Questions in category: 级数 (Infinite Series)

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## 21. 设数列 $na_n$ 收敛, 且级数 $\sum\limits_{n=2}^{\infty}n(a_n-a_{n-1})$ 收敛, 证明级数 $\sum\limits_{n=1}^{\infty}a_n$ 也是收敛的.

Posted by haifeng on 2019-12-19 14:50:00 last update 2019-12-19 14:50:46 | Answers (1) | 收藏

References:

## 22. 设级数 $\sum\limits_{n=1}^{\infty}a_n$ 的部分和为 $S_n$. 如果 $S_{2n}\rightarrow S$, 且 $a_n\rightarrow 0$, 则 $\sum\limits_{n=1}^{\infty}a_n$ 收敛.

Posted by haifeng on 2019-12-17 21:04:24 last update 2019-12-17 21:04:24 | Answers (0) | 收藏

References:

## 23. 将 $\arctan x$ 展开成关于 $x$ 的幂级数

Posted by haifeng on 2019-12-17 20:18:22 last update 2020-01-17 11:13:59 | Answers (1) | 收藏

$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots$

References:

## 24. 使用 Machin 公式计算 $\pi$

Posted by haifeng on 2019-12-17 19:49:58 last update 2019-12-17 20:08:16 | Answers (0) | 收藏

Machin 公式

$\pi=16\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)5^{2n+1}}-4\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)239^{2n+1}}$

Reference:

>> setprecision(200)

>> sum((-1)^n/((2*n+1)*5^(2*n+1)),n,0,100)
in> sum((0-1)^n/((2*n+1)*5^(2*n+1)),n,0,100)

out> 0.19739555984988075837004976519479029344758510378785210151768894024103396997824378573269782803728804411262811807369136010445647988679423935574756610436426519783815939958213242811317823855439513093348319

------------------------

>> sum((-1)^n/((2*n+1)*239^(2*n+1)),n,0,100)
in> sum((0-1)^n/((2*n+1)*239^(2*n+1)),n,0,100)

out> 0.00418407600207472386453821495928545274104806530763195082701961288718177834142289327378260581362290945497545066644486375605245839478931186505892212883309280084627196233077337594763460331847341457033195

------------------------

>> 16*0.19739555984988075837004976519479029344758510378785210151768894024103396997824378573269782803728804411262811807369136010445647988679423935574756610436426519783815939958213242811317823855439513093348319-4*0.00418407600207472386453821495928545274104806530763195082701961288718177834142289327378260581362290945497545066644486375605245839478931186505892212883309280084627196233077337594763460331847341457033195
in> 16*0.19739555984988075837004976519479029344758510378785210151768894024103396997824378573269782803728804411262811807369136010445647988679423935574756610436426519783815939958213242811317823855439513093348319-4*0.00418407600207472386453821495928545274104806530763195082701961288718177834142289327378260581362290945497545066644486375605245839478931186505892212883309280084627196233077337594763460331847341457033195

out> 3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172536915449587196202546254399102534602031340359642843665440324

3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253

## 25. $\arctan\frac{1+x}{1-x}$ 与 $\arctan x$ 的关系

Posted by haifeng on 2019-12-17 13:57:08 last update 2022-04-27 08:16:33 | Answers (4) | 收藏

(1) 当 $x > 1$ 时,

$\arctan x-\arctan\frac{1+x}{1-x}=\frac{3\pi}{4},$

(2) 当 $x < 1$ 时,

$\arctan\frac{1+x}{1-x}-\arctan x=\frac{\pi}{4}.$

## 26. 求下列级数的和

Posted by haifeng on 2019-12-17 13:38:35 last update 2023-05-01 09:53:19 | Answers (2) | 收藏

(1)

$\sum_{n=1}^{+\infty}\frac{1}{n(2n-1)}$

(2)

$\sum_{n=1}^{+\infty}(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n})$

(3)

$\sum_{n=1}^{+\infty}\frac{1}{n(n+3)}$

(4)

$\sum_{n=1}^{+\infty}\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}$

(5)

$\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}}{n}$

(6)

$\sum_{n=1}^{+\infty}\frac{1}{n(n+1)(n+2)}$

## 27. 研究下列级数的敛散性

Posted by haifeng on 2019-12-10 08:31:48 last update 2019-12-10 08:34:17 | Answers (1) | 收藏

(1)  $\sum_{n=1}^{\infty}\frac{n}{2n-1}$

(2)  $\sum_{n=1}^{\infty}\frac{n}{2^n}$

(3) $\sum_{n=1}^{\infty}\frac{1}{\sqrt[n]{n^2}}$

(4) $\sum_{n=1}^{\infty}\frac{\sin^2 n}{n(n+1)}$

(5) $\sum_{n=1}^{\infty}\ln\bigl(1+\frac{1}{n}\bigr)$

(6) $\sum_{n=1}^{\infty}\biggl[\frac{1}{2n+1}+\frac{(-1)^n}{2n+2}\biggr]$

## 28. 判断级数 $\sum_{n=1}^{\infty}\frac{n!}{n^n}$ 的敛散性.

Posted by haifeng on 2019-07-12 16:16:47 last update 2019-07-12 16:16:47 | Answers (1) | 收藏

## 29. 设 $a_n=\sum_{k=1}^{n}\frac{1}{k}$, 求最小的正实数 $\lambda$, 使得对所有 $n\geqslant 2$, 都有 $a_n^2 < \lambda\sum_{k=1}^{n}\frac{a_k}{k}$.

Posted by haifeng on 2017-06-26 22:53:39 last update 2017-07-11 00:16:35 | Answers (0) | 收藏

$a_n^2 < \lambda\sum_{k=1}^{n}\frac{a_k}{k}.$

## 30. 将下列函数展开成关于 $x$ 的幂级数, 并求展开式成立的区间.

Posted by haifeng on 2015-08-24 23:05:47 last update 2015-08-24 23:05:47 | Answers (1) | 收藏

(1) $\frac{1}{1+x^2}$.

(2) $\frac{1}{x^2-5x+6}$.

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