逆神的数学分析题答案总算补全了

9月5号逆神在数学竞赛交流群里给了一份试题，建模结束后自己才真正仔细思考起来，经过各位大神的指教，终于能够把所有试题的答案给补全，难免存在错误，联系2609480070@qq.cm进行纠正.

数学分析练习题

来源：逆蝶

整理：1729

2014年9月5日

下面的习题均来自大学生数学竞赛群群友逆蝶提供的数学分析练习题，难免存在错误，请联系（2609480070）进行纠正！

1.已知$a_1=a_2=1,a_{n+2}=2a_{n+1}+3a_n,n=1,2,\ldots$，求幂级数$\sum\limits_{n=1}^\infty{a_n x^n}$的收敛半径，收敛域以及和函数.

解.\begin{align*}&{a_{n + 2}} + {a_{n + 1}} = 3\left( {{a_{n + 1}} + {a_n}} \right) \\\Rightarrow &{a_{n + 1}} + {a_n} = \left( {{a_2} + {a_1}} \right) \times {3^{n - 1}} = 2 \times {3^{n - 1}}\\&{a_{n + 1}} - \frac{1}{2} \times {3^n} = \left( { - 1} \right)\left( {{a_n} - \frac{1}{2} \times {3^{n - 1}}} \right) \\\Rightarrow &{a_n} - \frac{1}{2} \times {3^{n - 1}} = \frac{1}{2}{\left( { - 1} \right)^{n - 1}}.\end{align*}

即

\begin{align*}\sum\limits_{n = 1}^\infty {{a_n}{x^n}} &= \frac{1}{2}\sum\limits_{n = 1}^\infty {\left( {{3^{n - 1}} + {{\left( { - 1} \right)}^{n - 1}}} \right){x^n}} \\&= \frac{1}{6}\sum\limits_{n = 1}^\infty {{{\left( {3x} \right)}^n}} - \frac{1}{2}\sum\limits_{n = 1}^\infty {{{\left( { - x} \right)}^n}} \\&= \frac{1}{2}\frac{x}{{1 - 3x}} + \frac{1}{2}\frac{x}{{1 + x}} = \frac{{x\left( {1 - x} \right)}}{{\left( {1 + x} \right)\left( {1 - 3x} \right)}}.\end{align*}

由\[\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{3^n} + {{\left( { - 1} \right)}^n}}}{{{3^{n - 1}} + {{\left( { - 1} \right)}^{n - 1}}}} = 3\]

知幂级数收敛半径为$\frac13$，收敛域为$\left(-\frac13,\frac13\right)$.

2.计算级数\[\sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)}}\left( {1 + \frac{1}{2} + \cdots + \frac{1}{n}} \right)} \]的和.

解.由Stolz公式

\[\mathop {\lim }\limits_{n \to \infty } \frac{{1 + \frac{1}{2} + \cdots + \frac{1}{{n + 1}}}}{{n + 1}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + 2}} = 0.\]

我们有

\begin{align*}&\sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)}}\left( {1 + \frac{1}{2} + \cdots + \frac{1}{n}} \right)} \\&= \sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)}}\left[ {\left( {n + 1} \right)\left( {1 + \frac{1}{2} + \cdots + \frac{1}{n}} \right) - n\left( {1 + \frac{1}{2} + \cdots + \frac{1}{n} + \frac{1}{{n + 1}}} \right) + \frac{n}{{n + 1}}} \right]} \\&= \sum\limits_{n = 1}^\infty {\left( {\frac{{1 + \frac{1}{2} + \cdots + \frac{1}{n}}}{n} - \frac{{1 + \frac{1}{2} + \cdots + \frac{1}{{n + 1}}}}{{n + 1}}} \right)} + \sum\limits_{n = 1}^\infty {\frac{1}{{{{\left( {n + 1} \right)}^2}}}} \\&= 1 - \mathop {\lim }\limits_{n \to \infty } \frac{{1 + \frac{1}{2} + \cdots + \frac{1}{{n + 1}}}}{{n + 1}} + \frac{{{\pi ^2}}}{6} - 1 = \frac{{{\pi ^2}}}{6}.\end{align*}

3.设$\alpha$是实数，计算\[\int_0^\infty {\frac{{dx}}{{\left( {1 + {x^2}} \right)\left( {1 + {x^\alpha }} \right)}}} .\]

解.令$x=\tan t$，我们知

\begin{align*}\int_0^\infty {\frac{{dx}}{{\left( {1 + {x^2}} \right)\left( {1 + {x^\alpha }} \right)}}} &= \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^\alpha }t}}{{{{\sin }^\alpha }t + {{\cos }^\alpha }t}}dt} = \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^\alpha }t}}{{{{\sin }^\alpha }t + {{\cos }^\alpha }t}}dt} \\&= \frac{1}{2}\left( {\int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^\alpha }t}}{{{{\sin }^\alpha }t + {{\cos }^\alpha }t}}dt} + \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^\alpha }t}}{{{{\sin }^\alpha }t + {{\cos }^\alpha }t}}dt} } \right)\\&= \frac{\pi }{4}.\end{align*}

4.设$f(x)$在$[0,\pi]$连续，求证：不能同时有\[\int_0^\pi {{{\left| {f\left( x \right) - \sin x} \right|}^2}dx} < \frac{\pi }{4},\int_0^\pi {{{\left| {f\left( x \right) - \cos x} \right|}^2}dx} < \frac{\pi }{4}.\]又问何时上面的两个不等式成为等式？

证.注意到

\[a^2+b^2\geq \frac{(a-b)^2}{2}.\]

我们有

\[\int_0^\pi {{{\left| {f\left( x \right) - \sin x} \right|}^2}dx} + \int_0^\pi {{{\left| {f\left( x \right) - \cos x} \right|}^2}dx} \ge \int_0^\pi {\frac{{{{\left( {\sin x - \cos x} \right)}^2}}}{2}dx} = \frac{\pi }{2}.\]

由抽屉原理知题给两式不能同时成立.由取等条件知当且仅当

\[f\left( x \right) - \sin x = \cos x - f\left( x \right) \Rightarrow f\left( x \right) = \frac{{\sin x + \cos x}}{2}\]

时取等成立.

5.设$f(x)$在$[0,\infty]$上有$n+1$阶连续导函数，且$f(0)\geq 0,f'(0)\geq0,\ldots,f^{(n)}(0)\geq0.$又对任意$x>0$，有$f(x)\leq f^{(n+1)}(x)$.求证：$f(x)\geq0$.

