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已知函数 ${\displaystyle f(x)={\displaystyle \frac{{\mathrm{e}}^x}{x+1},g(x)=\frac{{\mathrm{e}}^x+{\mathrm{e}}^{-x}}{2},h(x)={\mathrm{e}}^{\frac{x^2}{2}}}}$.

1. 求函数 ${\displaystyle f(x)}$ 在 ${\displaystyle x=1}$ 处的切线方程;
2. 当 ${\displaystyle x>0}$ 时, 试比较 ${\displaystyle f(x),g(x),h(x)}$ 的大小关系，并说明理由;
3. 设 ${\displaystyle n\in {\mathbb{N}}^{\ast }}$, 求证: $${\displaystyle 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots +\frac{1}{2n-1}-\frac{1}{2n}<\ln 2<1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots +\frac{1}{2n-1}.}$$

1. 解 ${\displaystyle {\displaystyle f^{\prime }(x)=\frac{x{\mathrm{e}}^x}{(x+1{)}^2}}}$，故 ${\displaystyle f^{\prime }(1)={\displaystyle \frac{\mathrm{e}}{4}}}$，故所求切线方程为 ${\displaystyle {\displaystyle y-\frac{\mathrm{e}}{2}=\frac{\mathrm{e}}{4}(x-1)\Rightarrow \mathrm{e}x-4y+\mathrm{e}=0}}$.

2. 解 当 ${\displaystyle x>0}$, 首先${\displaystyle {\displaystyle f(x)-g(x)=\frac{(1-x){\mathrm{e}}^x-(x+1){\mathrm{e}}^{-x}}{2(x+1)}}}$,
   记右侧分子为${\displaystyle u_1(x)}$，则${\displaystyle u_1^{\prime }(x)=-x({\mathrm{e}}^x+{\mathrm{e}}^{-x})<0,u_1(x)<u_1(0)=0}$,故${\displaystyle f(x)<g(x)}$.

   其次，${\displaystyle {\displaystyle \frac{g(x)}{h(x)}=\frac{1}{2}({\mathrm{e}}^{x-\frac{x^2}{2}}+{\mathrm{e}}^{-x-\frac{x^2}{2}})}}$, 记右侧函数为 ${\displaystyle u_2(x)}$，则 ${\displaystyle {\displaystyle u_2^{\prime }(x)=\frac{1}{2}{\mathrm{e}}^{-\frac{x^2}{2}}{u}_1(x)<0,u_2(x)<u_2(0)=1}}$, 故 ${\displaystyle g(x)<h(x)}$.

   综上，${\displaystyle f(x)<g(x)<h(x)}$.

3. 证明 注意到左式$$\begin{array}{r}\sum _{i=1}^{2n}\frac{(-1{)}^{i-1}}{i}=\sum _{i=1}^{2n}\frac{1}{i}-2\sum _{i=1}^n\frac{1}{2i}=\sum _{i=n+1}^{2n}\frac{1}{i}=\sum _{i=n+1}^{2n}(1-\frac{i-1}{i})<\sum _{i=n+1}^n\ln \frac{i}{i-1}=\ln 2,\end{array}$$
   利用了熟知的不等式 ${\displaystyle {\displaystyle \ln x\ge 1-\frac{1}{x}}}$. 而借助上面的结论可得右式为$$\begin{array}{r}\sum _{i=1}^{2n-1}\frac{(-1{)}^{i-1}}{i}=\sum _{i=n+1}^{2n}\frac{1}{i}+\frac{1}{2n}=\sum _{i=n}^{2n-1}\frac{1}{i}=\sum _{i=n}^{2n-1}(\frac{i+1}{i}-1)>\sum _{i=n}^{2n-1}\ln \frac{i+1}{n}=\ln 2,\end{array}$$
   利用了熟知的 ${\displaystyle \ln x\le x-1}$. 这样整个不等式证明完毕.