Sep 9

## 看似普通但又有技巧的极限题

$\lim_{n\rightarrow \infty}{\frac{1^n+2^n+\cdots+n^n}{n^n}}$
Solution:
$\sum_{k=1}^{n}\left(\frac{k}{n}\right)^{n}=\sum_{k=0}^{n-1}{\left(\frac{n-k}{n}\right)^{n}}=\sum_{k=0}^{n-1}{\left(1-\frac{k}{n}\right)^{n}}$
$\left(1-\frac{k}{n}\right)^{n}=e^{n\ln(1-\frac{k}{n})}\leq e^{-k}$
Thus.
$\varlimsup_{n\rightarrow \infty}{\frac{1^n+2^n+\cdots+n^n}{n^n}}\leq\sum_{k=0}^{\infty}{e^{-k}}=\frac{e}{e-1}$

for any $N\leq n-1$

$\sum_{k=0}^{N}{\left(\frac{n-k}{n}\right)^{n}}\leq \sum_{k=0}^{n-1}{\left(\frac{n-k}{n}\right)^{n}}$

$\sum_{k=0}^{N}{\left(\frac{n-k}{n}\right)^{n}}\sim \sum_{k=0}^{N}e^{-k}=\frac{e^{-1}(1-e^{N+1})}{1-e^{-1}}$

hence

$\varliminf_{n\rightarrow \infty}{\frac{1^n+2^n+\cdots+n^n}{n^n}}\geq \frac{e^{-1}(1-e^{N+1})}{1-e^{-1}}$

hold for any $N$,let $N\to \infty$,we have

$\lim_{n\rightarrow \infty}{\frac{1^n+2^n+\cdots+n^n}{n^n}} =\frac{e}{e-1}$
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$\lim_{n\to\infty}n\cdot\left(\frac{e}{e-1}-\sum_{i=1}^{n}\left(\frac{i}{n}\right)^{n}\right)=\frac{e(e+1)}{2(e-1)^3}$

Aug 25

## 一道有难度的极限题

$\lim_{n\rightarrow\infty}{\left(1-\frac{1}{1\cdot 2}\right)\left(1-\frac{1}{2\cdot 3}\right)\cdots\left(1-\frac{1}{n(n+1)}\right)}$

Solution：

We write

$P=\prod_{n=1}^\infty \left(1-\frac{1}{n(n+1)}\right)=\prod_{n=1}^\infty \left(\frac{n^2+n-1}{n^2+n}\right)=\prod_{n=1}^\infty\frac{(n-a_1)(n-a_2)}{(n-b_1)(n-b_2)}$,

where  $a_1=\frac{-1+\sqrt{5}}{2},\;a_2=\frac{-1-\sqrt{5}}{2},\;b_1=0$, and $b_2=-1.$

Using the [i]Weierstrass product[/i] for the gamma function, $\frac{1}{\Gamma(x)}=xe^\gamma x\prod_{n=1}^\infty \left(1+\frac{x}{n} \right)e^{-x/n}$, where $\gamma$ denotes Euler's constant, and the fundamental relation $\Gamma(x+1)=x\Gamma(x),$ we can deduce that

$P= \prod_{j=1}^2 \frac{\Gamma(1-b_j)}{\Gamma(1-a_j)}=\frac{\Gamma(1)\Gamma(2)}{\Gamma\left(\frac{3-\sqrt{5}}{2}\right)\Gamma\left(\frac{3+\sqrt{5}}{2}\right)}=\frac{1}{\Gamma\left(\frac{3-\sqrt{5}}{2}\right)\Gamma\left(\frac{3+\sqrt{5}}{2}\right)}\approx 0.2966751356.$

[This standard method of determining certain infinite products is explained, for example, in Section 12.13 of [i]A Course of Modern Analysis [/i]([i]Fourth Edition[/i]) by E. T. Whittaker and G. N. Watson.]

