1 (每小题6分，共48分)
(1) 求$limlimits_{x o 0+}x^x;$
解答: $$egin{eqnarray*} extrm{ 原式} & = & limlimits_{x o 0+}e^{xln x} = limlimits_{x o 0+}e^{cfrac{ln x}{1/x}} = e^{limlimits_{x o 0+}cfrac{ln x}{1/x}}stackrel{L'Hospital}{=} e^{limlimits_{x o 0+}cfrac{1/x}{-1/x^2}} \ & = & e^{-limlimits_{x o 0+}x} = e^0 = 1.end{eqnarray*}$$
(2) 求$int sqrt{x}sinsqrt{x} d x;$
解答: $$egin{eqnarray*}{ extrm{ 原式}} & stackrel{t = sqrt{x}}{=} & 2int t^2sin t d t = -2int t^2 d (cos t) = -2t^2cos t + 4int tcos t d t \ & = & -2t^2 cos t + 4int t d (sin t) = -2t^2cos t + 4tsin t -4int sin t d t \ & = & -2t^2cos t +4tsin t + 4cos t + C \ & = & -2xcossqrt{x} + 4sqrt{x}sinsqrt{x} + 4 cossqrt{x} + C ; ({ extrm{其中}}C{ extrm{是任意常数}}).end{eqnarray*}$$
(3) 求$int_{1}^{e}cfrac{ d x}{x(2 + ln^2 x)};$
解答: $$egin{eqnarray*}{ extrm{ 原式}} & stackrel{t = ln x}{=} & int_{0}^{1}cfrac{ d t}{2 + t^2} = cfrac{1}{2}int_{0}^{1}cfrac{ d t}{1 + cfrac{t^2}{2}} = cfrac{sqrt{2}}{2}int_{0}^{1}cfrac{ d (cfrac{t}{sqrt{2}})}{1 + (cfrac{t}{sqrt{2}})^2} \ & = & cfrac{sqrt{2}}{2}arctan(cfrac{t}{sqrt{2}})|_{0}^{1} = cfrac{sqrt{2}}{2}arctancfrac{sqrt{2}}{2}.end{eqnarray*}$$
(4) 求$int_{0}^{+ infty}cfrac{xe^{-x}}{(1 + e^{-x})^2} d x;$
解答: $$egin{eqnarray*}{ extrm{ 原式}}& = & int_{0}^{+ infty}cfrac{xe^{x}}{(e^x + 1)^2} d x = -int_{0}^{+ infty}x d (cfrac{1}{e^x + 1}) \ & = & -xcfrac{1}{e^x + 1}|_{0}^{+infty} + int_{0}^{+infty}cfrac{1}{e^x + 1} d x \ & = & -limlimits_{x o +infty}cfrac{x}{e^x + 1} + int_{0}^{+infty}cfrac{1}{e^x + 1} d x \ &stackrel{L'Hospital}{=} & -limlimits_{x o +infty}cfrac{1}{e^x} + int_{0}^{+infty}cfrac{1}{e^x + 1} d x = int_{0}^{+infty}cfrac{1}{e^x + 1} d x \ &stackrel{e^{x} = t}{=}& int_{1}^{+infty}cfrac{1}{t(t + 1)} d t = int_{1}^{+infty}left(cfrac{1}{t} - cfrac{1}{t + 1} ight) d t \ & = & left[ln t- ln(t + 1) ight]|_{1}^{+infty} = lnleft(cfrac{t}{t + 1} ight)|_{1}^{+infty} \ & = & limlimits_{t o +infty}lnleft(cfrac{t}{t + 1} ight) - lncfrac{1}{2} = ln 2.end{eqnarray*}$$
(5) 方程$z = f(x,xy) + varphi(y + z)$确定函数$z = z(x,y)$, 求全微分$ d 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$$ d z = cfrac{f_1(x,xy) + y f_2(x,xy)}{1 - varphi'(y + z)} d x + cfrac{xf_2(x,xy) + varphi'(y + z)}{1 - varphi'(y + z)} d y.$$
