习题6.3

第 1 题

求下列幂级数的收敛半径与收敛域.

(1) \(\sum\limits_{n = 1}^{\infty}\dfrac{x^n}{n^n}\);

(3) \(\sum\limits_{n = 1}^{\infty}\dfrac{x^{3n + 1}}{(2n - 1)2^n}\);

(5) \(\sum\limits_{n = 1}^{\infty}\dfrac{\ln n}{n}x^n\);

(7) \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{n^p}(x - 1)^n(p>0)\);

(9) \(\sum\limits_{n = 1}^{\infty}2^n(x+a)^{2n}\);

解

(1) 由于 \(q = \lim\limits_{n \to \infty}\sqrt[n]{\dfrac{1}{n^n}} = \lim\limits_{n \to \infty} \dfrac{1}{n} = 0\), 故 \(R = \dfrac{1}{q} = +\infty\). 收敛域为 \(\mathbb{R}\).

(3) 由于 \(q = \lim\limits_{n \to \infty}\dfrac{u_{n + 1}}{u_n} = \lim\limits_{n \to \infty}\dfrac{x^3(2n - 1)}{2(2n + 1)} = \dfrac{x^3}{2}\), 当 \(x \in (-\sqrt[3]{2}, \sqrt[3]{2})\) 时级数收敛. 故 \(R = \sqrt[3]{2}\). 当 \(x = -\sqrt[3]{2}\) 时为 Leibniz 级数, 收敛; 当 \(x = \sqrt[3]{2}\) 时可化为 \(\sum\limits_{n = 1}^{\infty}\dfrac{\sqrt[3]{2}}{2n - 1}\), 发散. 故收敛域为 \([-\sqrt[3]{2}, \sqrt[3]{2})\).

(5) 由于 \(q = \lim\limits_{n \to \infty}\dfrac{u_{n + 1}}{u_n} = \lim\limits_{n \to \infty}\dfrac{n\ln(n+1)}{(n+1)\ln n} = 1\), 故 \(R = \dfrac{1}{q} = 1\). 当 \(x = -1\) 时为 Leibniz 级数, 收敛; 当 \(x = 1\) 时可化为 \(\sum\limits_{n = 1}^{\infty}\dfrac{\ln n}{n} \ge -1 + \sum\limits_{n = 1}^{\infty}\dfrac{1}{n}\), 发散. 故收敛域为 \([-1, 1)\).

(7) 由于 \(q = \lim\limits_{n \to \infty}\sqrt[n]{\dfrac{1}{n^p}} = 1\), 故 \(R = \dfrac{1}{q} = 1\). 当 \(x = 0\) 时为该级数为 Leibniz 级数, 收敛. 当 \(x = 2\) 时, \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{n^p}(x - 1)^{n} = \sum\limits_{n = 1}^{\infty}\dfrac{1}{n^p}\). 若 \(p > 1\), 则级数收敛; 若 \(0 < p \le 1\), 则级数发散.

综上所述, 当 \(p > 1\) 时, 收敛域为 \([0, 2]\); 当 \(0 < p \le 1\) 时, 收敛域为 \([0, 2)\).

(9) 由于 \(q = \lim\limits_{n \to \infty}\dfrac{u_{n + 1}}{u_n} = \lim\limits_{n \to \infty}2(x+a)^2 = 2(x+a)^2\), 当 \(x \in \left(-a - \dfrac{\sqrt{2}}{2}, -a + \dfrac{\sqrt{2}}{2}\right)\) 时级数收敛. 故 \(R = \dfrac{\sqrt{2}}{2}\). 当 \(x = -a \pm \dfrac{\sqrt{2}}{2}\) 时级数均发散. 故收敛域为 \(\left(-a - \dfrac{\sqrt{2}}{2}, -a + \dfrac{\sqrt{2}}{2}\right)\).

第 2 题

求下列幂级数的收敛域与和函数.

