解
      由于\[\begin{align*}\displaystyle\sum_{n=1}^{\infty }\frac{1}{(n+1)}x^{n}
&=\frac{1}{x}\displaystyle\sum_{n=1}^{\infty }\frac{1}{(n+1)}x^{n+1} \\
&=\frac{1}{x}\int\limits(\displaystyle\sum_{n=1}^{\infty }\frac{1}{(n+1)}x^{n+1})'dx \\
&=\frac{1}{x}\int\limits\displaystyle\sum_{n=1}^{\infty }x^{n}dx \\
&=\frac{1}{x}\int\limits\frac{x}{(1-x)}dx \\
&=\frac{1}{x}[x-\ln(1-x)] .
\end{align*}\]令$x=\frac{1}{2}$，代入上式，得\[\displaystyle\sum_{n=1}^{\infty }\frac{1}{(n+1)2^n}=1+2\ln2.\]

解    由于\[\sqrt{1+\tan x}\backsim 1+\frac{1}{2}\tan x=1+\frac{1}{2}\sin x,\]\[\sqrt{1-\sin x}\backsim 1-\frac{1}{2}\sin x,\]所以\[\lim_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1-\sin x}}{x}=\lim_{x\to0}\frac{1+\frac{1}{2}\sin x-1+\frac{1}{2}\sin x}{x}=1.\]

解
         由于\[\frac{\mathrm{d}z}{\mathrm{d}x}=2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}.\]\[2x+y+x\frac{\mathrm{d}y}{\mathrm{d}x}+2y\frac{\mathrm{d}y}{\mathrm{d}x}=0,\]\[\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{2x+y}{x+2y}.\]\[\therefore  \frac{\mathrm{d}z}{\mathrm{d}x}=2x-2y\frac{2x+y}{x+2y}=\frac{2x^2-2y^2}{x+2y}.\]\[\frac{\mathrm{d}^2z}{\mathrm{d} x^2}=(\frac{2x^2-2y^2}{x+2y})'=\cdots .\]

解  由\[\begin{cases}
y^2=2x\\
x^2+y^2=8
\end{cases}\]\[解得两曲线的交点(2,2),(2,-2).设第一象限内两曲线共有区域的面积为\varDelta _1，\]\[圆被抛物线分割的另一部分面积为\varDelta _2.则\]\[\begin{align*}\varDelta _1
&=\int_{0}^{2}[\sqrt{8-y^2}-\frac{1}{2}y^2]dy \\
&=\int_{0}^{2}\sqrt{8-y^2}dy-\int_{0}^{2}\frac{1}{2}y^2dy \\
&=2+\pi-\frac{4}{3} =\frac{2}{3}+\pi .
\end{align*}\]\[左右面积之比=\frac{8\pi -2\varDelta _1}{2\varDelta _1}=\frac{6\pi -\frac{4}{3}}{2\pi +\frac{4}{3}}.\]


      

解  用参数方程\[x=a\cos \theta ,y=b\sin \theta .0\le \theta \le \frac{\pi }{2}.\]从而对弧长的曲线积分\[\begin{align*}I
&=\int_{L}xyds \\
&=ab\int_{0}^{\frac{\pi }{2}}\cos \theta \sin \theta\sqrt{a^2\sin ^2\theta+b^2\cos ^2\theta}d\theta  \\
&=ab\int_{0}^{\frac{\pi }{2}}\frac{\sqrt{a^2\sin ^2\theta+b^2\cos ^2\theta}d[(a^2-b^2)\sin^2 \theta+b^2 ]}{2(a^2-b^2)}\\
&=\frac{ab(a^2+ab+b^2)}{3(a+b)}.
\end{align*}\]

解 \[直线z=y,x=0绕z轴一圈的空间方程为：z^2=x^2+y^2.\]\[\Omega :z^2=x^2+y^2,z=1\]\[\begin{align*}I
&=\iiint\limits_{\Omega }(x^2+y^2+z)dxdydz \\
&=\iint\limits_{D} dxdy\int_{\sqrt{x^2+y^2}}^{1}(x^2+y^2+z)dz\\
&=\iint\limits_{D}(x^2+y^2)(1-\sqrt{x^2+y^2}) dxdy+\frac{1}{2}\iint\limits_{D}(1-(x^2+y^2))dxdy\\
&=\int_{0}^{2\pi }d\theta \int_{0}^{1}r^2(1-r^2)rdr+\frac{1}{2}\int_{0}^{2\pi }d\theta \int_{0}^{1}(1-r^2)rdr\\
&=\frac{\pi }{10} +\frac{\pi }{4}\\
&=\frac{7}{20}\pi .
\end{align*}\]

