[家里蹲大学数学杂志]第034期中山大学2008年数学分析考研试题参考解答

1  (每小题6分，共48分)

 (1) 求$\lim\limits_{x \to 0+}x^x;$

解答: 

$$\begin{eqnarray*}\textrm{ 原式} & = & \lim\limits_{x \to 0+}e^{x\ln x} = \lim\limits_{x \to 0+}e^{\cfrac{\ln x}{1/x}} = e^{\lim\limits_{x \to 0+}\cfrac{\ln x}{1/x}}\stackrel{L'Hospital}{=} e^{\lim\limits_{x \to 0+}\cfrac{1/x}{-1/x^2}} \\ & = & e^{-\lim\limits_{x \to 0+}x} = e^0 = 1.\end{eqnarray*}$$  

(2) 求$\int \sqrt{x}\sin\sqrt{x}\rd x;$

解答:  

$$\begin{eqnarray*}{\textrm{ 原式}} & \stackrel{t = \sqrt{x}}{=} & 2\int t^2\sin t \rd t = -2\int t^2 \rd (\cos t) = -2t^2\cos t + 4\int t\cos t \rd t \\ & = & -2t^2 \cos t + 4\int t \rd (\sin t) = -2t^2\cos t + 4t\sin t -4\int \sin t \rd t \\ & = & -2t^2\cos t +4t\sin t + 4\cos t + C \\ & = & -2x\cos\sqrt{x} + 4\sqrt{x}\sin\sqrt{x} + 4 \cos\sqrt{x} + C \; ({\textrm{其中}}C{\textrm{是任意常数}}).\end{eqnarray*}$$

 (3) 求$\int_{1}^{e}\cfrac{\rd x}{x(2 + \ln^2 x)};$

解答: 

$$\begin{eqnarray*}{\textrm{ 原式}} & \stackrel{t = \ln x}{=} & \int_{0}^{1}\cfrac{\rd t}{2 + t^2} = \cfrac{1}{2}\int_{0}^{1}\cfrac{\rd t}{1 + \cfrac{t^2}{2}} = \cfrac{\sqrt{2}}{2}\int_{0}^{1}\cfrac{\rd (\cfrac{t}{\sqrt{2}})}{1 + (\cfrac{t}{\sqrt{2}})^2} \\ & = & \cfrac{\sqrt{2}}{2}\arctan(\cfrac{t}{\sqrt{2}})|_{0}^{1} = \cfrac{\sqrt{2}}{2}\arctan\cfrac{\sqrt{2}}{2}.\end{eqnarray*}$$  

(4) 求$\int_{0}^{+ \infty}\cfrac{xe^{-x}}{(1 + e^{-x})^2}\rd x;$

解答:  

$$\begin{eqnarray*}{\textrm{ 原式}}& = & \int_{0}^{+ \infty}\cfrac{xe^{x}}{(e^x + 1)^2} \rd x = -\int_{0}^{+ \infty}x\rd (\cfrac{1}{e^x + 1}) \\ & = & -x\cfrac{1}{e^x + 1}|_{0}^{+\infty} + \int_{0}^{+\infty}\cfrac{1}{e^x + 1}\rd x \\ & = & -\lim\limits_{x \to +\infty}\cfrac{x}{e^x + 1} + \int_{0}^{+\infty}\cfrac{1}{e^x + 1}\rd x \\ &\stackrel{L'Hospital}{=} & -\lim\limits_{x \to +\infty}\cfrac{1}{e^x} + \int_{0}^{+\infty}\cfrac{1}{e^x + 1}\rd x = \int_{0}^{+\infty}\cfrac{1}{e^x + 1}\rd x \\ &\stackrel{e^{x} = t}{=}& \int_{1}^{+\infty}\cfrac{1}{t(t + 1)}\rd t = \int_{1}^{+\infty}\left(\cfrac{1}{t} - \cfrac{1}{t + 1}\right)\rd t \\ & = & \left[\ln t- \ln(t + 1)\right]|_{1}^{+\infty} = \ln\left(\cfrac{t}{t + 1}\right)|_{1}^{+\infty} \\ & = & \lim\limits_{t \to +\infty}\ln\left(\cfrac{t}{t + 1}\right) - \ln\cfrac{1}{2} = \ln 2.\end{eqnarray*}$$  