证.1. （gg）若$f(x)$在$x=0$的某个领域$(0,\xi)$内，满足$f(x)>0$.不妨设存在某个$x>0$，有$f(x)<0$，则此时由连续性，存在某个$x_1>0$，使得$f(x_1)=0$.当$x\in(0,x_1)$时，有$f(x)>0$，则$f^{n+1}(x)\geq f(x)>0$.易推得$f(x)$在$(0,x_1)$上为增函数，$f(x_1)>0$，故此时假设不成立；

2. 若$f(x)$在$x=0$的某个邻域$(0,\xi)$内，满足$f(x)<0$，下证矛盾.构造$g\left( x \right) = \sum\limits_{k = 0}^n {{f^{\left( k \right)}}\left( x \right)} /{e^x}$，则$g(x)$为增函数，所以$g\left( x \right) \ge g\left( 0 \right) \ge 0 \Rightarrow \sum\limits_{k = 0}^n {{f^{\left( k \right)}}\left( x \right)} \ge 0$.由于$f(x)$在$x=0$的某个邻域$(0,\xi)$内，满足$f(x)<0$，则必存在某个$\xi_1$，使得当$x\in(0,\xi_1)$时，有$f(x)<0$.对$k=1,2,\ldots,n$，均存在$\xi_k>0$，使得当$x\in(0,\xi_k)$时，使得$f^{(k)}(x)<0$.取$\eta=\min\{\xi_1,\xi_2,\ldots,\xi_n\}$.当$x\in(0,\eta)$时，有$\sum\limits_{k = 0}^n {{f^{\left( k \right)}}\left( x \right)}<0$，矛盾；

3. 若$f(x)$在$x=0$的某个邻域$(0,\xi)$内，满足$f(x)=0$，且$f(x)$不恒为0，易知此时可推得$(0,\xi)$内，有$f^{(k)}(x)=0,k=1,2,3,\ldots,n$.可转化为在$x=\xi$为初始点的情况，这时我们可采用类似1.,2.的讨论；

4. 若$f(x)=0$，则命题得证.

综上，命题成立.

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（逆蝶）对于$x\in(0,1]$，由Taylor公式，存在$x_1\in(0,1)$使\[f\left( x \right) = f\left( 0 \right) + f'\left( 0 \right)x + \cdots + \frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n} + \frac{{{f^{\left( {n + 1} \right)}}\left( {{x_1}} \right)}}{{\left( {n + 1} \right)!}}{x^{n + 1}}.\]根据条件得\[f\left( x \right) \ge f\left( {{x_1}} \right)\frac{{{x^{n + 1}}}}{{\left( {n + 1} \right)!}}.\]同样将$f(x_1)$展开，可得$x_2\in(0,x_1)$使得\[f\left( {{x_1}} \right) \ge f\left( {{x_2}} \right)\frac{{x_1^{n + 1}}}{{\left( {n + 1} \right)!}}.\]继续这个过程，可得$(0,x)$中严格递减序列$\{x_k\}$使得\[f\left( {{x_k}} \right) \ge f\left( {{x_{k + 1}}} \right)\frac{{x_k^{n + 1}}}{{\left( {n + 1} \right)!}}.\]于是\[f\left( x \right) \ge f\left( {{x_{k + 1}}} \right)\frac{{{x^{n + 1}}}}{{\left( {n + 1} \right)!}}\frac{{x_1^{n + 1}}}{{\left( {n + 1} \right)!}} \cdots \frac{{x_k^{n + 1}}}{{\left( {n + 1} \right)!}}.\]因为$x$及$x_k$都在$[0,1]$中，上式右端当$k\to+\infty$时趋于0，于是对于$x\in[0,1]$有$f(x)\geq0$.由此\[f'\left( x \right) = f'\left( 0 \right) + f''\left( 0 \right)x + \cdots + \frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{\left( {n - 1} \right)!}}{x^{n - 1}} + \frac{{{f^{\left( {n + 1} \right)}}\left( \xi \right)}}{{n!}}{x^n} \ge \frac{{f\left( \xi \right)}}{{n!}}{x^n} \ge 0,\]其中$\xi\in(0,x)$.归纳可证$f^{(k)}\geq0,x\in[0,1],k=1,2,\ldots,n+1$.对函数$g(x)=f(x+1)$重复以上过程可知$f(x)\geq0,x\in[1,2]$.用归纳法可证对任意自然数$m$，$f(x)$在$[m,m+1]$上非负.于是结论得证.

6. 设$f(x)$是$[0,+\infty)$上连续函数，满足$0<f(x)<1$，而且无穷积分在$\int_0^{+\infty}{f(x)\, dx}$和$\int_0^{+\infty}{xf(x)\, dx}$都收敛.求证：\[\int_0^{ + \infty } {xf\left( x \right)\, dx} > \frac{1}{2}{\left( {\int_0^{ + \infty } {f\left( x \right)\, dx} } \right)^2}.\]

证.令\[g\left( y \right) = \int_0^y {xf\left( x \right)dx} - \frac{1}{2}{\left( {\int_0^y {f\left( x \right)dx} } \right)^2},y>0.\]

我们得到\[g'\left( y \right) = yf\left( y \right) - f\left( y \right)\int_0^y {f\left( x \right)dx} = f\left( y \right)\left( {y - \int_0^y {f\left( x \right)dx} } \right).\]

又$0<f(x)<1$，我们得到\[0 < \int_0^y {f\left( x \right)dx} < \int_0^y {dx} = y.\]因此我们有$g'(y)>0,g(y)>g(0)=0.$再令$y\to+\infty$即可.

7. 设$0<\alpha\leq1,\beta>0,\alpha+\beta>1$，$f(x)$是$[1,+\infty)$的正函数，且$\int_1^{+\infty}{f(x)\, dx}$收敛.求证：$\int_1^{+\infty}{\frac{{(f(x))}^\alpha}{x^\beta}\, dx}$收敛.

证.当$\alpha=1$时，由\[0 < \int_1^{ + \infty } {\frac{{f\left( x \right)}}{{{x^\beta }}}dx} \le \int_1^{ + \infty } {f\left( x \right)dx} .\]可知积分收敛.

当$0<\alpha<1$时.

