1
(每小题6分，共48分)
(1) 求limx→0+
x
x
;
解答:
原式=
=
lim
x→0+
e
xlnx
=lim
x→0+
e
lnx
1/x
=e
lim
x→0+
lnx
1/x
=
L
′
Hospital
e
lim
x→0+
1/x
?1/x
2
e
?lim
x→0+
x
=e
0
=1.
(2)
求∫x√
sinx
√
dx;
解答:
原式=
t=x
√
=
=
=
2∫t
2
sintdt=?2∫t
2
d(cost)=?2t
2
cost+4∫tcostdt
?2t
2
cost+4∫td(sint)=?2t
2
cost+4tsint?4∫sintdt
?2t
2
cost+4tsint+4cost+C
?2xcosx
√
+4x
√
sinx
√
+4cosx
√
+C(其中C是任意常数).
(3) 求∫e
1
dx
x(2+ln
2
x)
;
解答:
原式=
t=lnx
=
∫
1
0
dt
2+t
2
=
1
2
∫
1
0
dt
1+
t
2
2
=
2
√
2
∫
1
0
d(
t
2
√
)
1+(
t
2
√
)
2
2
√
2
arctan(
t
2
√
)|
1
0
=
2
√
2
arctan
2
√
2
.
(4) 求∫+∞
0
xe
?x
(1+e
?x
)
2
dx;
解答:
原式=
=
=
=
L
′
Hospital
=
e
x
=t
=
=
∫
+∞
0
xe
x
(e
x
+1)
2
dx=?∫
+∞
0
xd(
1
e
x
+1
)
?x
1
e
x
+1
|
+∞
0
+∫
+∞
0
1
e
x
+1
dx
?lim
x→+∞
x
e
x
+1
+∫
+∞
0
1
e
x
+1
dx
?lim
x→+∞
1
e
x
+∫
+∞
0
1
e
x
+1
dx=∫
+∞
0
1
e
x
+1
dx
∫
+∞
1
1
t(t+1)
dt=∫
+∞
1
(
1
t
?
1
t+1
)dt
[lnt?ln(t+1)]|
+∞
1
=ln(
t
t+1
)|
+∞
1
lim
t→+∞
ln(
t
t+1
)?ln
1
2
=ln2.
(5) 方程z=f(x,xy)+φ(y+z)
确定函数z=z(x,y)
, 求全微分dz
;
解答: 在方程z=f(x,xy)+φ(y+z)
左右两边分别关于x,y
求偏导，可得
zx
=f
1
(x,xy)+yf
2
(x,xy)+φ
′
(y+z)z
x
?z
x
=
f
1
(x,xy)+yf
2
(x,xy)
1?φ
′
(y+z)
,
zy
=xf
2
(x,xy)+φ
′
(y+z)(1+z
y
)?z
y
=
xf
2
(x,xy)+φ
′
(y+z)
1?φ
′
(y+z)
,
从而全微分
dz=f
1
(x,xy)+yf
2
(x,xy)
1?φ
′
(y+z)
dx+
xf
2
(x,xy)+φ
′
(y+z)
1?φ
′
(y+z)
dy.
(6) 求曲线y2
=x
2
(4?x)
所围图形的面积;
解答: 由分析可知，曲线y2
=x
2
(4?x)
关于x
轴对称;又由于x≤4,y(0)=y(4)=0,
当x≤0
时，y(x)
单调递减且limx→?∞
y(x)=+∞
，故所求面积
S=
=
2∫
4
0
x
2
(4?x)
?
?
?
?
?
?
?
?
√
dx=2∫
4
0
x4?x
?
?
?
?
√
dx=
4?x
√
=t
4∫
2
0
(4?t
2
)t
2
dt
4(
4
3
t
3
?
1
5
t
5
)|
2
0
=
256
15
.
(7) 计算二重积分∫∫D
(
x
2
a
2
+
y
2
b
2
)dxdy,
其中D={(x,y)|x2
+y
2
≤1};
解答:
原式=
x=ρcosθ,y=ρsinθ
=
=
=
=
∫
1
0
∫
2π
0
(
ρ
2
cos
2
θ
a
2
+
ρ
2
sin
2
θ
b
2
)ρdρdθ
∫
1
0
ρ
3
dρ?∫
2π
0
(
ρ
2
cos
2
θ
a
2
+
ρ
2
sin
2
θ
b
2
)dθ
1
4
ρ
4
|
1
0
?∫
2π
0
(
1+cos2θ
2a
2
+
1?cos2θ
2b
2
)dθ
1
4
[(
1
2a
2
+
1
2b
2
)θ+
1
2
(
1
2a
2
?
1
2b
2
)sin2θ]|
2π
0
π
4
(
1
a
2
+
1
b
2
).
(8) 判别级数∑n=1
∞
u
n
的敛散性，其中un
=
1!+2!+3!+?+n!
(2n)!
,n=1,2,?
.
解答: 方法一：由于
un
=
1!+2!+3!+?+n!
(2n)!
≤
n?n!
(2n)!
<
n?n!
n!?n?n?n
=
1
n
2
,
而级数∑n=1
∞
1
n
2
收敛，因而由正项级数的比较原则可知，级数∑n=1
∞
u
n
收敛.
方法二：
limn→∞
u
n+1
u
n
=
=
=
=
=
1!+2!+3!+?+n!+(n+1)!
(2(n+1))!
?
(2n)!
1!+2!+3!+?+n!
lim
n→∞
1!+2!+3!+?+n!+(n+1)!
(2n+1)(2n+2)(1!+2!+3!+?+n!)
lim
n→∞
1!+2!+3!+?+n!
(2n+1)(2n+2)(1!+2!+3!+?+n!)
+
lim
n→∞
(n+1)!
(2n+1)(2n+2)(1!+2!+3!+?+n!)
lim
n→∞
1
(2n+1)(2n+2)
+lim
n→∞
n!
2(2n+1)(1!+2!+3!+?+n!)
0+0=0<1,
因而由正项级数的d‘Alembert判别法或比式判别法可知，级数∑n=1
∞
u
n
收敛.
注记: 由于
0≤
→
n!
2(2n+1)(1!+2!+3!+?+n!)
<
n!
2(2n+1)(n!)
=
1
2(2n+1)
0,(n→∞),
因此，由夹逼准则可知
limn→∞
n!
2(2n+1)(1!+2!+3!+?+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)=?xex?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)=xex?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
x→1
+
f(x)?f(1)
x?1
=lim
x→1
+
xe
?(x?1)
?1
x?1
=lim
x→1
+
(1?x)e
1?x
1
=0,
f′
?
(1)=lim
x→1
?
f(x)?f(1)
x?1
=lim
x→1
?
xe
(x?1)
?1
x?1
=lim
x→1
?
(x+1)e
x?1
1
=2,
从而f‘_{-}(1) \ne f‘_{+}(1),f′
?
(1)≠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.