(1) \(\sum\limits_{n = 2}^{\infty}\dfrac{x^n}{n(n-1)}\);

(3) \(\sum\limits_{n = 1}^{\infty}(2n+1)x^{2n+1}\);

(5) \(\sum\limits_{n = 1}^{\infty}\dfrac{n(n+1)}{2}x^{n-1}\);

解

(1) 由于 \(q = \lim\limits_{n \to \infty}\sqrt[n]{\dfrac{1}{n(n - 1)}} = 1\), 故 \(R = \dfrac{1}{q} = 1\). 当 \(x = \pm1\) 时级数均收敛, 故收敛域为 \([-1, 1]\). 设和函数为 \(S(x)\), 则当 \(x \neq 1\) 时 \(S''(x) = \sum\limits_{n = 0}^{\infty}x^n = \dfrac{1}{1 - x}\). 积分可得 \(S(x) = x + (1 - x)\ln(1 - x)\). 当 \(x = 1\) 时 \(S(x) = 1\). 因此

\[ S(x) = \begin{cases} 1, & x = 1, \\ x + (1 - x)\ln(1 - x), & x \in [-1, 1). \end{cases} \]

(3) 由于 \(q = \varlimsup\limits_{n \to \infty}a_n = \lim\limits_{n \to \infty}\sqrt[n]{2n+1} = 1\), 故 \(R = \dfrac{1}{q} = 1\). 当 \(x = \pm1\) 时级数均发散, 故收敛域为 \((-1, 1)\). 设和函数为 \(S(x)\). 因此

\[ \begin{aligned} S(x) & = \dfrac{\text{d}{}}{\text{d}{x}}\left(\sum\limits_{n = 1}^{\infty}x^{2n + 2}\right)-\left(\sum\limits_{n = 1}^{\infty}x^{2n + 1}\right) \\ & = \dfrac{\text{d}{}}{\text{d}{x}}\dfrac{x^4}{1 - x^2} - \dfrac{x^3}{1 - x^2} \\ & = \dfrac{4x^3 - 2x^5}{(1 - x^2)^2} - \dfrac{x^3}{1 - x^2} \\ & = \dfrac{3x^3 - x^5}{(1 - x^2)^2} \quad x \in (-1, 1) \end{aligned} \]

(5) 由于 \(q = \lim\limits_{n \to \infty}\sqrt[n]{\dfrac{n(n + 1)}{2}} = 1\), 故 \(R = \dfrac{1}{q} = 1\). 当 \(x = \pm1\) 时级数均发散, 故收敛域为 \((-1, 1)\). 设和函数为 \(S(x)\). 因此

\[ \begin{aligned} S(x) & = \dfrac{1}{2}\dfrac{\text{d}{}}{\text{d}{x^2}}\left(\sum\limits_{n = 1}^{\infty}x^{n+1}\right) \\ & = \dfrac{1}{2}\dfrac{\text{d}{}}{\text{d}{x^2}}\left(\dfrac{1}{1 - x} - 1 - x\right) \\ & = \dfrac{1}{2}\dfrac{\text{d}{}}{\text{d}{x}}\left(\dfrac{1}{(1 - x)^2} - 1\right) \\ & = \dfrac{1}{(1 - x)^3} \quad x \in (-1, 1) \end{aligned} \]

第 3 题

将下列函数在 \(x_0\) 点展成幂级数, 并求收敛域.

(1) \(\cos x, x_0 = \dfrac{\pi}{4}\);

(3) \(\ln(1 + x), x_0 = 2\);

(5) \(\sin x^2, x_0 = 0\);

(7) \(\dfrac{1}{x - 1}, x_0 = -1\);

(9) \(\dfrac{x}{(x - 1)(x + 3)}, x_0 = 0\);

(11) \(\ln(x + \sqrt{x^2 + 1}), x_0 = 0\);

解

(1) \(f(x) = \cos x, f^{(n)}(x) = \cos\left(x + \dfrac{n\pi}{2}\right)\). 因此

\[ f(x) = \sum\limits_{n = 0}^{\infty}\dfrac{\cos\left(\frac{\pi}{4} + \frac{n\pi}{2}\right)}{n!}\left(x - \dfrac{\pi}{4}\right)^n \]

收敛域为 \(\mathbb{R}\).

