设$f:[0,1]\to \mathbb{R}$是一个可微函数具有连续导数，且$f(1)=0$,证明:
\[ 4\int_{0}^{1}x^2|f'(x)|^{2}dx\geq \int_{0}^{1}|f(x)|^2dx+\left(\int_{0}^{1}|f(x)|dx\right)^{2} \]
证明：为了方便起见，我们令
\[ A=\int_{0}^{1}|f(x)|dx \]
由Cauchy-Schwarz不等式，我们有
\[ LHS=4\left(\int_{0}^{1}x^{2}|f'(x)|^{2}dx\right)\left(\int_{0}^{1}(|f(x)|+A)^{2}dx\right)\geq 4\left(\int_{0}^{1}x|f'(x)|\cdot|f(x)|dx+A\int_{0}^{1}x|f'(x)|dx \right)^{2}\]
这时，注意到
\[ \int_{0}^{1}x|f'(x)|\cdot|f(x)|dx\geq \left|\int_{0}^{1}xf'(x)dx\right|=\frac{1}{2}\int_{0}^{1}|f(x)|^{2}dx \]
和
\[ \int_{0}^{1}|f(x)|dx=\int_{0}^{1}\left|\int_{t}^{1}f'(x)dx \right|\leq \int_{0}^{1}\int_{t}^{1}|f'(x)|dx= \int_{0}^{1}x|f'(x)|dx \]
\[ LHS\geq \left(\int_{0}^{1}|f(x)|^{2}dx+2A\int_{0}^{1}|f(x)|dx\right)^{2}\]
于是，只要证明
\[ \left(\int_{0}^{1}|f(x)|^{2}dx+2A\int_{0}^{1}|f(x)|dx\right)^{2}\geq \left[\int_{0}^{1}|f(x)|^2dx+\left(\int_{0}^{1}|f(x)|dx\right)^{2} \right]\left(\int_{0}^{1}(|f(x)|+A)^{2}dx\right)\]
经过简单的化简运算，就是
\[ \left(\int_{0}^{1}|f(x)|dx\right)^{4}\geq 0 \]
显然成立。Hence we are done!
问题：$x_{1},x_{2}>0$,$\{x_{n}\}$满足
\[ x_{n+1}=x_{1}^{\frac{1}{n}}+x_{2}^{\frac{1}{n}}+\cdots+x_{n}^{\frac{1}{n}}\]
求
\[ \lim_{n\to\infty}\frac{x_{n}-n}{\ln{n}}\]
解：设$a=\max\{x_{1},x_{2},4\}$,由Holder不等式。可以得到
\[ x_{n+1}^{n}\leq \left(\sum_{k=1}^{n}x_{k}\right)\cdot n^{n-1}\]
\[ x_{n+1}\leq \left(\sum_{k=1}^{n}x_{k}\right)^{\frac{1}{n}}\cdot n^{\frac{n-1}{n}}\]
这样，用数学归纳法可以证明
\[ 0<a_{n}<an \]
而Taylor公式告诉我们
\[ x_{k}^{\frac{1}{n}}=1+\frac{1}{n}\ln x_{k}+o\left(\frac{1}{n}\right)\]
于是
\[ x_{n}=n-1+\frac{1}{n-1}\ln \prod_{k=1}^{n-1}x_{k}+o\left(\frac{1}{n-1}\right)\]
\begin{align*}
\lim_{n\to\infty}\frac{x_{n}-n}{\ln{n}}&=\lim_{n\to\infty}\frac{\frac{1}{n-1}\ln \prod_{k=1}^{n-1}x_{k}-1 }{\ln{n}}\\