(5) 方程$z = f(x,xy) + \varphi(y + z)$确定函数$z = z(x,y)$, 求全微分$\rd z$;

解答: 在方程$z = f(x,xy) + \varphi(y + z)$左右两边分别关于$x,y$求偏导，可得

$$z_x = f_1(x,xy) + yf_2(x,xy) + \varphi'(y + z)z_x \Longrightarrow z_x = \cfrac{f_1(x,xy) + y f_2(x,xy)}{1 - \varphi'(y + z)},$$ $$z_y = xf_2(x,xy) + \varphi'(y + z)(1 + z_y) \Longrightarrow z_y = \cfrac{xf_2(x,xy) + \varphi'(y + z)}{1 - \varphi'(y + z)},$$ 从而全微分 $$\rd z = \cfrac{f_1(x,xy) + y f_2(x,xy)}{1 - \varphi'(y + z)}\rd x + \cfrac{xf_2(x,xy) + \varphi'(y + z)}{1 - \varphi'(y + z)}\rd y.$$  

(6) 求曲线$y^2 = x^2(4-x)$所围图形的面积;

解答: 由分析可知，曲线$y^2 = x^2(4-x)$关于$x$轴对称;又由于$x \le 4, y(0) = y(4) = 0,$ 当$x \le 0$时，$y(x)$单调递减且$\lim\limits_{x \to -\infty}y(x) = +\infty$，故所求面积 

$$\begin{eqnarray*}S & = & 2\int_{0}^{4}\sqrt{x^2(4 - x)}\rd x = 2\int_{0}^{4}x\sqrt{4 - x}\rd x \stackrel{\sqrt{4 - x} = t}{=} 4\int_{0}^{2}(4 - t^2)t^2\rd t \\ & = & 4(\cfrac{4}{3}t^3 - \cfrac{1}{5}t^5)|_{0}^{2} = \cfrac{256}{15}.\end{eqnarray*}$$  

(7) 计算二重积分$\int\!\!\!\int_{D} (\cfrac{x^2}{a^2} + \cfrac{y^2}{b^2})\rd x\rd y,$其中$D = \{{(x,y)|x^2 + y^2 \le 1}\};$

解答:  

$$\begin{eqnarray*}{\textrm{ 原式}} & \stackrel{x = \rho\cos\theta, y = \rho\sin\theta}{=} & \int_{0}^{1}\int_{0}^{2\pi}\left(\cfrac{\rho^2\cos^2\theta}{a^2} + \cfrac{\rho^2\sin^2\theta}{b^2}\right)\rho \rd \rho \rd \tt \\ & = & \int_{0}^{1}\rho^3\rd \rho\cdot\int_{0}^{2\pi}\left(\cfrac{\rho^2\cos^2\theta}{a^2} + \cfrac{\rho^2\sin^2\theta}{b^2}\right)\rd \tt \\ & = & \cfrac{1}{4}\rho^4|_{0}^{1}\cdot\int_{0}^{2\pi}\left(\cfrac{1 + \cos2\theta}{2a^2} + \cfrac{1 - \cos2\theta}{2b^2}\right)\rd \tt \\ & = & \cfrac{1}{4}\left[\left(\cfrac{1}{2a^2} + \cfrac{1}{2b^2}\right)\theta + \cfrac{1}{2}\left(\cfrac{1}{2a^2} - \cfrac{1}{2b^2}\right)\sin2\theta\right]|_{0}^{2\pi} \\ & = & \cfrac{\pi}{4}\left(\cfrac{1}{a^2} + \cfrac{1}{b^2}\right).\end{eqnarray*}$$  

(8) 判别级数$\sum\limits_{n=1}^{\infty}u_n$的敛散性，其中$u_n = \cfrac{1! + 2! + 3! + \cdots + n!}{(2n)!}, n = 1,2,\cdots$.