（Holder积分不等式）若函数$f(x)$与$g(x)$在区间$[a,b]$上连续非负，且$p>1,\frac1p+\frac1q=1$，则有不等式\[\int_a^b {f\left( x \right)g\left( x \right)dx} \le {\left( {\int_a^b {{{\left[ {f\left( x \right)} \right]}^p}dx} } \right)^{\frac{1}{p}}}{\left( {\int_a^b {{{\left[ {g\left( x \right)} \right]}^q}dx} } \right)^{\frac{1}{q}}}.\]

取$p=\frac1\alpha,q=\frac1{1-\alpha},f(x)={[f(x)]}^\alpha,g(x)=\left(\frac1x\right)^\beta$，对于任给的正数$A$，我们有

\[0 < \int_1^A {\frac{{{{\left( {f\left( x \right)} \right)}^\alpha }}}{{{x^\beta }}}dx} \le {\left( {\int_1^A {f\left( x \right)dx} } \right)^\alpha }{\left( {\int_1^A {\frac{1}{{{x^{\frac{\beta }{{1 - \alpha }}}}}}dx} } \right)^{1 - \alpha }} \le {\left( {\int_1^{ + \infty } {f\left( x \right)dx} } \right)^\alpha }{\left( {\int_1^{ + \infty } {\frac{1}{{{x^{\frac{\beta }{{1 - \alpha }}}}}}dx} } \right)^{1 - \alpha }}.\]

再注意到$\int_1^{ + \infty } {f\left( x \right)dx} ,\int_1^{ + \infty } {\frac{1}{{{x^{\frac{\beta }{{1 - \alpha }}}}}}dx} \left( {\frac{\beta }{{1 - \alpha }} > 1} \right)$均收敛即可得证.

8. 设$\{a_n\}$是正的递增数列.求证：级数$\sum\limits_{n=1}^\infty \left(\frac{a_{n+1}}{a_n}-1\right)$收敛的充分必要条件是$\{a_n\}$有界.

证.1. ($\Rightarrow$)注意到\[\ln(1+x)<x,x>0.\]我们有\[A = \sum\limits_{n = 1}^\infty {\left( {\frac{{{a_{n + 1}}}}{{{a_n}}} - 1} \right)} > \sum\limits_{n = 1}^\infty {\ln \frac{{{a_{n + 1}}}}{{{a_n}}}} = \mathop {\lim }\limits_{n \to \infty } \ln \frac{{{a_n}}}{{{a_1}}}.\]故\[\mathop {\lim }\limits_{n \to \infty } {a_n} = B \le {a_1}{e^A}.\]

2. ($\Leftarrow$)又\[\sum\limits_{n = 1}^\infty {\left( {\frac{{{a_{n + 1}}}}{{{a_n}}} - 1} \right)} = \sum\limits_{n = 1}^\infty {\frac{{{a_{n + 1}} - {a_n}}}{{{a_n}}}} < \sum\limits_{n = 1}^\infty {\frac{{{a_{n + 1}} - {a_n}}}{{{a_1}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_n} - {a_1}}}{{{a_1}}} = \frac{1}{{{a_1}}}\mathop {\lim }\limits_{n \to \infty } {a_n} - 1.\]故$\sum\limits_{n=1}^\infty \left(\frac{a_{n+1}}{a_n}-1\right)$收敛.

9. 设$\alpha>0,\{a_n\}$是递增正数列.求证：级数$\sum\limits_{n=1}^\infty \frac{a_{n+1}-a_n}{a_{n+1}a_n^\alpha}$收敛.

证.1. 当$0<\alpha<1$时，有\[\sum\limits_{n = 1}^\infty {\frac{{{a_{n + 1}} - {a_n}}}{{{a_{n + 1}}a_n^\alpha }}} \le \sum\limits_{n = 1}^\infty {\frac{1}{\alpha }\left( {\frac{1}{{a_n^\alpha }} - \frac{1}{{a_{n + 1}^\alpha }}} \right)} = \frac{1}{{\alpha a_1^\alpha }} - \frac{1}{\alpha }\mathop {\lim }\limits_{n \to \infty } \frac{1}{{a_n^\alpha }}.\]

事实上，由Lagrange中值定理，我们有\[\frac{{a_{n + 1}^\alpha - a_n^\alpha }}{{{a_{n + 1}} - {a_n}}} = \alpha {\xi ^{\alpha - 1}} > \alpha a_{n + 1}^{\alpha - 1},\]其中$\xi\in(a_n,a_{n+1})$.

因此，根据$a_n$单调递增的性质，对于其有界和无界两种情况，不等式右端的级数都是收敛的，由此得证.

2. 当$\alpha\geq1$时，又有

\begin{align*}\sum\limits_{n = 1}^\infty {\frac{{{a_{n + 1}} - {a_n}}}{{{a_{n + 1}}a_n^\alpha }}} &= \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{a_n^\alpha }} - \frac{{a_n^{1 - \alpha }}}{{{a_{n + 1}}}}} \right)} \le \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{a_n^\alpha }} - \frac{{a_{n + 1}^{1 - \alpha }}}{{{a_{n + 1}}}}} \right)} \\&= \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{a_n^\alpha }} - \frac{1}{{a_{n + 1}^\alpha }}} \right)} = \frac{1}{{a_1^\alpha }} - \mathop {\lim }\limits_{n \to \infty } \frac{1}{{a_n^\alpha }}.\end{align*}

同理，不等式右端的级数亦是收敛的，证毕.

10. 设$0<\alpha<1$，证明数列\[{a_n} = \frac{1}{{1 + {n^\alpha }}} + \frac{1}{{2 + {n^\alpha }}} + \cdots + \frac{1}{{n + {n^\alpha }}},n = 1,2, \cdots \]发散.

证.注意到\[x>\ln(x+1).\]我们有\[\frac{1}{{k + {n^\alpha }}} > \ln \frac{{k + 1 + {n^\alpha }}}{{k + {n^\alpha }}}.\]

因此\[\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{1}{{k + {n^\alpha }}}} > \mathop {\lim }\limits_{n \to \infty } \ln \frac{{n + 1 + {n^\alpha }}}{{1 + {n^\alpha }}} \to \infty .\]

所以此数列发散.

11. 计算\[\mathop {\lim }\limits_{n \to \infty } \left( {\frac{n}{{{1^2} + \sqrt n + {n^2}}} + \frac{n}{{{2^2} + 2\sqrt n + {n^2}}} + \cdots + \frac{n}{{{n^2} + n\sqrt n + {n^2}}}} \right).\]

解.注意到

\begin{align*}&\mathop {\lim }\limits_{n \to \infty } \left[ {\sum\limits_{k = 1}^n {\frac{n}{{{k^2} + {n^2}}}} - \sum\limits_{k = 1}^n {\frac{n}{{{k^2} + k\sqrt n + {n^2}}}} } \right] = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{{kn\sqrt n }}{{\left( {{k^2} + k\sqrt n + {n^2}} \right)\left( {{k^2} + {n^2}} \right)}}} \\&\le \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{{kn\sqrt n }}{{{n^2} \cdot {n^2}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{n\sqrt n }}{{{n^2} \cdot {n^2}}} \cdot \frac{{n\left( {n + 1} \right)}}{2} = 0.\end{align*}