&=\lim_{n\to\infty}\frac{\ln \prod_{k=1}^{n-1}x_{k}-(n-1)}{(n-1)\ln{n}}\\
&=\lim_{n\to\infty}\frac{\ln x_{n}-1}{n\ln(n+1)-(n-1)\ln{n}}\qquad (O.Stolz)\\
&=\lim_{n\to\infty}\frac{\ln x_{n}-1}{\ln(n+1)}\\
&=\lim_{n\to\infty}\frac{\ln \frac{x_{n+1}}{x_{n}}}{\frac{1}{n+1}}\qquad (O.Stolz)\\
&=1
\end{align*}
最后一步只要注意到
\[ \frac{x_{n+1}}{x_{n}}=\frac{n+\frac{1}{n}\ln\prod_{k=1}^{n}x_{k}+o\left(\frac{1}{n}\right)}{n-1+\frac{1}{n-1}\ln \prod_{k=1}^{n-1}x_{k}+o\left(\frac{1}{n-1}\right)}\]
利用上面的放缩，可以看到
\[ \ln\left(\frac{x_{n+1}}{x_{n}}\right)\sim \ln\left(1+\frac{1}{n-1}\right)\quad (n\to+\infty) \]
设$k$是一个大于$1$的整数，且$x_0>0$,满足$x_{n+1}=x_{n}+\frac{1}{\sqrt[k]{x_{n}}}$,求
\[ \lim_{x\to\infty}\frac{x_{n}^{k+1}}{n^{k}}\]
求证
\[ \lim_{n\to\infty}\frac{n}{\ln{n}}\left(\frac{x_{n}^{k+1}}{n^{k}}-\left(\frac{k+1}{k}\right)^{k}\right)=\frac{1}{2}\left( \frac{k+1}{k}\right)^{k-1}\]
证明：
我们将要证明
\[ \lim_{n\to\infty}\frac{x^{k+1}_{n}}{n^{k}}=\left(\frac{k+1}{k}\right)^{k}\]
为此，只要证明
\[ \lim_{n\to\infty}\frac{x^{1+\frac{1}{k}}_{n}}{n}=\frac{k+1}{k}\]
而归纳得到
\[ x_{n+1}>x_{n},x_{n}>0\]
因此，$x_{n}$的单调递增序列。设$A=\lim_{n\to+\infty}x_{n}$,则有$A=+\infty$.说明$x_{n}\to+\infty$,$n\to+\infty$.
\[ (x_{n+1}^{1+\frac{1}{k}}-x_{n}^{1+\frac{1}{k}})=x_{n}^{1+\frac{1}{k}}\left(\left(\frac{x_{n+1}}{x_{n}} \right)^{1+\frac{1}{k}}-1\right)=x_{n}^{1+\frac{1}{k}}\left(\exp\left[\left(1+\frac{1}{k}\right)\ln\left(1+\frac{1}{x_{n}^{1+\frac{1}{k}}} \right) \right]-1 \right)\sim \frac{k+1}{k}(n\to+\infty)\]
所以，由O.Stolz定理
\[ \lim_{n\to\infty}\frac{x^{1+\frac{1}{k}}_{n}}{n}=\frac{k+1}{k}\]
对于加强版本，我们先看一个事实，对任意一个收敛的序列，比如说
\[ \lim_{n\to\infty}a_{n}=A \]
那么，
\[ \lim_{n\to\infty}\frac{(a_{n}^{k}-A^{k})}{a_{n}-A}=kA^{k-1}\]
这样，设$a_{n}=\frac{x_{n}^{\frac{k+1}{k}}}{n}$,就有$\displaystyle \lim_{n\to\infty}a_{n}=\frac{k+1}{k}=A$.