解答: 方法一：由于

$$u_n = \cfrac{1! + 2! + 3! + \cdots + n!}{(2n)!} \le \cfrac{n\cdot n!}{(2n)!} < \cfrac{n\cdot n!}{n!\cdot n \cdot n \cdot n} = \cfrac{1}{n^2},$$ 而级数$\sum\limits_{n=1}^{\infty}\cfrac{1}{n^2}$ 收敛，因而由正项级数的比较原则可知，级数$\sum\limits_{n=1}^{\infty}u_n$收敛.

方法二：

$$\begin{eqnarray*}\lim\limits_{n \to \infty}\cfrac{u_{n + 1}}{u_n} & = & \cfrac{1! + 2! + 3! + \cdots + n! + (n + 1)!}{(2(n+1))!}\cdot \cfrac{(2n)!}{1! + 2! + 3! + \cdots + n!}\\ & = & \lim\limits_{n \to \infty}\cfrac{1! + 2! + 3! + \cdots + n! + (n + 1)!}{(2n + 1)(2n + 2)(1! + 2! + 3! + \cdots + n!)}\\ & = & \lim\limits_{n \to \infty}\cfrac{1! + 2! + 3! + \cdots + n!}{(2n + 1)(2n + 2)(1! + 2! + 3! + \cdots + n!)} + \\ & & \lim\limits_{n \to \infty}\cfrac{(n + 1)!}{(2n + 1)(2n + 2)(1! + 2! + 3! + \cdots + n!)} \\ & = & \lim\limits_{n \to \infty}\cfrac{1}{(2n + 1)(2n + 2)} + \lim\limits_{n \to \infty}\cfrac{n!}{2(2n + 1)(1! + 2! + 3! + \cdots + n!)}\\ & = & 0 + 0 = 0 < 1,\end{eqnarray*}$$ 因而由正项级数的d'Alembert判别法或比式判别法可知，级数$\sum\limits_{n=1}^{\infty}u_n$收敛.

注记: 由于

$$\begin{eqnarray*} 0 &\leq& \cfrac{n!}{2(2n + 1)(1! + 2! + 3! + \cdots + n!)} < \cfrac{n!}{2(2n + 1)(n!)} = \cfrac{1}{2(2n+1)}\\ &\to& 0, (n \to \infty), \end{eqnarray*}$$ 因此，由夹逼准则可知 $$\lim\limits_{n \to \infty}\cfrac{n!}{2(2n + 1)(1! + 2! + 3! + \cdots + n!)} = 0.$$

 

2 (16分) 求函数$f(x) = |x|e^{-|x - 1|}$的导函数，以及函数$f(x)$的极值.

解答: 由题意可知

1) 当$x < 0$时，此时$f(x) = -xe^{x - 1}$, 从而$f'(x) = -e^{x - 1} - xe^{x - 1} = -(x + 1)e^{x - 1};$

2) 当$0 < x < 1$时，此时$f(x) = xe^{x - 1}$, 从而$f'(x) = e^{x - 1} + xe^{x - 1} = (x + 1)e^{x - 1};$

3) 当$x > 1$时，此时$f(x) = xe^{-(x - 1)}$, 从而$f'(x) = e^{-(x - 1)} - xe^{-(x - 1)} = (1 - x)e^{1 - x};$

4) 当$x = 1$时，此时

$$f'_{+}(1) = \lim\limits_{x \to 1^{+}}\cfrac{f(x) - f(1)}{x - 1} = \lim\limits_{x \to 1^{+}}\cfrac{xe^{-(x - 1)} - 1}{x - 1} = \lim\limits_{x \to 1^{+}}\cfrac{(1 - x)e^{1 - x}}{1} = 0,$$ $$f'_{-}(1) = \lim\limits_{x \to 1^{-}}\cfrac{f(x) - f(1)}{x - 1} = \lim\limits_{x \to 1^{-}}\cfrac{xe^{(x - 1)} - 1}{x - 1} = \lim\limits_{x \to 1^{-}}\cfrac{(x + 1)e^{x - 1}}{1} = 2,$$ 从而$f'_{-}(1) \ne f'_{+}(1),$ 即$f(x)$在$x = 1$时导数不存在;