解 \[f(x)为偶函数，周期2\pi .b_n=0.(n=1,2,\cdots ).\]\[\begin{align*}a_0
&=\frac{2}{\pi }\int_{0}^{\pi }\arcsin (\cos x)dx \\
&=\frac{2}{\pi }\int_{0}^{\pi }\arcsin (\sin (\frac{\pi }{2}-x))dx \\
&=\frac{2}{\pi }\int_{0}^{\pi }(\frac{\pi }{2}-x)dx\\
&=0.
\end{align*}\]\[\begin{align*}a_n
&=\frac{2}{\pi }\int_{0}^{\pi }\arcsin (\cos x)\cos nx dx \\
&=\frac{2}{\pi }\int_{0}^{\pi }\arcsin (\sin (\frac{\pi }{2}-x))\cos nx dx \\
&=\frac{2}{\pi }\int_{0}^{\pi } (\frac{\pi }{2}-x)\cos nx dx \\
&=\frac{2}{n\pi }[ (\frac{\pi }{2}-x)\sin nx-\frac{1}{n} \cos nx]|_0^\pi  \\
&=\begin{cases}
0,& x=2k\\
\frac{4}{(2k-1)^2\pi },& n=2k-1
\end{cases}(k=1,2,\cdots )
\end{align*}\]
由狄氏收敛定理，得\[f(x)=\arcsin (\cos x)=\frac{4}{\pi }\displaystyle\sum_{n=1}^{\infty }\frac{1}{(2n-1)^2}\cos (2n-1)x.\]

解 由积分区域的对称性可知：\[\iint_{S}xdS=\iint_{S}ydS=0.\]所以\[\begin{align*}I &=\iint_{S}(x+y+z)dS= \iint_{S}zdS\\
&=\iint_{D}\sqrt{a^2-x^2-y^2}\sqrt{1+\frac{x^2}{a^2-x^2-y^2}+\frac{y^2}{a^2-x^2-y^2}}dxdy \\
&=\iint_{D}\sqrt{a^2-x^2-y^2}\cdotp \frac{a}{\sqrt{a^2-x^2-y^2}} dxdy\\
&=a\iint_{D}dxdy=a\cdotp \pi a^2=\pi a^3.
\end{align*}\]




   

证明  \[\because \lim_{n\to\infty }a_n=a.\]\[\therefore \forall \varepsilon > 0,\exists N\in \mathbb{N},n> N,s.t.|a_n-a|< \varepsilon .\]对于上面的$N$,取\[\varepsilon =1.此时有a-1< a_n< a+1,\]而\[\sqrt[n]{a-1}< \sqrt[n]{a_n}< \sqrt[n]{a+1}.\]又因为有\[\lim_{n\to\infty }\sqrt[n]{a}=1,\]故由夹逼法得到\[\lim_{n\to\infty }\sqrt[n]{a_n}=1。\]

证明 \[\because \lim_{x\to+\infty }f(x)存在.\]\[\therefore \forall \varepsilon > 0，\exists X,x',x''> X,(x',x''\in [X+1,+\infty )),对于\delta > 0,0< ｜x'-x''｜< \delta，\]\[ s.t.|f(x')-f(x'')|< \varepsilon . 一致连续.\]又因为\[f(x)\in C[a,+\infty ).所以对于上面的\varepsilon ，\forall x',x''\in [a,X+1],对于\delta > 0,0< ｜x'-x''｜< \delta，\]\[ s.t.|f(x')-f(x'')|< \varepsilon . 一致连续.\]从而有\[f(x)在[a,+\infty )上一致连续.\]
        [注]此结论。相当有用。可以总结为：在某开区间内I连续的函数，且在I的左、右或左右两个端点极限存在，那么。函数在该区间I内一致连续。

证明 由\[f(x)=x^3e^{-x^2}.\]得\[f'(x)=3x^2e^{-x^2}-2x^4e^{-x^2}=0.x=\pm \sqrt{\frac{3}{2}}.\]\[f''(x)=6xe^{-x^2}-6x^3e^{-x^2}-8x^3e^{-x^2}+4x^5e^{-x^2}.f''(\frac{3}{2})< 0,f''(-\frac{3}{2})>  0.\]得到\[f(-\frac{3}{2})\le f(x)\le f(\frac{3}{2}).\]计算函数极大极小值 \[f(\frac{3}{2})=\frac{3}{2}\sqrt{\frac{3}{2}}e^{-\frac{3}{2}},\]\[f(-\frac{3}{2})=-\frac{3}{2}\sqrt{\frac{3}{2}}e^{-\frac{3}{2}},\]于是\[|f(x)|\le \frac{3}{2}\sqrt{\frac{3}{2}}e^{-\frac{3}{2}},函数f(x)有界.\]