因此

\[\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{n}{{{k^2} + k\sqrt n + {n^2}}}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{n}{{{k^2} + {n^2}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{{{\left( {\frac{k}{n}} \right)}^2} + 1}}} = \int_0^1 {\frac{1}{{{x^2} + 1}}dx} = \frac{\pi }{4}.\]

12. 设$f(x)$是$[0,2\pi]$上可导的凸函数，$f'(x)$有界.试证\[{a_n} = \frac{1}{\pi }\int_0^{2\pi } {f\left( x \right)\cos nx\, dx} \ge 0.\]

证.由分部积分，我们有

\[{a_n} = \frac{1}{\pi }\int_0^{2\pi } {f\left( x \right)\cos nxdx} = \frac{1}{\pi }\int_0^{2\pi } {f\left( x \right)d\frac{{\sin nx}}{n}} = - \frac{1}{{n\pi }}\int_0^{2\pi } {f'\left( x \right)\sin nxdx} .\]

又由第二积分中值定理，我们有

\begin{align*}\int_0^{2\pi } {f'\left( x \right)\sin nxdx} &= f'\left( {0 + } \right)\int_0^\xi {\sin nxdx} + f'\left( {2\pi - } \right)\int_\xi ^{2\pi } {\sin nxdx} \\&= \frac{{1 - \cos n\xi }}{n}\left[ {f'\left( {0 + } \right) - f'\left( {2\pi - } \right)} \right] \le 0.\end{align*}

故\[{a_n} = - \frac{1}{{n\pi }}\int_0^{2\pi } {f'\left( x \right)\sin nxdx} \ge 0.\]

13. 设$\{a_n\}$是正数列使得$\sum\limits_{n=1}^\infty{\frac{1}{a_n}}$收敛.求证\[\sum\limits_{n = 1}^\infty {\frac{n}{{{a_1} + {a_2} + \cdots + {a_n}}}} \le 2\sum\limits_{n = 1}^\infty {\frac{1}{{{a_n}}}} ,\]而且上式右端的系数2是最佳的.

证.由柯西不等式我们得

\[\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} \sum\limits_{m = 1}^n {{a_m}}  \ge {\left( {1 + 2 +  \cdots  + n} \right)^2} = \frac{1}{4}{n^2}{\left( {n + 1} \right)^2},\]

即\[\frac{n}{{{a_1} + {a_2} +  \cdots  + {a_n}}} \le \frac{4}{{n{{\left( {n + 1} \right)}^2}}}\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} .\]

因此

\begin{align*}\sum\limits_{n = 1}^\infty  {\frac{n}{{{a_1} + {a_2} +  \cdots  + {a_n}}}}  &\le 4\sum\limits_{n = 1}^\infty  {\frac{1}{{n{{\left( {n + 1} \right)}^2}}}\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} }  = 4\sum\limits_{m = 1}^\infty  {\frac{{{m^2}}}{{{a_m}}}\sum\limits_{n = m}^\infty  {\frac{1}{{n{{\left( {n + 1} \right)}^2}}}} } \\&\le 4\sum\limits_{m = 1}^\infty  {\frac{{{m^2}}}{{{a_m}}}\sum\limits_{n = m}^\infty  {\frac{1}{2}\left[ {\frac{1}{{{n^2}}} - \frac{1}{{{{\left( {n + 1} \right)}^2}}}} \right]} }  = 2\sum\limits_{m = 1}^\infty  {\frac{1}{{{a_m}}}} .\end{align*}

这里用到了\[\frac{1}{{n{{\left( {n + 1} \right)}^2}}} \le \frac{1}{2}\frac{{2n + 1}}{{{n^2}{{\left( {n + 1} \right)}^2}}} = \frac{1}{2}\left[ {\frac{1}{{{n^2}}} - \frac{1}{{n{{\left( {n + 1} \right)}^2}}}} \right].\]

注意到$a_n=n^\alpha,\alpha>1$时有

\[\mathop {\lim }\limits_{\alpha \to 1} \frac{{\sum\limits_{n = 1}^\infty {\frac{n}{{{a_1} + {a_2} + \cdots + {a_n}}}} }}{{\sum\limits_{j = 1}^\infty {\frac{1}{{{a_j}}}} }} = \mathop {\lim }\limits_{\alpha \to 1} \mathop {\lim }\limits_{N \to \infty } \frac{{\sum\limits_{n = 1}^N {\frac{n}{{{1^\alpha } + {2^\alpha } + \cdots + {n^\alpha }}}} }}{{\sum\limits_{n = 1}^N {\frac{1}{{{n^\alpha }}}} }} = \mathop {\lim }\limits_{\alpha \to 1} \mathop {\lim }\limits_{N \to \infty } \frac{{\frac{N}{{\frac{1}{{\alpha + 1}}{N^{\alpha + 1}} + O\left( {{N^\alpha }} \right)}}}}{{\frac{1}{{{N^\alpha }}}}} = 2.\]

14. 设$f(x)$是$[0,+\infty)$上正的连续函数，且$\int_0^{+\infty}{\frac{1}{f(x)}\, dx}$收敛.记$F(x)=\int_0^x{f(t)\, dt}$.求证\[\int_0^{ + \infty } {\frac{x}{{F\left( x \right)}}dx} < 2\int_0^{ + \infty } {\frac{1}{{f\left( x \right)}}dx} ,\]且上式右端的系数2是最佳的.

证.（陈洪葛）由$Cauchy-Schwarz$不等式，得到\[\left( {\int_0^x {f\left( t \right)dt} } \right) \cdot \left( {\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} } \right) \ge {\left( {\int_0^x {tdt} } \right)^2} = \frac{1}{4}{x^4}.\]

所以

\[\int_0^{ + \infty } {\frac{x}{{F\left( x \right)}}dx} \le \int_0^{ + \infty } {\frac{4}{{{x^3}}}\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} dx} .\]

注意到\[\mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{{{x^2}}}\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} = \mathop {\lim }\limits_{x \to 0} \frac{x}{{2f\left( x \right)}} = 0.\]

以及\[\frac{1}{{{x^2}}}\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} = \int_0^\xi {\frac{1}{{f\left( t \right)}}dt} < \int_0^{ + \infty } {\frac{1}{{f\left( t \right)}}dt} .\]

故\begin{align*}&\int_0^A {\frac{4}{{{x^3}}}\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} dx} = \int_0^A {\left( {\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} } \right)d\left( { - \frac{2}{{{x^2}}}} \right)} \\&= \int_0^A {\frac{2}{{f\left( x \right)}}dx} - \frac{2}{{{A^2}}}\int_0^A {\frac{{{t^2}}}{{f\left( t \right)}}dt} < \int_0^A {\frac{2}{{f\left( x \right)}}dx} - 2\int_\xi ^A {\frac{{{t^2}}}{{f\left( t \right)}}dt}.\end{align*}

