a+b \ge 2\sqrt{ab}\)

例2
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(A选项主要根据基本不等式:\(\left(\frac{a + b}{2}\right)^2 \le \frac{a^2+b^2}{2}\)以及\(\left(\frac{a - b}{2}\right)^2 \le \frac{a^2+b^2}{2}\)

例3
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例4
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例5
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(此题也可以用比值审敛法做)

例6
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例7
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例8
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例9
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例10
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极限审敛法/比较审敛法的极限形式,以及等价/同阶/高阶无穷小的应用
例1
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例2
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例3
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例4
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例5
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例6
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例7
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例8
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例9
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例10
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例11
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例12 此题需要功底
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例13 此题需要功底
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比值审敛法

例1
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例2
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根值审敛法

例1
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例2
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交错级数审敛

例1
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例2
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例3
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例4
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求级数收敛范围

求p级数收敛范围

例1
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求幂级数的收敛域

例1
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例2
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例3
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例4
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例5
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例6
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例7
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例8
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例9
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例10
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例11
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求级数的和函数或和

根据级数收敛的定义,求级数的和

例1
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例2
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幂级数求和函数

幂级数求和函数之前,要先确定收敛域。

常见函数的麦克劳林级数:

函数展开成麦克劳林级数(在\(x_0=0\)附近展开成级数)收敛域
\(e^x\)\(=1+x+ \frac{x^2}{2!} + ... + \frac{x^n}{n!} + ..\\ = \Sigma_{n=0}^{\infty} \frac{x^n}{n!}\)\(-\infty<x<+\infty\)
\(\sin x\)\(= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} +...\\ = \Sigma_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1}\)\(-\infty<x<+\infty\)
\(\cos x\)\(=1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ... \\ = \Sigma_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} x^{2n}\)\(-\infty<x<+\infty\)
\(\frac{1}{1-x}\)\(=1+x+x^2+x^3+... \\ = \Sigma_{n=0}^{\infty} x^n\)\(-1<x<1\)
\(\frac{1}{1+x}\)\(= \Sigma_{n=0}^{\infty} (-1)^n x^n\)\(-1<x<1\)
\(ln(1+x)\)\(=x-\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ... \\ = \Sigma_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n\)
注意这个级数从n=1开始
\(-1<x \le 1\)
\(-ln(1-x)\)\(=x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + ... \\ = \Sigma_{n=1}^{\infty} \frac{x^n}{n}\)
注意这个级数从n=1开始
\(-1 \le x<1\)
\((1+x)^{m}\)\(=1+m x+\frac{m(m-1)}{2 !} x^{2}+\cdots+\frac{m(m-1) \cdots(m-n+1)}{n !} x^{n}+\cdots\)
这个展开叫做二项展开式,也是代数学中的二项展开定理
\((-1<x<1)\)
例1
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例2
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例3
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例4
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例5
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例6
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例7
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例8
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例9
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例10
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例11
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例12
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例13
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例14
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例15
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例16
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例17
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例18 此题需要功底
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例19
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例20
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幂级数求和函数,再求级数的和

幂级数求和函数之前,要先确定收敛域。

例1
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例2
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例3
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例4
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例5 此题需要功底
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求傅里叶级数的和

例1
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例2
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例3
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函数展开成级数

函数展开成幂级数

函数展开成幂级数,有公式法(泰勒展开,展成泰勒级数)和间接法(借助基本函数的幂级数展开、逐项可导、可积性质的方法)。一般用间接法。展开成幂级数的同时一般要确定收敛域。

例1
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例2
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例3
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例4
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例5
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例6
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例7
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例8
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函数展开成傅里叶级数

例1
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例2
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例3
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