证明 由于\[A> 0,x\in [\delta ,+\infty ).|\int_{0}^{A}\sin (xy)dy|\le |\frac{1-\cos (\delta A)}{x}|\le \frac{2}{\delta }.\]\[\frac{1}{y}\downarrow ,\rightarrow 0.\]由Dirichlet判别法，\[\int_{0}^{+\infty }\frac{\sin (xy)}{y}dy在[\delta ,+\infty )上一致收敛.\]

      \[对M> 0,总能找到A_1,A_2> M.比如A_1=\frac{\pi }{3}M,A_2=\frac{2\pi }{3}M.x=\frac{1}{M}\in (0,+\infty ).\]由柯西收敛定理\[|\int_{A_1}^{A_2}\frac{\sin (xy)}{y}dy|\geqslant \frac{1}{2}|\int_{\frac{\pi }{3}M}^{\frac{2\pi }{3}M}\frac{1}{y}dy|=\frac{\ln2}{2}=\varepsilon _0> 0.\]因此\[\int_{0}^{+\infty }\frac{\sin (xy)}{y}dy在[0 ,+\infty )上非一致收敛.\]

证明  由于\[\mathrm{E}有界，\forall x_n\in \mathrm{E},a=M_1\le x_n\le M_2=b.即x_n\in [a,b]=[a_0,b_0].\]\[
中分[a,b]，得到两个区间，则必有一个部分包含\mathrm{E}的无穷多个点x_n,设这个区间为[a_1,b_1].\]\[如此下去，得到[a_2,b_2]，\cdots ，[a_n,b_n].\]由取法，显然有\[[a,b]=[a_0,b_0]\supset [a_1,b_1]\supset [a_2,b_2]\supset \cdots \supset [a_n,b_n].\]从而有\[\lim_{n\to\infty }(b_n-a_n)=0.\]\[\exists x\in \mathrm{E},s.t.\lim_{n\to\infty }b_n=\lim_{n\to\infty }a_n=x.\]
同时也可知，\[至少存一个无穷子列\{a_{n_k}\}(或\{b_{n_k}\})趋于x,x即为聚点.\]

证明 由法平面与积分元的关系，有\[ \cos \alpha dS=dydz,\cos \beta dS=dzdx,\cos \gamma dS=dxdy.\]\[\begin{align*}\therefore I
&=\frac{1}{3}\iint\limits_{S} (x\cos \alpha+y\cos \beta +z\cos \gamma )dS\\
&=\frac{1}{3}\iint\limits_{S} xdydz+ydzdx+zdxdy\\
&=\frac{1}{3}\iiint\limits_{\Omega }(1+1+1)dV \\
&=\varDelta V.
\end{align*}\]

解\[\begin{align*}(x^3-x^2+\frac{x}{2})e^{\frac{1}{x}}
&=(x^3-x^2+\frac{x}{2})(1+\frac{1}{x}+\frac{1}{2}\frac{1}{x^2}+\frac{1}{6}\frac{1}{x^3}+o(\frac{1}{x^3})) \\
&=(x^3-x^2+\frac{x}{2})+(x^2-x+\frac{1}{2})+\frac{1}{2}(x-1+\frac{1}{2x})+\frac{1}{6}(1-\frac{1}{x}+\frac{1}{2x^2}) \\
&=x^3+\frac{1}{6}+\frac{1}{12x}+\frac{1}{12x^2}.
\end{align*}\]\[\sqrt{1+x^6}=x^3\sqrt{1+x^{-6}}=x^3(1+\frac{1}{2}x^{-6})=x^3+\frac{1}{2}x^{-3}.\]\[\lim_{x\to\infty }[(x^3-x^2+\frac{x}{2})e^{\frac{1}{x}}-\sqrt{1+x^6}]=\lim_{x\to\infty }[x^3+\frac{1}{6}+\frac{1}{12x}+\frac{1}{12x^2}-(x^3+\frac{1}{2}x^{-3})]=\frac{1}{6}.\]