(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$limlimits_{x o -infty}y(x) = +infty$，故所求面积 $$egin{eqnarray*}S & = & 2int_{0}^{4}sqrt{x^2(4 - x)} d x = 2int_{0}^{4}xsqrt{4 - x} d x stackrel{sqrt{4 - x} = t}{=} 4int_{0}^{2}(4 - t^2)t^2 d t \ & = & 4(cfrac{4}{3}t^3 - cfrac{1}{5}t^5)|_{0}^{2} = cfrac{256}{15}.end{eqnarray*}$$
(7) 计算二重积分$int!!!int_{D} (cfrac{x^2}{a^2} + cfrac{y^2}{b^2}) d x d y, $其中$D = {{(x,y)|x^2 + y^2 le 1}};$
解答: $$egin{eqnarray*}{ extrm{ 原式}} & stackrel{x = hocos heta, y = hosin heta}{=} & int_{0}^{1}int_{0}^{2pi}left(cfrac{ ho^2cos^2 heta}{a^2} + cfrac{ ho^2sin^2 heta}{b^2} ight) ho d ho d t \ & = & int_{0}^{1} ho^3 d hocdotint_{0}^{2pi}left(cfrac{ ho^2cos^2 heta}{a^2} + cfrac{ ho^2sin^2 heta}{b^2} ight) d t \ & = & cfrac{1}{4} ho^4|_{0}^{1}cdotint_{0}^{2pi}left(cfrac{1 + cos2 heta}{2a^2} + cfrac{1 - cos2 heta}{2b^2} ight) d t \ & = & cfrac{1}{4}left[left(cfrac{1}{2a^2} + cfrac{1}{2b^2} ight) heta + cfrac{1}{2}left(cfrac{1}{2a^2} - cfrac{1}{2b^2} ight)sin2 heta ight]|_{0}^{2pi} \ & = & cfrac{pi}{4}left(cfrac{1}{a^2} + cfrac{1}{b^2} ight).end{eqnarray*}$$
(8) 判别级数$sumlimits_{n=1}^{infty}u_n$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$$u_n = cfrac{1! + 2! + 3! + cdots + n!}{(2n)!} le cfrac{ncdot n!}{(2n)!} < cfrac{ncdot n!}{n!cdot n cdot n cdot n} = cfrac{1}{n^2},$$ 而级数$sumlimits_{n=1}^{infty}cfrac{1}{n^2}$ 收敛，因而由正项级数的比较原则可知，级数$sumlimits_{n=1}^{infty}u_n$收敛.
方法二： $$egin{eqnarray*}limlimits_{n o infty}cfrac{u_{n + 1}}{u_n} & = & cfrac{1! + 2! + 3! + cdots + n! + (n + 1)!}{(2(n+1))!}cdot cfrac{(2n)!}{1! + 2! + 3! + cdots + n!}\ & = & limlimits_{n o infty}cfrac{1! + 2! + 3! + cdots + n! + (n + 1)!}{(2n + 1)(2n + 2)(1! + 2! + 3! + cdots + n!)}\ & = & limlimits_{n o infty}cfrac{1! + 2! + 3! + cdots + n!}{(2n + 1)(2n + 2)(1! + 2! + 3! + cdots + n!)} + \ & & limlimits_{n o infty}cfrac{(n + 1)!}{(2n + 1)(2n + 2)(1! + 2! + 3! + cdots + n!)} \ & = & limlimits_{n o infty}cfrac{1}{(2n + 1)(2n + 2)} + limlimits_{n o infty}cfrac{n!}{2(2n + 1)(1! + 2! + 3! + cdots + n!)}\ & = & 0 + 0 = 0 < 1,end{eqnarray*}$$ 因而由正项级数的d'Alembert判别法或比式判别法可知，级数$sumlimits_{n=1}^{infty}u_n$收敛.