令$A\to+\infty$，得到

\[\int_0^{ + \infty } {\frac{4}{{{x^3}}}\int_0^x {\frac{{{t^2}}}{{f\left( t \right)}}dt} dx} \le \int_0^{ + \infty } {\frac{2}{{f\left( x \right)}}dx} - 2\int_\xi ^{ + \infty } {\frac{{{t^2}}}{{f\left( t \right)}}dt} < \int_0^{ + \infty } {\frac{2}{{f\left( x \right)}}dx}.\]

另外，当我们取$f(x)=x^a+1(a>1)$时，有$\int_0^{ + \infty } {\frac{{dx}}{{{x^a} + 1}}} = \frac{\pi }{{a\sin \frac{\pi }{a}}}$收敛.此时有

\[\mathop {\lim }\limits_{a \to 1} \frac{{\int_0^{ + \infty } {\frac{x}{{F\left( x \right)}}dx} }}{{\int_0^{ + \infty } {\frac{1}{{f\left( x \right)}}dx} }} = \mathop {\lim }\limits_{a \to 1} \frac{{\int_0^{ + \infty } {\frac{x}{{{x^{a + 1}}/\left( {a + 1} \right) + x}}dx} }}{{\int_0^{ + \infty } {\frac{{dx}}{{{x^a} + 1}}} }} = \mathop {\lim }\limits_{a \to 1} \frac{{\int_0^{ + \infty } {\frac{1}{{{x^a}/\left( {a + 1} \right) + 1}}dx} }}{{\int_0^{ + \infty } {\frac{{dx}}{{{x^a} + 1}}} }} = \mathop {\lim }\limits_{a \to 1} {\left( {a + 1} \right)^{\frac{1}{a}}} = 2.\]

15. 设$f(x)$在$\mathbb{R}$上有二阶导函数，$f(x),f'(x),f''(x)$都大于零，假设存在正数$a,b$使得$f''(x)\leq af(x)+bf'(x)$对一切$x\in\mathbb{R}$成立.

1. 求证：$\mathop {\lim }\limits_{x \to - \infty } f'\left( x \right) = 0$;

2. 求证：存在常数$c$使得$f'(x)\leq cf(x)$;

3. 求使上面不等式成立的最小常数$c$.

证.1. 显然由单调有界定理知$\mathop {\lim }\limits_{x \to - \infty } f\left( x \right),\mathop {\lim }\limits_{x \to - \infty } f'\left( x \right)$均存在，不妨设$\mathop {\lim }\limits_{x \to - \infty } f\left( x \right)=c_1,\mathop {\lim }\limits_{x \to - \infty } f'\left( x \right)=c_2$.因此我们有$f(x)\geq c_1\geq0,f'(x)\geq c_2\geq0$.又\[f\left( 0 \right) = \int_x^0 {f'\left( x \right)dx} + f\left( x \right) \ge - {c_2}x + f\left( x \right) \Rightarrow f\left( x \right) \le {c_2}x + f\left( 0 \right).\]我们知$c_2=0$，否则令$x\to-\infty$，矛盾.

2. 令

\begin{align*}&g\left( x \right) = \left( {\frac{{b + \sqrt {4a + {b^2}} }}{2}f\left( x \right) - f'\left( x \right)} \right){e^{\frac{{ - b + \sqrt {4a + {b^2}} }}{2}x}},\\&g'\left( x \right) = \left( {af\left( x \right) + bf'\left( x \right) - f''\left( x \right)} \right){e^{\frac{{ - b + \sqrt {4a + {b^2}} }}{2}x}} \ge 0.\end{align*}

又$\lim\limits_{n\to-\infty} g(x)=0$.故$g(x)\geq g(-\infty)=0$.即\[f'\left( x \right) \le \frac{{b + \sqrt {4a + {b^2}} }}{2}f\left( x \right).\]故存在常数$c={\frac{{b + \sqrt {4a + {b^2}} }}{2}}$使得$f'(x)\leq cf(x)$成立.

3. 我觉得应该把“求使上面不等式成立的最小常数$c$”改成“求使上面不等式成立的最大常数$c$”.事实上，我们取：

\[\left\{ \begin{array}{l}h\left( x \right) = {e^{cx}}\\h'\left( x \right) = c{e^{cx}}\\h''\left( x \right) = {c^2}{e^{cx}}\end{array} \right.\left( {c > 0} \right).\]

只需保证\[c^2\leq bc+a\Rightarrow 0<c\leq {\frac{{b + \sqrt {4a + {b^2}} }}{2}}.\]

故常数$c$的最大值为${\frac{{b + \sqrt {4a + {b^2}} }}{2}}$.

16. 设$f(x)$是$\mathbb{R}$上有下界或者有上界的连续函数且存在正数$a$使得\[f\left( x \right) + a\int_{x - 1}^x {f\left( t \right)dt} \]为常数.求证：$f(x)$必为常数.

证.（Slade）由题意，$f(x)$必存在上下界，否则，在等式\[f\left( x \right) + a\int_{x - 1}^x {f\left( t \right)dt}=C\]两端同时取$x\to \infty$，等式左端无界，而右端为常数，矛盾.

再之，注意到$\int_{x-1}^x f(t)\,dt$是可微的，故$f$也是可微的.又\[f'\left( x \right) = a\left( {f\left( {x - 1} \right) - f\left( x \right)} \right) \Rightarrow \left| {f'\left( x \right)} \right| \le a\left( {\left| {f\left( {x - 1} \right)} \right| + \left| {f\left( x \right)} \right|} \right).\]由于$f$有界，故$f'$亦有界.记$N=[a]+1$，则$0<\frac aN<1$.令$x=\frac x N$，我们得到\[f'\left( {\frac{x}{N}} \right) = a\left( {f\left( {\frac{x}{N} - 1} \right) - f\left( {\frac{x}{N}} \right)} \right).\]

固定$x$，由Lagrange中值定理，存在${a_1} \in \left( {\frac{x}{N} - 1,\frac{x}{N}} \right)$使得\[f\left( {\frac{x}{N} - 1} \right) - f\left( {\frac{x}{N}} \right) = - \frac{1}{N}f'\left( {{a_1}} \right),\]即$f'\left( {\frac{x}{N}} \right) = - \frac{a}{N}f'\left( {{a_1}} \right)$.类似地，我们有数列$\{a_n\}$使得\[f'\left( {\frac{x}{N}} \right) = - \frac{a}{N}f'\left( {{a_1}} \right) = {\left( { - \frac{a}{N}} \right)^2}f'\left( {{a_2}} \right) = \cdots = {\left( { - \frac{a}{N}} \right)^n}f'\left( {{a_n}} \right).\]令$n\to\infty$，我们有$f'\left( {\frac{x}{N}} \right) = 0 $即$f'\left( x \right) = 0$.故$f(x)$必为常数.

