József Wildt International Mathematical Competition

简介

József Wildt International Mathematical Competition（约瑟夫·维尔德国际数学竞赛）是一个面向高中生的国际数学竞赛，以纪念匈牙利数学家约瑟夫·维尔德（József Wildt）。它通常在罗马尼亚特尔古穆列什举办，旨在鼓励学生对数学的兴趣并挑战他们的解决问题的能力。

各届问题及其解答

2019 年

W1. The Pell numbers $P_n$ satisfy $P_0=0$, $P_1=1$, and $P_n=2P_{n-1}+P_{n-2}$ for $n\ge2$. Find

$$\sum_{n=1}^{\infty}\left(\arctan\frac1{P_{2n}}+\arctan\frac1{P_{2n+2}}\right)\arctan\frac2{P_{2n+1}}.$$

提示

依次证明:
(1). $P_{n}P_{n+2}+\left(-1\right)^{n}=P_{n+1}^{2}$;
(2). $\arctan\frac{1}{P_{2n}}-\arctan\frac{1}{P_{2n+2}}=\arctan\frac{2}{P_{2n+1}}$;
(3). $\sum\limits_{n=1}^{\infty}\left(\arctan\frac{1}{P_{2n}}+\arctan\frac{1}{P_{2n+2}}\right)\arctan\frac{2}{P_{2n+1}}=\arctan^{2}\frac{1}{2}$.

W2. If $0<a\le c\le b$ then:

$$\frac{\left(b^{30}-a^{30}\right)\left(b^{30}-c^{30}\right)}{36b^{10}}\le\frac{\left(b^{25}-a^{25}\right)\left(b^{25}-c^{25}\right)}{25}\le\frac{\left(b^{30}-a^{30}\right)\left(b^{30}-c^{30}\right)}{36\left(ac\right)^{5}}$$

提示

考虑

$$\int_{a}^{b}\int_{c}^{b}t^{24}s^{24}\,ds\,dt\le\frac{1}{(ac)^{5}}\int_{a}^{b}\int_{c}^{b}t^{29}s^{29}\,ds\,dt$$

和

$$\frac{1}{b^{10}}\int_{a}^{b}\int_{c}^{b}t^{29}s^{29}\,ds\,dt\le\int_{a}^{b}\int_{c}^{b}t^{24}s^{24}\,ds\,dt.$$

W3. Compute

$$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\cos x+1-x^{2}}{\left(1+x\sin x\right)\sqrt{1-x^{2}}}dx.$$

提示

$$\begin{align*} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\cos x+1-x^{2}}{\left(1+x\sin x\right)\sqrt{1-x^{2}}}dx & =2\int_{0}^{\frac{\pi}{4}}\frac{\cos x+1-x^{2}}{\left(1+x\sin x\right)\sqrt{1-x^{2}}}dx\\ & =\left.-4\arctan\left(\frac{\cos\left(\frac{\arcsin x-x}{2}\right)}{\sin\left(\frac{\arcsin x+x}{2}\right)}\right)\right|_{0}^{\frac{\pi}{4}}\\ & =2\pi-4\arctan\left(\frac{\cos\left(\frac{\pi-4\arcsin\frac{\pi}{4}}{8}\right)}{\sin\left(\frac{\pi+4\arcsin\frac{\pi}{4}}{8}\right)}\right) \end{align*}$$

W4. If $x,y,z,t>1$ then:

$$\left(\log_{zxt}x\right)^{2}+\left(\log_{xyt}y\right)^{2}+\left(\log_{xyz}z\right)^{2}+\left(\log_{yzt}t\right)^{2}>\frac{1}{4}.$$

提示

$$\begin{align*} \sum_{cyc}\left(\frac{a}{a+c+d}\right)^{2} & >\frac{1}{4}\\ \sum_{cyc}1\sum_{cyc}\left(\frac{a}{a+c+d}\right)^{2} & \ge\left(\sum_{cyc}\frac{a}{a+c+d}\right)^{2}>\left(\sum_{cyc}\frac{a}{a+b+c+d}\right)^{2}=1. \end{align*}$$

W5. Let $n\ge 1$. Find a set of distincts real numbers $(x_j)_{1\le j\le n}$ such that for any bijections $f:\{1,2,\cdots, n\}^2\to\{1,2,\cdots,n\}^2$ the matrix $\left(x_{f(i,j)}\right)_{1\le i,j\le n}$ is invertible.