注记: 由于 $$egin{eqnarray*} 0 &leq& cfrac{n!}{2(2n + 1)(1! + 2! + 3! + cdots + n!)} < cfrac{n!}{2(2n + 1)(n!)} = cfrac{1}{2(2n+1)}\ & o& 0, (n o infty), end{eqnarray*}$$ 因此，由夹逼准则可知$$limlimits_{n o infty}cfrac{n!}{2(2n + 1)(1! + 2! + 3! + cdots + n!)} = 0.$$
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$$f'_{+}(1) = limlimits_{x o 1^{+}}cfrac{f(x) - f(1)}{x - 1} = limlimits_{x o 1^{+}}cfrac{xe^{-(x - 1)} - 1}{x - 1} = limlimits_{x o 1^{+}}cfrac{(1 - x)e^{1 - x}}{1} = 0,$$ $$f'_{-}(1) = limlimits_{x o 1^{-}}cfrac{f(x) - f(1)}{x - 1} = limlimits_{x o 1^{-}}cfrac{xe^{(x - 1)} - 1}{x - 1} = limlimits_{x o 1^{-}}cfrac{(x + 1)e^{x - 1}}{1} = 2,$$ 从而$f'_{-}(1) e f'_{+}(1),$ 即$f(x)$在$x = 1$时导数不存在;
5) 当$x = 0$时，此时$$f'_{+}(0) = limlimits_{x o 0^{+}}cfrac{f(x) - f(0)}{x - 0} = limlimits_{x o 0^{+}}cfrac{xe^{(x - 1)} - 0}{x - 0} = limlimits_{x o 0^{+}}e^{x - 1} = e^{-1},$$ $$f'_{-}(0) = limlimits_{x o 0^{-}}cfrac{f(x) - f(0)}{x - 0} = limlimits_{x o 0^{-}}cfrac{-xe^{(x - 1)} - 0}{x - 0} = -limlimits_{x o 0^{-}}e^{x - 1} = -e^{-1},$$ 从而$f'_{-}(0) e f'_{+}(0),$ 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$f'(x) = left{ egin{array}{ll} (1 - x)e^{1 - x}, & x > 1 \ extrm{不存在}, & x = 1 \ (x + 1)e^{x - 1}, & 0 < x < 1 \ extrm{不存在}, & x = 0 \ -(x + 1)e^{x - 1}, & x < 0 end{array} ight.$; $1\,^{circ}$, $x > 1, f'(x) < 0,$ $ 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$M = maxlimits_{0 le x le 1}|f'(x)|,$ 求证：$|int_{0}^{1}f(x) d x|le cfrac{1}{4}M.$
证明:
方法一: $$egin{eqnarray*}|int_{0}^{1}f(x) d x| & le & |int_{0}^{cfrac{1}{2}}f(x) d x| + |int_{cfrac{1}{2}}^{1}f(x) d x|\ & = & |int_{0}^{cfrac{1}{2}}[f(x) - f(0)] d x| + |int_{cfrac{1}{2}}^{1}[f(x) -f(1)] d x| \ & = & |int_{0}^{cfrac{1}{2}}f'(xi)(x-0) d x| + |int_{cfrac{1}{2}}^{1}f'(eta)(x-1) d x| \ & le & int_{0}^{cfrac{1}{2}}|f'(xi)(x-0)| d x + int_{cfrac{1}{2}}^{1}|f'(eta)(x-1)| d x \ & le & Mint_{0}^{cfrac{1}{2}}x d x + Mint_{cfrac{1}{2}}^{1}(1 - x) d x \ & = & Mcfrac{x^2}{2}|_{0}^{cfrac{1}{2}} + M(x - cfrac{x^2}{2})|_{cfrac{1}{2}}^{1} = cfrac{1}{4}M, ( extrm{其中}xi in (0,cfrac{1}{2}), eta in (cfrac{1}{2},1)). end{eqnarray*}$$
方法二： $$egin{eqnarray*}|int_{0}^{1}f(x) d x| & stackrel{t = x - cfrac{1}{2}}{=} & |int_{-cfrac{1}{2}}^{cfrac{1}{2}}f(t + cfrac{1}{2}) d t| stackrel{ extrm{ 分步积分}}{=} |t f(t+cfrac{1}{2})|_{cfrac{1}{2}}^{cfrac{1}{2}} - int_{-cfrac{1}{2}}^{cfrac{1}{2}}t f'(t + cfrac{1}{2}) d t|\ & = & |int_{-cfrac{1}{2}}^{cfrac{1}{2}}t f'(t + cfrac{1}{2}) d t| le int_{-cfrac{1}{2}}^{cfrac{1}{2}}|t f'(t + cfrac{1}{2})| d t\ & le & Mint_{-cfrac{1}{2}}^{cfrac{1}{2}}|t| d t = 2Mint_{0}^{cfrac{1}{2}}|t| d t = 2Mint_{0}^{cfrac{1}{2}}t d t \ & = & 2Mcdot cfrac{t^2}{2}|_{0}^{cfrac{1}{2}} = cfrac{1}{4}M.end{eqnarray*}$$