17. 设$f:[0,+\infty)\to [0,+\infty)$且对任意$x\geq0$有$f\circ f(x)=af(x)+bx$，其中$a<0,b>0$.求$f(x).$

证.先证明几个引理：

引理1.设$f\in C^0(\mathbb{R},\mathbb{R})$是方程$f(f(x))=af(x)+bx$的解.若方程中的常数$b\neq0$，则$f:\mathbb{R}\to\mathbb{R}$即单射又是满射，即是一个一一映射.

引理2.设$f\in C^0(\mathbb{R},\mathbb{R})$是方程$f(f(x))=af(x)+bx$的解.若$b\neq0$，且$\lambda$的多项式$\lambda^2-a\lambda-b$的两个根$r$与$s$不相等，则对任意$x\in\mathbb{R}$及任意$n\in\mathbb{Z}$均有

\[{f^n}\left( x \right) = \frac{{{s^n}\left( {f\left( x \right) - rx} \right) + {r^n}\left( {sx - f\left( x \right)} \right)}}{{s - r}}.\]

事实上，当$n$为非正整数时下一式成立：\[{f^n}\left( x \right) = \frac{{{s^n}\left( {{f^{ - 1}}\left( x \right) - \frac{x}{r}} \right) + {r^n}\left( {\frac{x}{s} - {f^{ - 1}}\left( x \right)} \right)}}{{\frac{1}{s} - \frac{1}{r}}}.\]以$f(x)$代替$x$，以$n-1$代替$n$，可推知$n$为非正整数时亦成立.

引理3.设$g_n\in C^0(\mathbb{R},\mathbb{R})$，且${(-1)}^n g_n(x)$对$x$单调递增（$n=0,1,2,\ldots$）.若极限$g(x)=\lim\limits_{n\to\infty}g_n(x)$对任$x\in \mathbb{R}$均存在，则极限函数$g$是个常值函数.

事实上，因$g(x)=\lim\limits_{n\to\infty}g_{2n}(x)$，而$g_{2n}(x)$对$x$递增.又因$g(x)=\lim\limits_{n\to\infty}g_{2n+1}(x)$，而$\lim\limits_{n\to\infty}g_{2n+1}(x)$对$x$递减，故$g(x)$对$x$递减.于是$g$只能是个常值函数.

回到原题，我们有定理：设$r$及$s$是$\lambda$的二次多项式$\lambda^2-a\lambda-b=0$的两个根.若$r<0<s$，且$r\neq -s$，则$f\in C^0(\mathbb{R},\mathbb{R})$是方程$f(f(x))=af(x)+bx$的解的充分必要条件是$f(x)=rx$（对任$x\in\mathbb{R}$），或者$f(x)=sx$（对任$x\in\mathbb{R}$），或者，当$s=1$时还可以是$f(x)=rx+c$（对任$x\in\mathbb{R}$，$c$可以是任一个给定的实数）.

定理的充分性部分是显然的.下面证明定理的必要性部分，设$f$是方程$f(f(x))=af(x)+bx$的一个解，据引理1知$f$严格单调.若$f$严格递增，可推出

\[sx - f\left( x \right) = \left\{ \begin{array}{l}\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {s - r} \right){f^n}\left( x \right)}}{{{r^n}}},\text{若}\left| r \right| > s,\\\mathop {\lim }\limits_{n \to - \infty } \frac{{\left( {s - r} \right){f^n}\left( x \right)}}{{{r^n}}},\text{若}\left| r \right| < s.\end{array} \right.\]

因${(-1)}^n(s-r)f^n(x)/r^n$是$x$的递增函数（任意$n\in\mathbb{Z}$），由引理3可知极限函数$sx-f(x)$恒取常值，即存在某$c\in\mathbb{R}$使得$f(x)=sx+c$对任意$x\in\mathbb{R}$成立，把$f(x)$的这一表达式代入得到$s^2x+sc+c=(r+s)(sx+c)-rsx$，推出$c=0$.因此，当$f$递增时$f(x)=sx$（对任意$x\in\mathbb{R}$）.

若$f$严格递减，可推出\[f\left( x \right) - rx = \left\{ \begin{array}{l}\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {s - r} \right){f^n}\left( x \right)}}{{{s^n}}},s > \left| r \right|,\\\mathop {\lim }\limits_{n \to - \infty } \frac{{\left( {s - r} \right){f^n}\left( x \right)}}{{{s^n}}},s < \left| r \right|.\end{array} \right.\]

因${(-1)}^n(s-r)f^n(x)/s^n$是$x$的递增函数（任意$n\in\mathbb{Z}$），由引理3可同样推出$f(x)=rx+c$（对任意$x\in\mathbb{R}$）.将此表达式代入得到$r^2x+rc+c=(r+s)(rx+c)-rsx$，可化简为$(s-1)c=0$.因此，当$s\neq1$时$c=0$，当$s=1$时$c$可为任意给定的实数.

另外：注意到$f:[0,+\infty)\to [0,+\infty)$这一条件，我们知$f(x)=sx(s>0)$.可参阅：

[1]关于迭代函数方程$f^2(x)=af(x)+bx$的通解，麦结华，数学研究与评论第17卷第1期83-90页，1997年2月.

[2]J.Matkowski and Zhang Weinian, Method of characteristics for functional equations in polynomial

form, Acta Math.Sinica, New Series.

18. 设$f(x)$在$(-\infty,+\infty)$上连续，且对任意$x$有$f(2x-f(x))=x$.求证：$f(x)\equiv x+c$，其中$c$为常数.

证.（逆蝶）令$g(x)=2x-f(x)$，则$f[g(x)]=x$，显然$g$是单射且为增函数.若$g$有上界，则$\lim\limits_{x\to+\infty}{g(x)}=a$存在.又$f$连续，$\lim\limits_{x\to+\infty}{f[g(x)]}=f(a)$，矛盾.从而$g$无上界也无下界.由于$g$连续，$\forall x\in R,\exists y$使得$x=g(y)$.从而$f(x)=f[g(y)]=y$，于是$g[f(x)]=g(y)=x$.由$g$递增知$f$递增.