5) 当$x = 0$时，此时

$$f'_{+}(0) = \lim\limits_{x \to 0^{+}}\cfrac{f(x) - f(0)}{x - 0} = \lim\limits_{x \to 0^{+}}\cfrac{xe^{(x - 1)} - 0}{x - 0} = \lim\limits_{x \to 0^{+}}e^{x - 1} = e^{-1},$$ $$f'_{-}(0) = \lim\limits_{x \to 0^{-}}\cfrac{f(x) - f(0)}{x - 0} = \lim\limits_{x \to 0^{-}}\cfrac{-xe^{(x - 1)} - 0}{x - 0} = -\lim\limits_{x \to 0^{-}}e^{x - 1} = -e^{-1},$$ 从而$f'_{-}(0) \ne f'_{+}(0),$ 即$f(x)$在$x = 0$时导数不存在;

综上可知，所求$f(x)$的导函数为$f'(x) = \left\{ \begin{array}{ll} (1 - x)e^{1 - x}, & x > 1 \\ \textrm{不存在}, & x = 1 \\ (x + 1)e^{x - 1}, & 0 < x < 1 \\ \textrm{不存在}, & x = 0 \\ -(x + 1)e^{x - 1}, & x < 0 \end{array} \right.$; $1\,^{\circ}$, $x > 1, f'(x) < 0,$ $0 < x < 1, f'(x) > 0$ $\Longrightarrow f(x)$ 在$x = 1$处取得极大值$f(1) = 1;$ $2\,^{\circ}$, $0 < x < 1, f'(x) > 0,$ $-1 < x < 0, f'(x) < 0$ $\Longrightarrow f(x)$ 在$x = 0$处取得极小值$f(0) = 0;$  $3\,^{\circ}$, $-1 < x < 0, f'(x) < 0,$ $x < -1, f'(x) > 0$ $\Longrightarrow f(x)$ 在$x = -1$处取得极大值$f(-1) = e^{-2};$

 综上可知，$f(x)$的极大值为1和$e^{-2}$, 极小值为0.

 

3 (10分) 设$f(x)$在$[0,1]$上有一阶连续导数，且$f(0) = f(1) = 0,$ 记$M = \max\limits_{0 \le x \le 1}|f'(x)|,$ 求证：$|\int_{0}^{1}f(x)\rd x|\le \cfrac{1}{4}M.$

证明:

方法一:

$$\begin{eqnarray*}|\int_{0}^{1}f(x)\rd x| & \le & |\int_{0}^{\cfrac{1}{2}}f(x)\rd x| + |\int_{\cfrac{1}{2}}^{1}f(x)\rd x|\\ & = & |\int_{0}^{\cfrac{1}{2}}[f(x) - f(0)]\rd x| + |\int_{\cfrac{1}{2}}^{1}[f(x) -f(1)]\rd x| \\ & = & |\int_{0}^{\cfrac{1}{2}}f'(\xi)(x-0)\rd x| + |\int_{\cfrac{1}{2}}^{1}f'(\eta)(x-1)\rd x| \\ & \le & \int_{0}^{\cfrac{1}{2}}|f'(\xi)(x-0)|\rd x + \int_{\cfrac{1}{2}}^{1}|f'(\eta)(x-1)|\rd x \\ & \le & M\int_{0}^{\cfrac{1}{2}}x \rd x + M\int_{\cfrac{1}{2}}^{1}(1 - x)\rd x \\ & = & M\cfrac{x^2}{2}|_{0}^{\cfrac{1}{2}} + M(x - \cfrac{x^2}{2})|_{\cfrac{1}{2}}^{1} = \cfrac{1}{4}M, (\textrm{其中}\xi \in (0,\cfrac{1}{2}), \eta \in (\cfrac{1}{2},1)). \end{eqnarray*}$$