4 (18分) 设函数$f(x,y) = left{ egin{array}{ll} x - y + cfrac{(xy)^2}{(x^2 + y^2)^{3/2}}, & (x,y) e (0,0) \ 0, & (x,y) = (0,0)end{array} ight.$, 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) $$egin{eqnarray*}mbox{原极限} & stackrel{x= hocos heta,y= hosin heta}{=} & limlimits_{ ho o 0}left[ hocos heta - hosin heta + cfrac{( hocos heta hosin heta)^2}{ ho^3} ight] \ & = & limlimits_{ ho o 0} holeft[cos heta - sin heta + (cos hetasin heta)^2 ight] \ & = & 0 = f(0,0),end{eqnarray*}$$从而$f(x,y)$在原点处连续;
(2) $$f_x(0,0) = limlimits_{x o 0}cfrac{f(x,0) - f(0,0)}{x - 0} = limlimits_{x o 0}cfrac{x}{x} = 1,$$ $$f_y(0,0) = limlimits_{y o 0}cfrac{f(0,y) - f(0,0)}{y - 0} = limlimits_{x o 0}cfrac{-y}{y} = -1,$$从而$f(x,y)$在原点的偏导数$f_x(0,0)$和$f_y(0,0)$存在;
(3) $$egin{eqnarray*}& & limlimits_{(Delta x,Delta y) o (0,0)}cfrac{f(Delta x,Delta y) - f(0,0) - f_x(0,0)Delta x - f_y(0,0)Delta y}{sqrt{(Delta x)^2 + (Delta y)^2}}\ & = & limlimits_{(Delta x,Delta y) o (0,0)}cfrac{Delta x - Delta y + cfrac{(Delta x Delta y)^2}{((Delta x)^2 + (Delta y)^2)^{3/2}} - Delta x + Delta y}{sqrt{(Delta x)^2 + (Delta y)^2}} \ & = & limlimits_{(Delta x,Delta y) o (0,0)}cfrac{(Delta xDelta y)^2}{((Delta x)^2 + (Delta y)^2)^2}\ & stackrel{Delta y=k Delta x}{=} & limlimits_{Delta x o 0}cfrac{(k(Delta x)^2)^2}{((Delta x)^2 + k^2(Delta x)^2)^2} \ & = & cfrac{k^2}{(1 + k^2)^2}( extrm{随着}k extrm{的值的变化而变化}),end{eqnarray*}$$从而极限$limlimits_{(Delta x,Delta y) o (0,0)}cfrac{f(Delta x,Delta y) - f(0,0) - f_x(0,0)Delta x - f_y(0,0)Delta y}{sqrt{(Delta x)^2 + (Delta y)^2}}$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$left{ egin{array}{l} F_x = 2x + lambda y = 0\ F_y = 2y + lambda x = 0\ F_z = 2z - lambda = 0\ F_{lambda} = xy - z - 1 =0end{array} ight.$ $Longrightarrow$ $left{ egin{array}{l} x = 0\ y = 0\ z = -1\ lambda = -2,end{array} ight.$ 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$vec{F} = cfrac{yvec{i} - xvec{j}}{x^2 + y^2},$ 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$int_{L}vec{F} d vec{s}.$
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$x^2 + y^2 = a, (0 < a < 1, a o 0)$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, $cfrac{partial P}{partial x} = cfrac{partial Q}{partial y} = cfrac{x^2 - y^2}{(x^2 + y^2)^2} extrm{且连续},$ 由此可知 $$egin{eqnarray*} int_{L}vec{F} d vec{s} &=& int_{(L_1 - L_3) + (L_2 - L_3) + 2 L_3}vec{F} d vec{s}\ &=& int_{L_1 - L_3}vec{F} d vec{s} + int_{L_2 - L_3}vec{F} d vec{s} + int_{2 L_3}vec{F} d vec{s}\ &equiv& I_1 + I_2 + I_3. end{eqnarray*}$$ 由格林公式立即可得$I_1 = I_2 = int!!!intleft[cfrac{partial P}{partial x} - cfrac{partial Q}{partial y} ight] d x d y = 0,$ $$I_3 = 2int_{L_3}vec{F} d vec{s} stackrel{x=acos heta,y=asin heta}{=} 2int_{0}^{2pi}cfrac{a^2cos^2 heta + a^2sin^2 heta}{a^2} d t = 2int_{0}^{2pi} d t = 4pi.$$ 从而$$int_{L}vec{F} d vec{s}= 4pi.$$