若$\exists x_1,x_2$使得$f(x_1)-x_1\neq f(x_2)-x_2$，则由于$f(x)-x$的连续性知其可以取遍$f(x_1)-x_1$及$f(x_2)-x_2$之间的任何数.

设$d,d'$在它们之间，且$d/d'>1$是正无理数，设$d=f(x)-x,d'=f(x')-x'$.令$x_0=x,x_{n+1}=2x_n-f(x_n)$，则$f(x_{n+1})=x_n$.归纳可得$x_n=x+nd$，同理$x'_n=x'+nd'$.

令$x_m>x'_n,x_{m+1}<x'_n+1$，即$x+md>x'+nd,x+d+md<x'+d'+nd'$.即$(x'-x)/d<(md')/d-n<(x'-x)/d+(d/d'-1)$.由$d/d'$为无理数知上式关于$m,n$有解，这与$f$递增矛盾.

下面是一个推广：对$\forall m\neq0$，若一个连续函数$f:\mathbb{R}\to\mathbb{R}$满足函数方程\[f\left(2x-\frac{f(x)}{m}\right)=mx,\]则有$f(x)=m(x-c)$.

证.（陈洪葛）我们设$g(x)=2x-\frac{f(x)}{m}$，显然$g(x)$是连续函数且有\[g(g(x))=2g(x)-x,\forall x\in\mathbb{R}.\]

若$g(x_1)=g(x_2)$，则有$g(g(x_1))=g(g(x_2))$，我们得到\[x_1=x_2.\]故$g(x)$是一个单射，而我们知道，若$g(x)$是一个连续的单射，则$g(x)$严格单调（关于这点可以用反证法证明）.因此，$g(x)$有2种情况，严格递增或者严格递减.下面证明$g(x)$只能严格递增.

（反证）设$g(x)$严格递减，则对于$x_1<x_2$，我们有$g(x_1)>g(x_2)$，接着又有$g(g(x_1))<g(g(x_2))$，二者等价于\[2g(x_1)-x_1<2g(x_2)-x_2\Leftrightarrow 2[g(x_1)-g(x_2)]<x_1-x_2.\]上面不可能成立，因为左边大于0而右边小于0，故$g(x)$只能严格递增.

改写$g(g(x))=2g(x)-x$为\[g(g(x))-g(x)=g(x)-x.\]递推后得到\[g^n(x)=ng(x)-(n-1)x,(n\geq1),\]这里$g^{(n)}(x)$表示$n$次复合.那么有\[g^n(x)-g^n(0)=n[g(x)-x-g(0)]+x\Leftrightarrow\frac{g^n(x)-g^n(0)}{n}=g(x)-x-g(0)+\frac{x}{n}.\]而$g(x)$严格递增，$g^n(x)$也严格递增，故对上式令$n\to\infty$，由$g(x)$的单调性，我们得到

\begin{align*}g(x)&\leq x+g(0),&&x<0\\g(x)&\geq x+g(0),&&x>0.\end{align*}

这样，我们得到$g(x)$的值域也是$\mathbb{R}$，故$g(x)$是一个一一映射，且$g^{-1}$存在.现在，用$x=g^{-1}(g^{-1}(y))$代入原来的方程，则有\[g^{-1}(g^{-1}(y))=2g^{-1}(y)-y.\]$g^{-1}(y)$同样满足这个方程，则用相同的手段，我们得到

\begin{align*}g^{-1}(y)&\leq y+g(0),&&y<0\\g^{-1}(y)&\geq y+g(0),&&y>0.\end{align*}

现在，用$x=g^{-1}(y)$代入\[g(g(x))-g(x)=g(x)-x\]得到\[g(y)-y=y-g^{-1}(y).\]令$y=0$得到$g^{-1}(0)=-g(0)$.

假设$g(0)\geq0$，则对$x>0$有$g(x)\geq x+g(0)>0$，则对$y=g(x)>0$有$x>g(x)+g^{-1}(0)=g(x)-g(0)$.故得到

\[g(x)=x+g(0),x>0\]同理可得\[g(x)=x+g(0),x<0.\]这样我们得到$f(x)=m(x-g(0))$对$x\in\mathbb{R}$成立.

19. 设$0<a<2$.求证：不存在$(-\infty,+\infty)$上连续的函数$f(x)$，使得对任意$x$有$f(ax-f(x))=x$.

证.我们设$g(x)=ax-f(x)$，显然$g(x)$是连续函数且有\[g(g(x))=ag(x)-x,\forall x\in\mathbb{R}.\]假设此方程有一连续函数解$g:\mathbb{R}\to\mathbb{R}$且多项式方程$r^2-ar+1=0$有一对复数特征根\[{r_1} = a - ib = S\exp \left( { - i\theta } \right),{r_2} = a + ib = S\exp \left( {i\theta } \right)，\]其中$a,b\in \mathbb{R},b>0,S>0$及$\theta\in(0,\pi)$.易知$f$是单调的且$f^2$是严格递增的.并且对于$x\neq0$有$f(x)\neq x$.因此当$f$是严格递增时，数列$\left\{ {{f^{n + 1}}\left( x \right)-{f^n}\left( x \right)} \right\}$对于任意固定的$x\neq0$同样是严格递增的.因此，我们有

\begin{align*}{f^n}\left( x \right) &= \frac{{r_2^n}}{{{r_2} - {r_1}}}\left( {f\left( x \right) - {r_1}x} \right) + \frac{{r_1^n}}{{{r_2} - {r_1}}}\left( {{r_2}x - f\left( x \right)} \right)\\&= \frac{1}{b}{S^n}\sin \theta \cdot f\left( x \right) - \frac{1}{b}{S^{n + 1}}\sin \left( {n - 1} \right)\theta \cdot x.\end{align*}

则

\[{f^{n + 1}}\left( x \right) - {f^n}\left( x \right) = r_2^nU\left( x \right) + r_1^nV\left( x \right),\]

其中$U\left( x \right) = \frac{{{r_2} - 1}}{{{r_2} - {r_1}}}\left( {f\left( x \right) - {r_1}x} \right),V\left( x \right) = \frac{{{r_1} - 1}}{{{r_2} - {r_1}}}\left( {{r_2}x - f\left( x \right)} \right)$.显然$\overline U\left( x \right) = V\left( x \right)$，故对于固定的$x\neq0$我们可令

\[U\left( x \right) = T\exp \left( {it} \right)\text{和}V\left( x \right) = T\exp \left( { - it} \right),\]

其中$T\geq0$且$t\in[0,2\pi]$.因此\[\begin{array}{l}{f^{n + 1}}\left( x \right) - {f^n}\left( x \right) = {S^n}T\left[ {\exp \left( {i\left( {n\theta + t} \right)} \right) + \exp \left( { - i\left( {n\theta + t} \right)} \right)} \right]\\= 2{S^n}T\cos \left( {n\theta + t} \right)\end{array}.\]

因为当$T>0$时$S>0$，与数列$\left\{ {{f^{n + 1}}\left( x \right)-{f^n}\left( x \right)} \right\}$的性质矛盾；当$T=0$时我们有$U(x)=V(x)=0$，因此对所有$x\neq0$我们有$f(x)=r_1x=r_2x$，由此我们得到$r_1=r_2$这一矛盾的结论.证毕.