\[ \lim_{n\to\infty}\frac{n}{\ln{n}}(a_{n}^{k}-A^{k})= \lim_{n\to\infty}\frac{n}{\ln{n}}(a_{n}-A)\cdot k\cdot \left(\frac{k+1}{k}\right)^{k-1}\]
于是，只要证
\[ \lim_{n\to\infty}\frac{n}{\ln{n}}(a_{n}-A)=\frac{1}{2k}\]
就是
\[ \lim_{n\to\infty}\frac{n}{\ln{n}}\left(\frac{x_{n}^{\frac{k+1}{k}}}{n}-A\right)=\frac{1}{2k}\]
\[ \frac{n}{\ln{n}}\left(\frac{x_{n}^{\frac{k+1}{k}}}{n}-A\right)=\frac{x_{n}^{\frac{k+1}{k}}-nA}{\ln{n}}\]
这时，可以用O.Stolz定理了。就有
\begin{align*}
&\lim_{n\to\infty}\frac{x_{n}^{\frac{k+1}{k}}-nA}{\ln{n}}\\
&\overbrace{=}^{O.Stolz}\lim_{n\to\infty}\frac{x_{n+1}^{1+\frac{1}{k}}-x_{n}^{1+\frac{1}{k}}-A}{\ln\left(1+\frac{1}{n} \right)}\\
&=\lim_{n\to\infty}n\left[x_{n}^{1+\frac{1}{k}}\left(\left(\frac{x_{n+1}}{x_{n}}\right)^{1+\frac{1}{k}}-1\right) -A\right]\\
&=\lim_{n\to\infty}n\left[x_{n}^{1+\frac{1}{k}}\left(e^{\left(1+\frac{1}{k}\right)\ln\left(1+\frac{1}{x_{n}^{1+\frac{1}{k}}}\right) } -1\right)-A\right]\\
&=\lim_{n\to\infty}n\left[x_{n}^{1+\frac{1}{k}}\left(e^{\left(1+\frac{1}{k}\right)\left(\frac{1}{x_{n}^{1+\frac{1}{k}}}-\frac{1}{2}\left(\frac{1}{x_{n}^{1+\frac{1}{k}}} \right)^{2}+o\left( \left(\frac{1}{x_{n}^{1+\frac{1}{k}}} \right)^{2}\right) \right) }-1 \right) -A\right]\\
&=\lim_{n\to\infty}n\left[x_{n}^{1+\frac{1}{k}}\left(\left(\frac{k+1}{k}\right)\left(\frac{1}{x_{n}^{1+\frac{1}{k}}}-\frac{1}{2}\left(\frac{1}{x_{n}^{1+\frac{1}{k}}} \right)^{2}\right)+\frac{1}{2}\left(\frac{k+1}{k}\right)^{2}\left(\frac{1}{x_{n}^{1+\frac{1}{k}}}-\frac{1}{2}\left(\frac{1}{x_{n}^{1+\frac{1}{k}}} \right)^{2}\right)^{2} \right)-A\right]\\
&=\lim_{n\to\infty}\frac{k+1}{2k^{2}}\frac{n}{x_{n}^{1+\frac{1}{k}}}\\
&=\frac{1}{2k}
\end{align*}
因此，命题得证。
Problems is the heart of Math!
_______________________________________________________
西西每日一题
（2006PTN）
1设$k$是一个大于$1$的整数，且$x_0>0$,满足$x_{n+1}=x_{n}+\frac{1}{\sqrt[k]{x_{n}}}$,求
\[ \lim_{x\to\infty}\frac{x_{n}^{k+1}}{n^{k}}\]
求证
\[ \lim_{n\to\infty}\frac{n}{\ln{n}}\left(\frac{x_{n}^{k+1}}{n^{k}}-\left(\frac{k+1}{k}\right)^{k}\right)=\frac{1}{2}\left( \frac{k+1}{k}\right)^{k-1}\]
2 正数序列$\{a_{n}\}$满足
\[ a_{n}\sum_{n=1}^{\infty}a_{n}^{p}=1\qquad n\geq 1 \]
求证：
\[ \lim_{n\to\infty}\frac{n}{\ln{n}}(1-n(p+1)a_{n}^{p+1})=\frac{p+2}{2(p+1)}\qquad p>-1 \]
3.熟悉的IMO预选题
\[\dfrac{1}{1+x_{1}}+\dfrac{1}{1+x_{1}+x_{2}}+\cdots+\dfrac{1}{1+x_{1}+x_{2}+\cdots+x_{n}}<\sqrt{\dfrac{1}{x_{1}}+\dfrac{1}{x_{2}}+\cdots+\dfrac{1}{x_{n}}}\]
加强到
\[\dfrac{1}{1+x_{1}}+\dfrac{1}{1+x_{1}+x_{2}}+\cdots+\dfrac{1}{1+x_{1}+x_{2}+\cdots+x_{n}}<\left(1-\dfrac{\ln{n}}{2n}\right)\sqrt{\dfrac{1}{x_{1}}+\dfrac{1}{x_{2}}+\cdots+\dfrac{1}{x_{n}}}
\]
4.