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, $cfrac{partial P}{partial x} = cfrac{partial Q}{partial y} = cfrac{x^2 - y^2}{(x^2 + y^2)^2} extrm{且连续},$ 由此可知 $$int_{L}vec{F} d vec{s} = int_{(L_2 - L_1) + 2 L_1}vec{F} d vec{s} = int_{L_2 - L_1}vec{F} d vec{s} + int_{2 L_1}vec{F} d vec{s} = I_1 + I_2,$$ 由格林公式立即可得$I_1 = int!!!intleft[cfrac{partial P}{partial x} - cfrac{partial Q}{partial y} ight] d x d y = 0,$ $$I_2 = 2int_{L_1}vec{F} d vec{s} stackrel{x=cos heta,y=sin heta}{=} 2int_{0}^{2pi}cfrac{cos^2 heta + sin^2 heta}{1} d t = 2int_{0}^{2pi} d t = 4pi.$$ 从而$int_{L}vec{F} d vec{s}= 4pi.$
7 (16分) 求函数项级数$sumlimits_{n=1}^{infty}cfrac{x^2}{(1 + x^2)^n}$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$sumlimits_{n=1}^{infty}cfrac{x^2}{(1 + x^2)^n}$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$$S(x) = limlimits_{n o infty}S_n(x) = limlimits_{n o infty}0 = 0;$$
(2) 当$x e 0$时，此时$$S_n(x) = cfrac{cfrac{x^2}{1 + x^2}left[1 - (cfrac{1}{1 + x^2})^n ight]}{1 - cfrac{1}{1 + x^2}},$$ 从而$$S(x) = limlimits_{n o infty}S_n(x) = limlimits_{n o infty}left[1 - (cfrac{1}{1 + x^2})^n ight] = 1;$$
综上可知, $$S(x) = left{ egin{array}{ll} 1, & x e 0 \ 0, & x = 0 end{array} ight. ;$$ 又由于$S(x)$不连续，而$f_n(x)(n = 1,2,cdots)$每一项都连续，故$sumlimits_{n=1}^{infty}cfrac{x^2}{(1 + x^2)^n}$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
$$a_1 = sqrt{2} = 2sincfrac{pi}{4} = 2coscfrac{pi}{4}, a_2 = sqrt{2 - sqrt{2}} = sqrt{2 - 2coscfrac{pi}{4}} = 2sincfrac{pi}{8},$$ $$ a_3 = sqrt{2 - sqrt{2 + sqrt{2}}} = sqrt{2 - sqrt{2 + 2coscfrac{pi}{4}}}= sqrt{ 2 - 2coscfrac{pi}{8}} = 2sincfrac{pi}{16},$$ $$a_4= sqrt{2 - sqrt{2 + sqrt{2 + sqrt{2}}}} = sqrt{2 - sqrt{2 + sqrt{2 + 2coscfrac{pi}{4}}}} = 2sincfrac{pi}{32},$$ $cdots, a_n = 2sincfrac{pi}{2^{n + 1}}$,于是猜想$a_n = 2sincfrac{pi}{2^{n + 1}}(n=1,2,cdots)$, 下面用数学归纳法来证明
(1) 当$n = 1$时，此时$a_1=2sincfrac{pi}{4}$显然成立;
(2) 假设$n = k$时，$a_k$成立，即$a_k = 2sincfrac{pi}{2^{k + 1}}$, 下面证明当$n = k + 1$时， $a_{k+1} = sqrt{2 - sqrt{2 + 2 - a_k^2}} = sqrt{2 - 2coscfrac{pi}{k+1}} = 2sincfrac{pi}{2^{k+2}},$ 可知当$n = k + 1$时也成立.
于是可得$a_n = 2sincfrac{pi}{2^{n + 1}}(n=1,2,cdots)$, 而显然可得$a_n le 2cfrac{pi}{2^{n + 1}} = cfrac{pi}{2^{n}}$, 而级数$sumlimits_{n=1}^{infty}cfrac{pi}{2^{n}}$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$n o infty$ 时有
$a = sqrt{2 + a}$, $Longrightarrow a = 2$, 即$limlimits_{n o infty}a_n = 2.$ 从而 $$egin{eqnarray*}limlimits_{n o infty}cfrac{sqrt{2 - a_{n+1}}}{sqrt{2 - a_n}} & = & limlimits_{n o infty}sqrt{cfrac{2 - a_{n+1}}{2 - a_n}} = limlimits_{n o infty}sqrt{cfrac{2 -sqrt{2 + a_n}}{2 - a_n}} \ & = & limlimits_{n o infty}sqrt{cfrac{(2 -sqrt{2 + a_n})(2 +sqrt{2 + a_n})}{(2 - a_n)(2 +sqrt{2 + a_n})}} \ & = & limlimits_{n o infty}sqrt{cfrac{2 - a_n}{(2 - a_n)(2 +sqrt{2 + a_n})}}\ & = & limlimits_{n o infty}sqrt{cfrac{1}{2 +sqrt{2 + a_n}}} = cfrac{1}{2} < 1.end{eqnarray*}$$ 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.