(3) \(f(x) = \ln(1 + x), f^{(n)}(x) = \begin{cases}\ln(1 + x), & n = 0, \\ \dfrac{(-1)^{n-1}(n-1)!}{(1 + x)^n}, & n \ge 1.\end{cases}\) 因此

\[ f(x) = \ln 3 + \sum\limits_{n = 1}^{\infty}\dfrac{(-1)^{n-1}}{n\cdot 3^n}(x - 2)^n \]

由 \(\left|x - 2\right| < 3\) 解得 \(x \in (-1, 5)\). 因为当 \(x = -1\) 时级数发散, 当 \(x = 5\) 时级数收敛, 故收敛域为 \((-1, 5]\).

(5) \(f(x) = \sin x^2\). 由于 \(\sin x = \sum\limits_{k = 0}^{\infty}\dfrac{(-1)^kx^{2k+1}}{(2k+1)!}\), 因此

\[ f(x) = \sum\limits_{k = 0}^{\infty}\dfrac{(-1)^kx^{4k+2}}{(2k+1)!} \]

收敛域为 \(\mathbb{R}\).

(7) \(f(x) = \dfrac{1}{x - 1}, f^{(n)}(x) = \dfrac{(-1)^nn!}{(x - 1)^{n + 1}}\). 因此

\[ f(x) = -\sum\limits_{n = 0}^{\infty}\dfrac{1}{2^{n+1}}(x + 1)^n \]

由 \(\left|x + 1\right| < 2\) 解得 \(x \in (-3, 1)\). 因为当 \(x = -3\) 时级数收敛, 当 \(x = 1\) 时级数发散, 故收敛域为 \([-3, 1)\).

(9) \(f(x) = \dfrac{x}{(x - 1)(x - 3)} = \dfrac{3}{2(x - 3)} + \dfrac{1}{2(x - 1)}, f^{(n)}(x) = \dfrac{3}{2}\cdot\dfrac{(-1)^nn!}{(x - 3)^{n + 1}} - \dfrac{1}{2}\cdot\dfrac{(-1)^nn!}{(x - 1)^{n + 1}}\). 因此

\[ f(x) = \sum\limits_{n = 0}^{\infty}\left(\dfrac{1}{2} - \dfrac{1}{2\cdot3^n}\right)x^n \]

由 \(\left|x\right| < 1\) 解得 \(x \in (-1, 1)\). 因为当 \(x = \pm1\) 时级数均发散, 故收敛域为 \((-1, 1)\).

(11) \(f(x) = \ln(x + \sqrt{x^2 + 1}), f'(x) = (1 + x^2)^{-\frac{1}{2}}\). 由于

\[ (1 + x^2)^{\frac{1}{2}} = 1 + \sum\limits_{n = 1}^{\infty}(-1)^n\dfrac{(2n-1)!!}{(2n)!!}x^{2n} \quad x \in [-1, 1] \]

因此

\[ \begin{aligned} \ln(x + \sqrt{x^2 + 1}) & = \int_{0}^{x}(1 + t^2)^{-\frac{1}{2}}\text{d}{t} \\ & = \int_{0}^{x}\left[1 + \sum\limits_{n = 1}^{\infty}(-1)^n\dfrac{(2n-1)!!}{(2n)!!}t^{2n}\right]\text{d}{t} \\ & = x + \sum\limits_{n = 1}^{\infty}(-1)^n\dfrac{(2n-1)!!}{(2n)!!(2n + 1)}x^{2n + 1} \quad x \in [-1, 1] \end{aligned} \]

收敛域为 \([-1, 1]\).

第 4 题

将下列函数在 \(x_0 = 0\) 点展到指定的项.

(1) \(e^{\sin x}\), 展到 \(x^3\) 项;

(3) \(\cos^3 x\), 展到 \(x^4\) 项;