设$a_{i}>0$,$i=1,2,\cdots,n$,$n\geq 3$,求证
\[\sum_{k=1}^{n}\dfrac{k}{a_{1}+a_{2}+\cdots+a_{k}}\le\left(2-\dfrac{7\ln{2}}{8\ln{n}}\right)\sum_{k=1}^{n}\dfrac{1}{a_{k}}
\]
5.已知实系数多项式$f(x)=\displaystyle\sum_{i=0}^{n}a_{i}x^i$,且满足对于任意$|x|\leq 1$,都有$|f(x)|\leq 1$.
证明：
（1）
\[|a_{n}|+|a_{n-1}|\le 2^{n-1}\]
（2）
\[ |a_{n}|+|a_{n-1}|+\cdots+|a_{1}|+|a_{0}| \le \frac{(1+\sqrt{2})^n+(1-\sqrt{2})^n}{2}\]
6.(代数，求行列式)
\[\left| {\begin{array}{*{20}{c}}
9&{26}&{24}&0& \cdots &0&0\\
1&9&{26}&{24}& \cdots &0&0\\
0&1&9&{26}& \cdots &0&0\\
0& \ddots &1&9& \ddots & \ddots & \vdots \\
\vdots & \ddots & \ddots & \ddots & \ddots & \ddots &{24}\\
0&0& \ddots & \ddots & \ddots &9&{26}\\
0&0& \cdots &0&0&1&9
\end{array}} \right|\]
\[\left| {\begin{array}{*{20}{c}}
x&1&{}&{}&{}&{}\\
{ - n}&{x - 2}&2&{}&{}&{}\\
{}&{ - \left( {n - 1} \right)}&{x - 4}& \ddots &{}&{}\\
{}&{}& \ddots & \ddots &{n - 1}&{}\\
{}&{}&{}&{ - 2}&{x - 2\left( {n - 1} \right)}&n\\
{}&{}&{}&{}&{ - 1}&{x - 2n}
\end{array}} \right|\]
其中未填的为0。(李炯生，线性代数上习题)
7 设矩阵\[A = \left| {\begin{array}{*{20}{c}}
{2012}&1&1& \cdots &1\\
1&{2012}&1& \cdots &1\\
1&1&{2012}& \cdots &1\\
\vdots & \vdots & \ddots & \cdots & \vdots \\
1&1&1& \cdots &{2012}
\end{array}} \right|\]
设$f(x)=\frac{1}{2}(x+1)^{2013}$,求$|f(A)|$.
8.设$f(x)$在$[0,1]$上连续，且$\int_{0}^{1}f(x)dx=1$,若$\lambda\in(0,1)$,证明存在不同的$\xi,\eta$,使得
\[ \lambda f(\xi)+(1-\lambda)f(\eta)=1 \]
9.设函数$f(x)=\frac{1}{5}(x^3+4x^2+6x-6)$,若数列$\{x_n\}$满足$x_{n+1}=f(x_n)$,$n=0,1,2,\cdots$.且$\lim_{n\to\infty}x_{n}$存在，求$x_0$范围，并求该极限.