方法二：

$$\begin{eqnarray*}|\int_{0}^{1}f(x)\rd x| & \stackrel{t = x - \cfrac{1}{2}}{=} & |\int_{-\cfrac{1}{2}}^{\cfrac{1}{2}}f(t + \cfrac{1}{2})\rd t| \stackrel{\textrm{ 分步积分}}{=} |t f(t+\cfrac{1}{2})|_{\cfrac{1}{2}}^{\cfrac{1}{2}} - \int_{-\cfrac{1}{2}}^{\cfrac{1}{2}}t f'(t + \cfrac{1}{2})\rd t|\\ & = & |\int_{-\cfrac{1}{2}}^{\cfrac{1}{2}}t f'(t + \cfrac{1}{2})\rd t| \le \int_{-\cfrac{1}{2}}^{\cfrac{1}{2}}|t f'(t + \cfrac{1}{2})|\rd t\\ & \le & M\int_{-\cfrac{1}{2}}^{\cfrac{1}{2}}|t|\rd t = 2M\int_{0}^{\cfrac{1}{2}}|t|\rd t = 2M\int_{0}^{\cfrac{1}{2}}t \rd t \\ & = & 2M\cdot \cfrac{t^2}{2}|_{0}^{\cfrac{1}{2}} = \cfrac{1}{4}M.\end{eqnarray*}$$

 

4 (18分) 设函数$f(x,y) = \left\{ \begin{array}{ll} x - y + \cfrac{(xy)^2}{(x^2 + y^2)^{3/2}}, & (x,y) \ne (0,0) \\ 0, & (x,y) = (0,0)\end{array} \right.$, 证明：

 (1) $f(x,y)$在原点处连续;

 (2) $f(x,y)$在原点的偏导数$f_x(0,0)$和$f_y(0,0)$存在;

 (3) $f(x,y)$在原点不可微.

解答:  (1)  

$$\begin{eqnarray*}\mbox{原极限} & \stackrel{x=\rho\cos\theta,y=\rho\sin\theta}{=} & \lim\limits_{\rho \to 0}\left[\rho\cos\theta - \rho\sin\theta + \cfrac{(\rho\cos\theta\rho\sin\theta)^2}{\rho^3}\right] \\ & = & \lim\limits_{\rho \to 0}\rho\left[\cos\theta - \sin\theta + (\cos\theta\sin\theta)^2\right] \\ & = & 0 = f(0,0),\end{eqnarray*}$$ 从而$f(x,y)$在原点处连续;

 (2)

$$f_x(0,0) = \lim\limits_{x \to 0}\cfrac{f(x,0) - f(0,0)}{x - 0} = \lim\limits_{x \to 0}\cfrac{x}{x} = 1,$$ $$f_y(0,0) = \lim\limits_{y \to 0}\cfrac{f(0,y) - f(0,0)}{y - 0} = \lim\limits_{x \to 0}\cfrac{-y}{y} = -1,$$ 从而$f(x,y)$在原点的偏导数$f_x(0,0)$和$f_y(0,0)$存在;

 (3)  

$$\begin{eqnarray*}& & \lim\limits_{(\Delta x,\Delta y) \to (0,0)}\cfrac{f(\Delta x,\Delta y) - f(0,0) - f_x(0,0)\Delta x - f_y(0,0)\Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}}\\ & = & \lim\limits_{(\Delta x,\Delta y) \to (0,0)}\cfrac{\Delta x - \Delta y + \cfrac{(\Delta x \Delta y)^2}{((\Delta x)^2 + (\Delta y)^2)^{3/2}} - \Delta x + \Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} \\ & = & \lim\limits_{(\Delta x,\Delta y) \to (0,0)}\cfrac{(\Delta x\Delta y)^2}{((\Delta x)^2 + (\Delta y)^2)^2}\\ & \stackrel{\Delta y=k \Delta x}{=} & \lim\limits_{\Delta x \to 0}\cfrac{(k(\Delta x)^2)^2}{((\Delta x)^2 + k^2(\Delta x)^2)^2} \\ & = & \cfrac{k^2}{(1 + k^2)^2}(\textrm{随着}k\textrm{的值的变化而变化}),\end{eqnarray*}$$ 从而极限$\lim\limits_{(\Delta x,\Delta y) \to (0,0)}\cfrac{f(\Delta x,\Delta y) - f(0,0) - f_x(0,0)\Delta x - f_y(0,0)\Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}}$不存在，故$f(x,y)$在原点不可微.