Using the relation $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin\pi x},$ for $x$ not an integer, it should be possible to deduce a closed form for the infinite product given by the computer algebra system [i]Maple[/i]:  $\; - \frac{\sin \pi\left(\frac{1+\sqrt{5}}{2}\right)}{\pi}.$

Aug 25

## 一道牛X的积分题

$\int_{0}^{1}{\frac{\arctan{\sqrt{x^{2}+2}}}{(x^{2}+1)\sqrt{x^{2}+2}}dx}$

Solution:

$\frac{\pi^{2}}{16}=\int_{0}^{1}{\int_{0}^{1}{\frac{dxdy}{(1+x^2)(1+y^2)}}}$
$=\int_{0}^{1}{\int_{0}^{1}{\frac{1}{(1+x^{2})(2+x^2+y^2)}+\frac{1}{(1+y^{2})(2+x^2+y^2)}dxdy}}$
$=2\int_{0}^{1}{\int_{0}^{1}{\frac{1}{(1+x^2)(2+x^2+y^2)}dydx}}$
$=2\int_{0}^{1}{\frac{1}{(1+x^2)\sqrt{2+x^2}}\arctan{\frac{1}{\sqrt{2+x^2}}}dx}$
$=2\int_{0}^{1}{\left(\frac{\pi}{2(1+x^2)\sqrt{2+x^2}}-\frac{\arctan{\sqrt{2+x^2}}}{(1+x^2)\sqrt{2+x^2}} \right)dx}$
$=\frac{\pi^{2}}{6}-2\int_{0}^{1}{\frac{\arctan{\sqrt{2+x^2}}}{(1+x^2)\sqrt{2+x^2}}dx}$
$\Rightarrow \int_{0}^{1}{\frac{\arctan{\sqrt{2+x^2}}}{(1+x^2)\sqrt{2+x^2}}dx}=\frac{5}{96}\pi^{2}$

Aug 25

## 几个数学分析试题。(proposed by tian275461)

1.设$f(x)$在$\left[0,\frac{\pi}{2}\right]$上连续可导，$f(0)=0,f(\frac{\pi}{2})=1,$求证：
$\int_{0}^{\frac{\pi}{2}}{|f(x)\cdot\sin{x}+f'(x)|dx}\geq 1$

$\Rightarrow F'(x)e^{\cos{x}}=f(x)\cdot\sin{x}+f'(x)$
$\Rightarrow \int_{0}^{\frac{\pi}{2}}{|f(x)\cdot\sin{x}+f'(x)|dx}=\int_{0}^{\frac{\pi}{2}}{|e^{\cos{x}}F'(x)|dx}$

2.若曲面$z=f(x,y)$的所有切平面均过原点，证明该曲面是锥面。

$z_{0}=f(x_{0},y_{0})$

$z-z_{0}=f'_{x}(x_{0},y_{0})(x-x_{0})+f'_{y}(x_{0},y_{0})(y-y_{0})$

$\Rightarrow z_{0}=x_{0}\cdot f'_{x}(x_{0},y_{0})+y_{0}\cdot f'_{y}(x_{0},y_{0})$

$\Rightarrow z=x\cdot f'_{x}(x,y)+y\cdot f'_{y}(x,y)$
$z=f(x,y)$是一次齐次函数
$\Rightarrow f(tx,ty)=tf(x,y)$

$z=f(x,y)=f(r\cos{\theta},r\sin{\theta})=rf(\cos{\theta},\sin{\theta})=\sqrt{x^2+y^2}f(\cos{\theta},\sin{\theta})$

3.设
$H=\{(x,y,z):x^2+y^2+z^2=1,z\geq 0 \}$
$C=\{(x,y,0):x^2+y^2=1\}$

$P=2\pi \int_{a}^{b}{y\sqrt{1+(y'_{x})^{2}}dx}$

$S_{H}+5I=\frac{1}{2}4\pi=2\pi$
$I=\frac{1}{2}\cdot 2\pi \int_{\sin{54^{o}}}^{1}{\sqrt{1-x^2}\sqrt{1+\frac{x^2}{1-x^2}}dx}=\pi(1-\sin{54^{o}})$
$\Rightarrow S_{H}=5\pi\cdot\sin{54^{o}}-3\pi$