10.对$x\neq 0$,且$|x|$充分小，证明：存在唯一的$\theta(x)\in[0,1]$使得
\[ \sin{x}=x-\frac{x^3}{6}\cos(x\theta(x)) \]
并求$\lim_{x\to 0}\theta(x)$
11.设$f(x)=\sum_{k=0}^{n}a_{k}\cos{kx}$.其中系数$a_i(i=0,1,\cdots,n)$都是常数，且$a_{n}>|a_{0}|+|a_{1}|+\cdots+|a_{n-1}|$,证明:$f^{n}(x)$在$(0,2\pi)$内至少有$n$个零点。
12.设$f(x)$满足$f''(x)>0$,$\int_{0}^{1}f(x)dx=0$,证明$\forall x\in[0,1]$,必有
\[ |f(x)|\leq \max\{f(0),f(1)\} \]
13.设数列$\{x_n\}$满足$0<x_0<\pi$,且$x_n=\sin{x_{n-1}}$,$n=1,2,\cdots$,求证
\[ x_{n}=\sqrt{\frac{3}{n}}\left[1-\frac{3}{10}\cdot\frac{\ln{n}}{n}-\frac{C}{2n}+\frac{271\ln^2{n}}{200n^2}+\frac{\beta\ln{n}+\gamma}{n^2}+O\left(\frac{\ln^3{n}}{n^3}\right) \right]\]
$C=C(x_0)$,且$\lim_{x\to0^{+}}C(x_0)=+\infty$,$\beta=\frac{9}{20}C-\frac{9}{50}.\gamma=\frac{3}{8}C^2-\frac{3}{10}C+\frac{79}{700}$
$x,y,z>0$,求证：
\[ (x^2+y^2+z^2)\left[\frac{x^2}{(x^2+yz)^2}+\frac{y^2}{(y^2+xz)^2}+\frac{z^2}{(z^2+xy)^2}\right]\geq \frac{9}{4}\]
14计算积分(西西博客)
\[\int_0^{+\infty} {\frac{{x\cos x}}{{\left( {1 + {x^2}} \right)\left( {2 + {x^2}} \right)}}dx} \]
15
求
\[ \sum_{n=1}^{\infty}\frac{\ln{2}-\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\right)}{n}\]
16
\[ \int_{0}^{1}\left\{\frac{1}{x}\right\}^{4}dx \]
17
设数列$\{x_{n}\}$满足$x_0>0$,且$x_1+x_2+\cdots+x_n=\frac{1}{\sqrt{x_{n+1}}}$,$n\geq 0$,求
\[ \lim_{n\to\infty}\left[\frac{n\left(n^2x_{n}^{3}-\frac{1}{9}\right)}{\ln{n}}\right] \]
18
若数列$\{a_{n}\}$满足$a_{n+3}=a_{n+1}+a_{n}$.
\[ A=\lim_{n\to\infty}\frac{\sqrt{a_{n+5}^2+a_{n+4}^2+a_{n+3}^2-a_{n+2}^2+a_{n+1}^2-a_{n}^2}}{a_{n+6}}\]
\[ B=\lim_{n\to\infty}\frac{3^{n}}{\prod_{k=1}^{n}\frac{\arcsin\left(\frac{9k+2}{\sqrt{27k^3+54k^2+36k+8}} \right)}{\arctan\left(\frac{1}{\sqrt{3k+1}}\right)}}\]
求$A+B$.