 

5 (16分) 求曲面$z = xy -1$上与原点最近的点的坐标.

解答: 首先构造拉格朗日函数$F(x,y,z,\lambda) = x^2 + y^2 + z^2 + \lambda(xy - z -1)$, 于是有$\left\{ \begin{array}{l} F_x = 2x + \lambda y = 0\\ F_y = 2y + \lambda x = 0\\ F_z = 2z - \lambda = 0\\ F_{\lambda} = xy - z - 1 =0\end{array} \right.$ $\Longrightarrow$ $\left\{ \begin{array}{l} x = 0\\ y = 0\\ z = -1\\ \lambda = -2,\end{array} \right.$ 由于$(0,0,-1)$是此问题的唯一驻点 (稳定点) ，而此问题一定有最小值，故$(0,0,-1)$为所求点.

 

6 (16分)  设$\vec{F} = \cfrac{y\vec{i} - x\vec{j}}{x^2 + y^2},$ 曲线$L$ 由圆$x^2 + y^2 = 1$ 和椭圆$\cfrac{x^2}{4} + y^2 = 1$组成，方向均为逆时针方向，求$\int_{L}\vec{F}\rd \vec{s}.$

解答: 方法一：记圆$x^2 + y^2 = 1$为曲线$L_1$, 椭圆$\cfrac{x^2}{4} + y^2 = 1$为曲线$L_2$, 于是$L = L_1 + L_2$,又设圆$x^2 + y^2 = a, (0 < a < 1, a \to 0)$为曲线$L_3$, 方向为逆时针方向，于是$L = (L_1 - L_3) + (L_2 - L_3) + 2 L_3$,再记$P(x,y) = \cfrac{-x}{x^2 + y^2}, Q(x,y) = \cfrac{y}{x^2 + y^2}$,于是在$(L_1 - L_3) + (L_2 - L_3)$上, $\cfrac{\partial P}{\partial x} = \cfrac{\partial Q}{\partial y} = \cfrac{x^2 - y^2}{(x^2 + y^2)^2}\textrm{且连续},$ 由此可知

$$\begin{eqnarray*} \int_{L}\vec{F}\rd \vec{s} &=& \int_{(L_1 - L_3) + (L_2 - L_3) + 2 L_3}\vec{F}\rd \vec{s}\\ &=& \int_{L_1 - L_3}\vec{F}\rd \vec{s} + \int_{L_2 - L_3}\vec{F}\rd \vec{s} + \int_{2 L_3}\vec{F}\rd \vec{s}\\ &\equiv& I_1 + I_2 + I_3. \end{eqnarray*}$$ 由格林公式立即可得$I_1 = I_2 = \int\!\!\!\int\left[\cfrac{\partial P}{\partial x} - \cfrac{\partial Q}{\partial y}\right]\rd x\rd y = 0,$ $$I_3 = 2\int_{L_3}\vec{F}\rd \vec{s} \stackrel{x=a\cos\theta,y=a\sin\theta}{=} 2\int_{0}^{2\pi}\cfrac{a^2\cos^2\theta + a^2\sin^2\theta}{a^2}\rd \tt = 2\int_{0}^{2\pi}\rd \tt = 4\pi.$$ 从而 $$\int_{L}\vec{F}\rd \vec{s}= 4\pi.$$