解

(1) 由 \(e^x = 1 + x + \dfrac{1}{2}x^2 + \dfrac{1}{6}x^3 + o(x^3)\) 以及 \(\sin x = x - \dfrac{1}{6}x^3 + o(x^3)\) 得

\[ \begin{aligned} e^{\sin x} & = 1 + \sin x + \dfrac{1}{2}\sin^2x + \dfrac{1}{6}\sin^3x + o(\sin^3 x) \\ & = 1 + \left(x - \dfrac{1}{6}x^3 + o(x^3)\right) + \dfrac{1}{2}\left(x - \dfrac{1}{6}x^3 + o(x^3)\right)^2 + \dfrac{1}{6}\left(x - \dfrac{1}{6}x^3 + o(x^3)\right)^3 + o(x^3) \\ & = 1 + x - \dfrac{1}{6}x^3 + \dfrac{1}{2}x^2 + \dfrac{1}{6}x^3 + o(x^3) \\ & = 1 + x + \dfrac{1}{2}x^2 + o(x^3) \end{aligned} \]

(3) 由 \(\cos x = 1 - \dfrac{1}{2}x^2 + o(x^2)\) 得

\[ \begin{aligned} \cos^3 x & = \left(1 - \dfrac{1}{2}x^2 + \dfrac{1}{24}x^4 + o(x^4)\right)^3 \\ & = 1 - \dfrac{3}{2}x^2 + \dfrac{3}{4}x^4 + \dfrac{1}{8}x^4 + o(x^4) \\ & = 1 - \dfrac{3}{2}x^2 + \dfrac{7}{8}x^4 + o(x^4) \end{aligned} \]

第 7 题

幂级数 \(\sum\limits_{n = 1}^{\infty}a_nx^n\) 在 \(x = 3\) 处条件收敛, 求 \(\sum\limits_{n = 1}^{\infty}na_n(x - 1)^{n + 1}\) 的收敛区间.

解

由题知 \(\sum\limits_{n = 1}^{\infty}a_nx^n\) 的收敛半径为 \(R_1 = 3\). 因此 \(q_1 = \varlimsup\limits_{n \to \infty}\sqrt[n]{a_n} = \dfrac{1}{R} = \dfrac{1}{3}\). 所以 \(q_2 = \varlimsup\limits_{n \to \infty}\sqrt[n]{na_n} = \dfrac{1}{3}\). 故 \(\sum\limits_{n = 1}^{\infty}na_n(x - 1)^{n + 1}\) 的收敛半径 \(R_2 = \dfrac{1}{q_2} = 3\), 即收敛区间为 \((-2, 4)\).

第 9 题

幂级数 \(\sum\limits_{n = 0}^{\infty}a_nx^n\) 与 \(\sum\limits_{n = 0}^{\infty}b_nx^n\) 的收敛半径分别为 \(R_1, R_2\), 证明:

(1) \(\sum\limits_{n = 0}^{\infty}(a_n+b_n)x^n\) 的收敛半径 \(R \ge \min\{R_1, R_2\}\).

(2) \(\sum\limits_{n = 0}^{\infty}a_nb_nx^n\) 的收敛半径 \(R \ge R_1 R_2\).

证明

(1) 由题知 \(\sum\limits_{n = 0}^{\infty}a_nx^n\) 与 \(\sum\limits_{n = 0}^{\infty}b_nx^n\) 在 \(\left|x\right|\le \min\{R_1, R_2\}\) 时同时收敛. 因此, 当 \(\left|x\right|\le \min\{R_1, R_2\}\) 时一定有 \(\sum\limits_{n = 0}^{\infty}(a_n+b_n)x^n\) 收敛. 故其收敛半径 \(R \ge \min\{R_1, R_2\}\).

(2) 由题知 \(q_1 = \varlimsup\limits_{n \to \infty}\sqrt[n]{\left|a_n\right|}, q_2 = \varlimsup\limits_{n \to \infty}\sqrt[n]{\left|b_n\right|}, q = \varlimsup\limits_{n \to \infty}\sqrt[n]{\left|a_nb_n\right|}\). 而 \(\varlimsup\limits_{n \to \infty}\sqrt[n]{\left|a_nb_n\right|} \le \varlimsup\limits_{n \to \infty}\sqrt[n]{\left|a_n\right|}\cdot\varlimsup\limits_{n \to \infty}\sqrt[n]{\left|b_n\right|}\). 故 \(q \le q_1q_2\). 由于 \(q_1 = \dfrac{1}{R_1}, q_2 = \dfrac{1}{R_2}, q = \dfrac{1}{R}\), 所以 \(R \ge R_1R_2\).