19证明
\[ \sum_{n=1}^{\infty}\arctan\left(\frac{1}{n^2+1}\right)=\arctan\left[\tan\left(\pi\sqrt{\frac{\sqrt{2}-1}{2}} \right)\cdot\frac{\exp\left(\pi\sqrt{\frac{\sqrt{2}+1}{2}}\right)+\exp\left(-\pi\sqrt{\frac{\sqrt{2}+1}{2}}\right)}{\exp\left(\pi\sqrt{\frac{\sqrt{2}+1}{2}}\right)-\exp\left(-\pi\sqrt{\frac{\sqrt{2}+1}{2}}\right)} \right]-\frac{\pi}{8} \]
20
设$a,b,c$是三个正数，且满足$abc=\frac{1}{16}$,求证
\[ \int_{0}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}dx\leq \pi \]
21
设$a_n=\sqrt{1+\sqrt{2+\cdots+\sqrt{n}}}$,求证
\[ \lim_{n\to\infty}\sqrt{n}\sqrt[n]{\left(\lim_{n\to\infty}a_{n}\right)-a_{n}}\]
存在，并计算该极限。
22
设$a_n=L_{n}-\frac{4}{\pi^2}\ln{n}$,且$\displaystyle L_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|\frac{\sin\left(n+\frac{1}{2}\right)}{\sin{\frac{x}{2}}}\right|$,证明
$\{a_{n}\}$是有界数列。
23
已知$a_1+a_2+a_3+\cdots+a_n=0$,并满足$a_i\neq \frac{\pi}{2}+k\pi,k\in \mathbb{Z}$,且$f(x)=\frac{\cos{2x}+\sin{2x}+(-1)^{n-1}\sin{2x}+(-1)^{n}\cos{2x}}{2}$
求\[D = \left| {\begin{array}{*{20}{c}}
1&1& \cdots & \cdots & \cdots &1\\
{\tan {a_1}}&{\tan {a_2}}& \cdots & \cdots & \cdots &{\tan {a_n}}\\
{{{\tan }^2}{a_1}}&{{{\tan }^2}{a_2}}& \cdots & \cdots & \cdots &{{{\tan }^2}{a_n}}\\
\vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\
{{{\tan }^{n - 2}}{a_1}}&{{{\tan }^{n - 2}}{a_2}}& \cdots & \cdots & \cdots &{{{\tan }^{n - 2}}{a_n}}\\
{f\left( {{a_1}} \right)}&{f\left( {{a_2}} \right)}& \cdots & \cdots & \cdots &{f\left( {{a_n}} \right)}
\end{array}} \right|\]
的取值范围。
24
设$f(x)$是多项式函数，若对于任意的$x\in \mathbb{R}$,只要有$f(x)+f'(x)-Kf''(x)\geq 0$ 恒成立，就一定有$f(x)\geq 0$,求$K$的最小值。
24
证明
\[ \int_{0}^{\frac{\pi}{4}}(\tan{x})^{a}\geq \frac{\pi}{4+2a\pi},a>0 \]
\[ \frac{\pi^2}{12-\sqrt{3}}\left(3^{2-\frac{\sqrt{3}}{6}\pi}-\frac{1}{3}\right)\leq \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{x}{\sin{x}}dx\leq \frac{7\pi^2\ln3}{72\ln3-36\ln 2}\]
25
设$a$是正常数，证明方程
\[ \frac{1}{x}+\frac{2x}{x^2-1^2}+\frac{2x}{x^2-2^2}+\cdots+\frac{2x}{x^2-n^2}=a \]
在$[n,+\infty)$有唯一的根，并求极限$\lim_{n\to\infty}\frac{x_n}{n}$.
26
设$\displaystyle f(n)=\int_{1}^{n}\frac{(x-1)^n}{x^{4}-2x^3+3x^2-2x+1}dx$,求当$f(n)$达到最小时。计算
\[ \int_{1}^{\sqrt{3}}x^{nx^2+1}+\ln\left(x^{nx^{nx^2+1}}\right)dx\]
的值.