方法二：记圆$x^2 + y^2 = 1$为曲线$L_1$, 椭圆$\cfrac{x^2}{4} + y^2 = 1$为曲线$L_2$, 于是$L = L_1 + L_2$,再记$P(x,y) = \cfrac{-x}{x^2 + y^2}, Q(x,y) = \cfrac{y}{x^2 + y^2}$,于是在$L_2 - L_1$上, $\cfrac{\partial P}{\partial x} = \cfrac{\partial Q}{\partial y} = \cfrac{x^2 - y^2}{(x^2 + y^2)^2}\textrm{且连续},$ 由此可知

$$\int_{L}\vec{F}\rd \vec{s} = \int_{(L_2 - L_1) + 2 L_1}\vec{F}\rd \vec{s} = \int_{L_2 - L_1}\vec{F}\rd \vec{s} + \int_{2 L_1}\vec{F}\rd \vec{s} = I_1 + I_2,$$ 由格林公式立即可得$I_1 = \int\!\!\!\int\left[\cfrac{\partial P}{\partial x} - \cfrac{\partial Q}{\partial y}\right]\rd x\rd y = 0,$ $$I_2 = 2\int_{L_1}\vec{F}\rd \vec{s} \stackrel{x=\cos\theta,y=\sin\theta}{=} 2\int_{0}^{2\pi}\cfrac{\cos^2\theta + \sin^2\theta}{1}\rd \tt = 2\int_{0}^{2\pi}\rd \tt = 4\pi.$$ 从而$\int_{L}\vec{F}\rd \vec{s}= 4\pi.$

 

7 (16分) 求函数项级数$\sum\limits_{n=1}^{\infty}\cfrac{x^2}{(1 + x^2)^n}$的和函数，并讨论在$x \in (-\infty,+\infty)$上的一致收敛性.

解答: 记$f_n(x) = \cfrac{x^2}{(1 + x^2)^n}$, 函数项级数$\sum\limits_{n=1}^{\infty}\cfrac{x^2}{(1 + x^2)^n}$的前$n$项部分和函数为$S_n(x)$, 和函数为$S(x)$, 于是有

(1) 当$x = 0$时，此时$S_n(x) = 0$, 从而

$$S(x) = \lim\limits_{n \to \infty}S_n(x) = \lim\limits_{n \to \infty}0 = 0;$$

 (2) 当$x \ne 0$时，此时

$$S_n(x) = \cfrac{\cfrac{x^2}{1 + x^2}\left[1 - (\cfrac{1}{1 + x^2})^n\right]}{1 - \cfrac{1}{1 + x^2}},$$ 从而 $$S(x) = \lim\limits_{n \to \infty}S_n(x) = \lim\limits_{n \to \infty}\left[1 - (\cfrac{1}{1 + x^2})^n\right] = 1;$$

综上可知,

$$S(x) = \left\{ \begin{array}{ll} 1, & x \ne 0 \\ 0, & x = 0 \end{array} \right. ;$$ 又由于$S(x)$不连续，而$f_n(x)(n = 1,2,\cdots)$每一项都连续，故$\sum\limits_{n=1}^{\infty}\cfrac{x^2}{(1 + x^2)^n}$不一致连续.

 

8 (10分) 研究级数$\sqrt{2} + \sqrt{2 - \sqrt{2}} + \sqrt{2 - \sqrt{2 + \sqrt{2}}} + \sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{2}}}} + \cdots$的敛散性.

解答: 方法一：设$a_1 = \sqrt{2}, a_2 = \sqrt{2 - \sqrt{2}}, a_3 = \sqrt{2 - \sqrt{2 + \sqrt{2}}}, \cdots,$ 从而可得

$$a_1 = \sqrt{2} = 2\sin\cfrac{\pi}{4} = 2\cos\cfrac{\pi}{4}, a_2 = \sqrt{2 - \sqrt{2}} = \sqrt{2 - 2\cos\cfrac{\pi}{4}} = 2\sin\cfrac{\pi}{8},$$ $$a_3 = \sqrt{2 - \sqrt{2 + \sqrt{2}}} = \sqrt{2 - \sqrt{2 + 2\cos\cfrac{\pi}{4}}}= \sqrt{ 2 - 2\cos\cfrac{\pi}{8}} = 2\sin\cfrac{\pi}{16},$$ $$a_4= \sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{2}}}} = \sqrt{2 - \sqrt{2 + \sqrt{2 + 2\cos\cfrac{\pi}{4}}}} = 2\sin\cfrac{\pi}{32},$$ $\cdots, a_n = 2\sin\cfrac{\pi}{2^{n + 1}}$,于是猜想$a_n = 2\sin\cfrac{\pi}{2^{n + 1}}(n=1,2,\cdots)$, 下面用数学归纳法来证明