解\[偶延拓，b_n=0.a_0=\frac{2}{\pi }\int_{0}^{\pi }\sin xdx=\frac{4}{\pi }.\]\[\begin{align*}a_n
&=\frac{2}{\pi }\int_{0}^{\pi }\sin x\cos nxdx \\
&=\frac{2}{\pi }\int_{0}^{\pi }[\sin (n+1)x+\sin (1- n)x]dx \\
&=\frac{1}{\pi } [\frac{1}{1+n}(1-\cos (n+1)\pi )+\frac{1}{1-n}(1-\cos (1-n)\pi )].
\end{align*}\]注意到\[n=2k+1,a_n=0.\]\[n=2k,a_n=\frac{4}{\pi }\frac{1}{1-4k^2}.\]所以\[f(x)=\sin x\backsim \frac{a_0}{2}+\displaystyle\sum_{n=1}^{\infty }a_n\cos nx=\frac{2}{\pi }+\frac{4}{\pi }\displaystyle\sum_{k=1}^{\infty }\frac{1}{1-4k^2}\cos 2kx.\]

解 变量变换，令\[u=x+y,v=y.0\le v\le u,0\le u\le 1.\]\[\frac{\partial(u,v)}{\partial (x,y)}=\begin{vmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
\end{vmatrix}=\begin{vmatrix}
1 & 1\\
0 & 1
\end{vmatrix}=1.\]\[I=\iint\limits_{D}e^{\frac{y}{x+y}}dxdy=\int_{0}^{1}du\int_{0}^{u}e^{\frac{v}{u}}dv=\int_{0}^{1}udu\int_{0}^{u}e^{\frac{v}{u}}d(\frac{v}{u})=\int_{0}^{1}u[e-1]du=\frac{1}{2}(e-1).\]

解\[\begin{align*}I
&=\oint_{L}\frac{(x+y)dx-(x-y)dy}{x^2+y^2} \\
&=\frac{1}{a^2}\oint_{L}(x+y)dx-(x-y)dy \\
&= \frac{1}{a^2}\iint_{D}(-1-1)dxdy,(格林公式)\\
&=-2\pi .
\end{align*}\]

解\[\begin{align*}I
&=\iint\limits_{S}xdydz+ydzdx+zdxdy \\
&=\iiint\limits_{\Omega }(1+1+1)dV \\
&=3\cdotp \frac{4}{3}\pi a^3 \\
&=4 \pi a^3.
\end{align*}\]

证明\[\because \lim_{n\to\infty }a_n=a.\]\[\therefore \forall \varepsilon > 0,\exists N_1\in \mathbb{N},n> N_1,s.t.a-\varepsilon < a_n< a+\varepsilon .\]同理，\[\forall \varepsilon > 0,\exists N_2\in \mathbb{N},n> N_2,s.t.b-\varepsilon < b_n< b+\varepsilon .\]令\[N=\max \{N_1.N_2\},\]由\c_n=\frac{a_1b_N+a_2b_{N-1}+\cdots +a_Nb_1}{n}+\frac{a_{N+1}b_{Nn}+\cdots +a_{n}b_{N+1}}{n},\]而\[\frac{a_1b_N+a_2b_{N-1}+\cdots +a_Nb_1}{n}+\frac{n(a-\varepsilon )(b-\varepsilon )}{n}< c_n< \frac{a_1b_N+a_2b_{N-1}+\cdots +a_Nb_1}{n}+\frac{n(a+\varepsilon )(b+\varepsilon )}{n}.\]\[\Rightarrow \lim_{n\to\infty }c_n=ab.\]

证明 由泰勒公式：\[f(x+h)=f(x)+f'(x)h+\frac{1}{2}f''(\xi )h^2,\xi \in (x,x+h)\]\[f(x-h)f(x)-f'(x)h+\frac{1}{2}f''(\eta )h^2,\eta \in (x-h,x)\]两式相减，得\[f(x+h)-f(x-h)=2M_1h+\frac{1}{2}f''(\xi )h^2-\frac{1}{2}f''(\eta )h^2,\]\[|2f'(x)h|\le |f(x+h)-f(x-h)|+\frac{1}{2}|f''(\xi )-f''(\eta )|h^2\le 2M_0+M_2h^2.\]\[M_1\le \frac{2M_0}{2h}+\frac{M_2h^2}{2h}=\frac{M_0}{h}+\frac{M_2h}{2},\]令\[h=2\sqrt{\frac{M_0}{M_2}},\]即有\[M_1^2\le 2M_0M_2.\]