27
设$x_1>0$,$x_{n+1}=\frac{1}{n^{x_{n}}}+\sqrt[n]{x_{n}}$,求
\[ \lim_{n\to\infty}\frac{\ln{n}}{x_{1}+x_{2}+x_{3}+\cdots+x_{n}} \]
28
设$\displaystyle a_{n}=\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin{nx}}{\sin{x}}\right)^4 dx$,证明
\[ \lim_{n\to\infty}\frac{a_{n}}{n^2}=\ln 2 \]
29.$\displaystyle \lim_{h\to 0^{+}}h^{n}\int_{c}^{\infty}\cdots\int_{c}^{\infty}\left(\int_{c}^{\infty}\cdots\int_{c}^{\infty}f(x_1,\cdots,x_n,y_1,\cdots,y_n)dy_1dy_2\cdots dy_n \right)^{p}dx_1\cdots dx_n $,其中
\[ f(x_1,\cdots,x_n,y_1,\cdots,y_n)=\left(\prod_{k=1}^{n}x_{k}y_{k}\right)^{-\frac{1}{4}}\exp\left(\sum_{k=1}^{n}\sqrt{x_{k}y_{k}}-\frac{\sum_{k=1}^{n}(x_{k}+y_{k})}{2}-\frac{h\sum_{k=1}^{n}y_{k}}{p} \right)\]
$c>0,p\geq 1$,$n$,$p$是常数。
30.设$a_{n+3}=a_{n+2}+\dfrac{a_{n}}{2n+6}$,且$a_0=a_1=a_2=1 $,求极限
\[ \lim_{n\to\infty}\frac{a_{n}}{\sqrt{n}} \]
同时证明
\[ \lim_{n\to\infty}n\left(\frac{a_{n}}{\sqrt{n}}-\frac{2}{\sqrt{\pi}\cdot\sqrt[4]{e^{3}}}\right)=\frac{7}{4\sqrt{\pi}\cdot\sqrt[4]{e^{3}}}\]
31.若$a,b>0$,求证
\[\left| {\begin{array}{*{20}{c}}
{\Gamma \left( a \right)}&{\Gamma \left( {a + b} \right)}&{\Gamma \left( {a + 2b} \right)}\\
{\Gamma \left( {a + b} \right)}&{\Gamma \left( {a + 2b} \right)}&{\Gamma \left( {a + 3b} \right)}\\
{\Gamma \left( {a + 2b} \right)}&{\Gamma \left( {a + 3b} \right)}&{\Gamma \left( {a + 4b} \right)}
\end{array}} \right| > 0\]
32.证明：$SU(l+1)$($A_{l}$代数),有任意的$\alpha$,满足
\[ g_{\alpha}\equiv\frac{\alpha(g,\alpha)}{(\alpha,\alpha)}=1 \]
设级数
\[ \sum_{n=1}^{\infty}2^{-n}a_{n+1}(a_{1}a_{2}\cdots a_{n}a_{n+1})^{-\frac{1}{2}}\]
收敛，证明数列
\[ x_{n}=\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}}\]
也收敛，这里$a_{1},a_{2},\cdots,a_{n},\cdots $为正实数。
证明：注意到$\{x_{n}\}$是单调递增的。而又有
\begin{align*}
&x_{n+1}-x_{n}\\
&=\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n+1}}}}-\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}}\\
&=\frac{\left(\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n+1}}}}-\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}} \right)\left(\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n+1}}}}+\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}} \right)}{\left(\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n+1}}}}+\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}} \right)}\\
&<\frac{\sqrt{a_{2}+\cdots+\sqrt{a_{n+1}}}-\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}}{2\sqrt{a_{1}}}\\
&<\frac{\sqrt{a_{3}+\cdots+\sqrt{a_{n+1}}}-\sqrt{a_{3}+\cdots+\sqrt{a_{n}}}}{2\sqrt{a_{1}}\cdot 2\sqrt{a_{2}}}\\
&<\cdots\\
&<\frac{\sqrt{a_{n+1}}}{2^{n}\sqrt{a_{1}a_{2}\cdots a_{n}}}\\
&=\frac{a_{n+1}}{2^{n}\sqrt{a_{1}a_{2}\cdots a_{n+1}}}
\end{align*}
而由于级数收敛，所以数列$\{x_{n}\}$收敛。