(1) 当$n = 1$时，此时$a_1=2\sin\cfrac{\pi}{4}$显然成立;

(2) 假设$n = k$时，$a_k$成立，即$a_k = 2\sin\cfrac{\pi}{2^{k + 1}}$, 下面证明当$n = k + 1$时， $a_{k+1} = \sqrt{2 - \sqrt{2 + 2 - a_k^2}} = \sqrt{2 - 2\cos\cfrac{\pi}{k+1}} = 2\sin\cfrac{\pi}{2^{k+2}},$ 可知当$n = k + 1$时也成立.

于是可得$a_n = 2\sin\cfrac{\pi}{2^{n + 1}}(n=1,2,\cdots)$, 而显然可得$a_n \le 2\cfrac{\pi}{2^{n + 1}} = \cfrac{\pi}{2^{n}}$, 而级数$\sum\limits_{n=1}^{\infty}\cfrac{\pi}{2^{n}}$收敛，由正项级数的比较原则可知，所求原级数收敛.

 

方法二：设$a_n = \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2}}},$ ($n$个根号) ，满足$a_{n + 1} = \sqrt{2 + a_n},$

现在用数学归纳法来证明数列$\{a_n\}$是有界的.

显然，$a_1 = \sqrt{2} \in (0,2);$

假设$n = k$时，$0 < a_k < 2,$

则当$n = k + 1$时，$0 < a_{k + 1} = \sqrt{2 + a_k} < \sqrt{2 + 2} = 2,$ 所以$0 < a_n < 2 (n = 1,2,\cdots),$ 数列$\{a_n\}$有界的. 由于

$$\cfrac{a_{n + 1}}{a_n} = \cfrac{\sqrt{2 + a_n}}{a_n} = \sqrt{\cfrac{2}{a_n^2} + \cfrac{1}{a_n}} >1,$$ 因此数列$\{a_n\}$单调递增.

由单调有界原理，数列$\{a_n\}$有极限，记为$a$.由于

$$a_{n + 1} = \sqrt{2 + a_n},$$ 运用数列极限的四则运算法则，当$n \to \infty$ 时有

$a = \sqrt{2 + a}$, $\Longrightarrow a = 2$, 即$\lim\limits_{n \to \infty}a_n = 2.$ 从而 

$$\begin{eqnarray*}\lim\limits_{n \to \infty}\cfrac{\sqrt{2 - a_{n+1}}}{\sqrt{2 - a_n}} & = & \lim\limits_{n \to \infty}\sqrt{\cfrac{2 - a_{n+1}}{2 - a_n}} = \lim\limits_{n \to \infty}\sqrt{\cfrac{2 -\sqrt{2 + a_n}}{2 - a_n}} \\ & = & \lim\limits_{n \to \infty}\sqrt{\cfrac{(2 -\sqrt{2 + a_n})(2 +\sqrt{2 + a_n})}{(2 - a_n)(2 +\sqrt{2 + a_n})}} \\ & = & \lim\limits_{n \to \infty}\sqrt{\cfrac{2 - a_n}{(2 - a_n)(2 +\sqrt{2 + a_n})}}\\ & = & \lim\limits_{n \to \infty}\sqrt{\cfrac{1}{2 +\sqrt{2 + a_n}}} = \cfrac{1}{2} < 1.\end{eqnarray*}$$ 因而由正项级数的d'Alembert判别法或比式判别法可知，所求原级数收敛.