证明  $ f(x)在(a,b)上一致收敛\Rightarrow 。$\[\forall \varepsilon > 0,\exists n> \mathrm{N}\in \mathbb{N},p> 0\delta > 0,0< |x_{n+p}-x_n|< \delta ,\]\[|f(x_{n+p})-f(x_{n})|< \varepsilon .(\{f(x_n)\}\subseteq \{f(x)\})\]即函数值列$\{f(x_n)\}也是cauchy列.$
       $ f(x)在(a,b)上一致收敛\Leftarrow 。$\[\forall \varepsilon > 0,\forall x_n.x_m\in (a,b),\exists n,m> \mathrm{N_1}\in \mathbb{N},\delta > 0,\]\[\forall x_n,x_m\in (a,b),0< |x_n-x_m|< \delta ,s.t.\]\[|f(x_n)-f(x_m)|< \varepsilon .\]
       $对于点x_n,x_m的邻域\cup ^0(x_n,\frac{\delta }{2}),\cup ^0(x_m,\frac{\delta }{2}),分别存在收敛于x_n和x_m的子列\{x_{n_k}\} 和\{x_{m_l}\},$\[\exists N_2\in \mathbb{N},n_k,m_l> N_2,|x_n-x_{n_k}|<\frac{\delta }{3} ,|x_m-x_{m_l}|<\frac{\delta }{3},|x_{n_k}-x_{m_l}|<\frac{\delta }{3} .\]\[由于\{f(x_n)\}为cauchy列，故必有|f(x_{n_k})-f(x_{m_l})|< \frac{\varepsilon }{3}.\]\[而子列\{x_{n_k}\} 和\{x_{m_l}\}分别收敛于x_n和x_m，\]\[所以有|f(x_n)-f(x_{n_k})|< \frac{\varepsilon }{3}，|f(x_m)-f(x_{m_l})|< \frac{\varepsilon }{3}.\]
        $令N=\max \{N_1,N_2\}.f(x_n),f(x_m)\in \{f(x_n)\}此时$\[|f(x_n)-f(x_m)|< |f(x_n)-f(x_{n_k})|+|f(x_{n_k})-f(x_{m_l})|+|f(x_m)-f(x_{m_l})|< \frac{\varepsilon }{3}+\frac{\varepsilon }{3}+\frac{\varepsilon }{3}=\varepsilon .\]也即\[f(x)在(a,b)上存在cauchy列，f(x)在(a,b)上一致收敛.\]

证明  令\[F(x)=\frac{1}{2}[\int_{0}^{x}f(t)dt]^2.\]则\[F'(x)=\varphi (x)=f(x)\int_{0}^{x}f(t)dt.\]显然有\[F'(0)=\varphi (0)=0.\]又\[\because \varphi (x)\downarrow ，\]\[\therefore x\in (-\infty ,+\infty ）,0\le F'(x)\le F'(0)=0.\]从而必有\[\int_{0}^{x}f(t)dt=0，\Rightarrow f(x)\equiv 0.(可以证明)\]

解  \[\because |\displaystyle\sum_{n=1}^{\infty }\frac{\sgn(x-x_n)}{2^n}|\le \displaystyle\sum_{n=1}^{\infty }|\frac{\sgn(x-x_n)}{2^n}|\le \displaystyle\sum_{n=1}^{\infty }\frac{1}{2^n}< \infty .\]\[f(x)在（0，1）上一致收敛.\]令\[u_n(x)=\frac{\sgn(x-x_n)}{2^n}.\]\[\because \lim_{x\to x_n^+}=\frac{|x-x_n|}{2^n}=0=\lim_{x\to x_n^-}=-\frac{|x-x_n|}{2^n}.\]\[\therefore u_n(x)在x=x_n时，连续.\]因此\[f(x)\in C(-\infty,+\infty).\]

解（1）、\[\because I(r)=\int_{0}^{+\infty }re^{-rx}dx=e^{-rx}|_0^{+\infty }=-1.收敛.\]\[\lim_{x\to+\infty }e^{-rx}=0.\]\[\limsup_{x\to+\infty } |e^{-ax}-0|=0.\]\[\therefore I(r)在[a,b]上一致收敛.（0<a<b）\]
   （2）、由于\[r\in [0,b],\because I(0)=\int_{0}^{+\infty }re^{-rx}dx=\int_{0}^{+\infty }e^{-rx}d(rx)=\infty.\]\[\therefore I(r)在[0,b]上非一致收敛.\]