20. 求证：不存在可微函数$f:(0,+\infty)\to(0,+\infty)$满足方程\[f'(x)=f\circ f(x),x\in(0,+\infty).\]

证.（刘畅）由$f(x)>0$知$f(f(x))>0$，从而有$f'(x)>0$，显然$\lim\limits_{x\to+\infty}f(x)=+\infty$（否则得到$\lim\limits_{x\to+\infty}f'(x)=0$，矛盾），因此$\lim\limits_{x\to+\infty}f'(x)=\lim\limits_{x\to+\infty}f(f(x))=+\infty$.当$x$充分大时，我们有$f'(x)>1$，故存在足够大的$n\in\mathbb{N_+}$使得$f(n)>n+1$，我们有\[f\left( {n + 1} \right) - f\left( n \right) = \int_n^{n + 1} {f'\left( x \right)dx} = \int_n^{n + 1} {f\left( {f\left( x \right)} \right)dx} > f\left( {n + 1} \right),\]矛盾.

21. 设正数列$\{a_n\}$满足$\varliminf\limits_{n\to+\infty}{a_n}=1,\varlimsup\limits_{n\to+\infty}{a_n}<+\infty,\lim\limits_{n\to+\infty}\sqrt[n]{a_1a_2\ldots a_n}=1.$求证：

\[\lim\limits_{n\to+\infty}\frac{a_1+a_2+\ldots+a_n}{n}=1.\]

证.（别人解答）记$x_n=\ln a_n$，由题设条件，

\begin{align*}&\varliminf\limits_{n\to+\infty}x_n=0,\varlimsup\limits_{n\to+\infty}x_n\leq A<+\infty(A>0),\\&\lim\limits_{n\to\infty}\frac1n\sum_{k=1}^n x_k=0.\end{align*}

假设所有的$x_n\geq0$，则当$x\leq\ln2$时，成立不等式$e^x\leq1+2x$.对于固定的$n$，记${S_n} = \left\{ {i \in \mathbb{Z}\left| {1 \le i \le n,{x_i} \le \ln 2} \right.} \right\},{T_n} = \left\{ {i \in \mathbb{Z}\left| {1 \le i \le n,{x_i} > \ln 2} \right.} \right\}$，则

\[\frac{1}{n}\sum\limits_{k = 1}^n {{x_k}} = \frac{1}{n}\sum\limits_{k \in {S_n}} {{x_k}} + \frac{1}{n}\sum\limits_{k \in {T_n}} {{x_k}} \ge \frac{{\left| {{T_n}} \right|}}{n}\ln 2 \ge 0.\]

由此即知\[\mathop {\lim }\limits_{n \to \infty } \frac{{\left| {{T_n}} \right|}}{n} = 0.\]

从而\[\frac{1}{n}\sum\limits_{k = 1}^n {{e^{{x_k}}}} = \frac{1}{n}\sum\limits_{k \in {S_n}} {{e^{{x_k}}}} + \frac{1}{n}\sum\limits_{k \in {T_n}} {{e^{{x_k}}}} \le 1 +\frac{{\left| {{T_n}} \right|}}{n}\left( {C+ {e^A}} \right) + \frac{2}{n}\sum\limits_{k = 1}^n {{x_k}} .\]而\[\frac{1}{n}\sum\limits_{k = 1}^n {{e^{{x_k}}}} \ge {e^{\frac{1}{n}\sum\limits_{k = 1}^n {{x_k}} }}.\]由迫敛性即得\[\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{e^{{x_k}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{a_k}} = 1.\]在一般情况下，作序列\[{z_n} = \left\{ \begin{array}{l}- {x_n},{x_n} < 0\\0,{x_n} \ge 0\end{array} \right.,n = 1,2, \cdots \]由题设可知\[\mathop {\lim }\limits_{n \to \infty } {z_n} = 0.\]故$y_n=x_n+z_n\geq0$，得到$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{y_k}} = 0.$从而$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{e^{{y_k}}}} = 1$，再由$\frac{1}{n}\sum\limits_{k = 1}^n {{e^{{x_k}}}} \le \frac{1}{n}\sum\limits_{k = 1}^n {{e^{{y_k}}}}$，可得\[\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{e^{{x_k}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{a_k}} = 1.\]

22. 设$f(x)$在$\mathbb{R}$上有二阶连续导数且满足方程\[f^3+{(f')}^3=1.\]求证：$f=1$.

证.（刘畅）显然$f=1$满足题意.当$f\neq1$时，我们有

\[f' = \sqrt[3]{{1 - {f^3}}},f'' = \frac{{ - {f^2}f'}}{{\sqrt[3]{{{{\left( {1 - {f^3}} \right)}^2}}}}} = \frac{{ - {f^2}}}{{\sqrt[3]{{1 - {f^3}}}}}.\]

若$\exists x_0\in\mathbb{R}$使$f(x_0)>1$，则$f'(x_0)<0$，故对$x\leq x_0$都有$f'(x)<0,f''(x)>0$,从而$f(-\infty)=+\infty$.若$\exists x_1>x_0$使得$f(x_1)<1$，则对$x\leq x_1$都有$f'(x)>0,f''(x)<0$，故$f(-\infty)=-\infty$，矛盾.

故对$x>x_0$，均有$f(x)\geq1$.

若$\exists a>x_0$使得$f(a)=1$（其中$a$为使得$f(x)=1$的最小实数），则我们有$x>a$均有$f(x)=1$，这时有$f''(x)$在$x=a$处不连续，矛盾.

若不$\exists b>x_0$使得$f(b)=1$，则对于$x\in\mathbb{R}$均有$f(x)>1$，设$f(+\infty)=1$（否则有$\mathop {\lim }\limits_{x \to + \infty } f'\left( x \right) < 0$，与$\mathop {\lim }\limits_{x \to + \infty } f'\left( x \right) = 0$矛盾），而\[\mathop {\lim }\limits_{x \to + \infty } f'\left( x \right) = 0,\mathop {\lim }\limits_{x \to + \infty } f''\left( x \right) = + \infty .\]亦矛盾.证毕.

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