解（1）、Abel公式：\[\displaystyle\sum_{k=1}^{n}a_nb_n=\displaystyle\sum_{k=1}^{n-1}A_k(b_k-b_{lk+1})+A_nb_n.\]其中\[A_k=\textstyle\sum_{k=1}^{n}a_n.\]
（2）、由\[\displaystyle\sum_{n=1}^{\infty }(b_{n+1}-b_n)\le |\displaystyle\sum_{n=1}^{\infty }(b_{n+1}-b_n)|< \infty .\]所以\[\displaystyle\sum_{n=1}^{\infty }(b_{n+1}-b_n)=\lim_{n\to\infty }\displaystyle\sum_{k=1}^{n }(b_{n+1}-b_n)=\lim_{n\to\infty }(b_{n+1}-b_1)=\displaystyle\sum_{n=1}^{\infty }b_{n+1}-b_1< \infty .\]
  （3）、设\[\displaystyle\sum_{n=1}^{\infty }a_n=A,\displaystyle\sum_{n=1}^{\infty }b_n=B.\]\[\begin{align*}\therefore  \lim_{n\to\infty }\displaystyle\sum_{k=1}^{n}a_nb_n &=\lim_{n\to\infty }[\displaystyle\sum_{k=1}^{n-1}A_k(b_k-b_{k+1})+A_nb_n \\
&\le M\lim_{n\to\infty }\displaystyle\sum_{k=1}^{n-1}(b_k-b_{k+1})+Ac,(A_k\le M有界,c=\lim_{n\to\infty }b_n有界) \\
&\le Mb_1-Mc+Ac \\
&< \infty .
\end{align*}\]

证明
           致密性定理也称威尔斯特拉斯(Weierstrass)聚点定理：有界数列必含有收敛子列。
           由条件知：\[R_f=\{f(x)|x\in [a,b]\}有界,f(x)连续.\]\[\because R_f=\{f(x)|x\in [a,b]\}有界,有上下确界.\]记\[\alpha =\inf R_f，\beta =\sup R_f.\]上下确界也是$R_f$的最大、小值.
           由上下确界的定义\[\forall x\in [a,b],f(x)\geqslant \alpha .\forall \varepsilon > 0,\exists x\in [a,b],f(x)< \alpha +\varepsilon .\] \[对于[a,b]存在有界\{x_n\}.由致密性定理，\{x_n\}一定存在收敛子列\{x_｛n_k｝\},\lim_{n_k\to\infty }x_{n_k}=c.\]\[对于x_{n_k}\in [a,b],\alpha \le f(x_{n_k})< \alpha +\varepsilon .\]\[取\varepsilon =\frac{1}{n_k}，则当n_k\to\infty 时，有f(c)=\alpha .即f(x)能取到最小值.\]\[同样的方法，可证f(x)能取到最大值.\]

解 对$x+y+z=e^z,$隐函数求偏导.\[1+\frac{\partial z}{\partial x}=e^z\frac{\partial z}{\partial x}.\frac{\partial z}{\partial x}=\frac{1}{e^z-1}.\]\[1+\frac{\partial z}{\partial y}=e^z\frac{\partial z}{\partial y}.\frac{\partial z}{\partial y}=\frac{1}{e^z-1}.\]\[\frac{\partial^2z}{\partial x \partial y}=e^z\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}+e^z\frac{\partial^2z}{\partial x \partial y}.\]\[\frac{\partial^2z}{\partial x \partial y}=\frac{e^z}{1-e^z}\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}=\frac{e^z}{(1-e^z)^3}.\]

解 \[\lim_{n\to\infty }\displaystyle\sum_{k=1}^{n}\frac{n}{n^2+k^2}\sin \frac{k}{n}=\lim_{n\to\infty }\frac{1}{n}\displaystyle\sum_{k=1}^{n}\frac{1}{1+(\frac{k}{n})^2}\sin \frac{k}{n}.\]\[\because \frac{1}{1+\frac{1}{n}}< \frac{1}{1+(\frac{k}{n})^2}< \frac{1}{1+\frac{1}{n^2}}.\]\[\therefore \lim_{n\to\infty }\frac{1}{n}\displaystyle\sum_{k=1}^{n}\frac{1}{1+(\frac{k}{n})^2}\sin \frac{k}{n}=\lim_{n\to\infty }\frac{1}{n}\displaystyle\sum_{k=1}^{n}\sin \frac{k}{n}=\int_{0}^{1}\sin xdx=1-\cos 1.\]