W6. Compute

$$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{(1+\ln x)\cos x+x\sin x\ln x}{\cos^{2}x+x^{2}\ln^{2}x}\ud x$$

提示

$$\begin{align*} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{(1+\ln x)\cos x+x\sin x\ln x}{\cos^{2}x+x^{2}\ln^{2}x}dx & =\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{(x\ln x)^{\prime}\cos x+\sin x\cdot x\ln x}{\cos^{2}x+x^{2}\ln^{2}x}dx\\ & =\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{1}{1+\left(\frac{x\ln x}{\cos x}\right)^{2}}d\left(\frac{x\ln x}{\cos x}\right)\\ & =\arctan\left(\frac{x\ln x}{\cos x}\right)|_{\pi/6}^{\pi/4}\\ & =\arctan\left(\frac{\sqrt{2}\pi\ln\frac{\pi}{4}}{4}\right)-\arctan\left(\frac{\sqrt{3}\pi\ln\frac{\pi}{6}}{9}\right) \end{align*}$$

W7. If

$$\Omega_n=\sum_{k=1}^n\left(\int_{-\frac{1}{k}}^{\frac{1}{k}}\left(2 x^{10}+3 x^8+1\right) \cdot \cos ^{-1}(k x) d x\right)$$

then find:

$$\Omega=\lim _{n \rightarrow \infty}\left(\Omega_n-\pi \cdot H_n\right)$$

提示

分部积分得到:

$$\int_{-\frac{1}{k}}^{\frac{1}{k}}x^{2n}\arccos\left(kx\right)\ud x=\frac{\pi}{(2n+1)k^{2n+1}},$$

代入,

$$\Omega_{n}=\sum_{k=1}^{n}\left(\int_{-\frac{1}{k}}^{\frac{1}{k}}\left(2x^{10}+3x^{8}+1\right)\cdot\arccos(kx)dx\right)=\sum_{k=1}^{n}\left(\frac{2\pi}{11k^{11}}+\frac{\pi}{3k^{9}}+\frac{\pi}{k}\right),$$

所以

$$\Omega=\frac{2\pi}{11}\zeta(11)+\frac{\pi}{3}\zeta(9).$$

W8. Let $(a_n)_{n \geq 1}$ be a positive real sequence given by $a_n=\sum_{k=1}^n \frac{1}{k}$. Compute

$$\lim _{n \to \infty} e^{-2 a_n} \sum_{k=1}^n\left[(\sqrt[2 k]{k!}+\sqrt[2(k+1)]{(k+1)!})^2\right]$$

where we denote by $[x]$ the integer part of $x$.

提示

由 Stirling 公式,

$$n!=\sqrt{2\pi n}\left(\frac{n}{\ue}\right)^{n}\ue^{O\left(\frac{1}{n}\right)},$$

所以

$$\sqrt[n]{n!}=\frac{n}{\ue}\cdot\ue^{\frac{\ln\left(2\pi n\right)}{2n}+O\left(\frac{1}{n^{2}}\right)}=\frac{n}{\ue}\left(1+\frac{\ln\left(2\pi n\right)}{2n}+o\left(\frac{1}{n}\right)\right),$$

又由于不等式

$$\sqrt[k]{k!}<\sqrt[k+1]{(k+1)!}\Longrightarrow\sqrt[2k]{k!}<\sqrt[2(k+1)]{(k+1)!},$$

所以有自然的不等式 (下界有稍微的不严谨, 不过影响不大, 这里忽略)

$$4\sqrt[k]{k!}<\left[\left(\sqrt[2k]{k!}+\sqrt[2(k+1)]{(k+1)!}\right)^{2}\right]<4\sqrt[k+1]{(k+1)!}.$$

所以所求极限化为

$$\lim_{n\to\infty}4\ue^{-2H_{n}}\sum_{k=1}^{n}\sqrt[k]{k!}<\lim_{n\to\infty}\ue^{-2H_{n}}\sum_{k=1}^{n}\left[\left(\sqrt[2k]{k!}+\sqrt[2(k+1)]{(k+1)!}\right)^{2}\right]<\lim_{n\to\infty}4\ue^{-2H_{n}}\sum_{k=1}^{n}\sqrt[k+1]{(k+1)!}.$$

由于 $\ue^{-2H_{n}}=\ue^{-2(H_{n}-\ln n)-2\ln n}\to\ue^{-2\gamma}\cdot n^{-2}$,
所以

$$\begin{align*} \lim_{n\to\infty}4\ue^{-2H_{n}}\sum_{k=1}^{n}\sqrt[k]{k!} & =\lim_{n\to\infty}\frac{4}{\ue^{2\gamma}}\cdot\frac{1}{n}\sum_{k=1}^{n}\frac{\sqrt[k]{k!}}{n}\\ & =\lim_{n\to\infty}\frac{4}{\ue^{2\gamma}}\cdot\frac{1}{n}\sum_{k=1}^{n}\frac{\frac{k}{\ue}\left(1+\frac{\ln\left(2\pi k\right)}{2k}+o\left(\frac{1}{k}\right)\right)}{n}\\ & =\frac{4}{\ue^{2\gamma}}\cdot\int_{0}^{1}\frac{x}{\ue}\ud x+\lim_{n\to\infty}\frac{4}{\ue^{2\gamma}}\cdot\frac{1}{n}\sum_{k=1}^{n}\left(\frac{\ln\left(2\pi k\right)}{2kn}+o\left(\frac{1}{kn}\right)\right)\\ & =\frac{2}{\ue^{2\gamma+1}}. \end{align*}$$

同理可证:

$$\lim_{n\to\infty}4\ue^{-2H_{n}}\sum_{k=1}^{n}\sqrt[k+1]{(k+1)!}=\frac{2}{\ue^{2\gamma+1}}.$$

所以

$$\lim_{n\to\infty}\ue^{-2H_{n}}\sum_{k=1}^{n}\left[\left(\sqrt[2k]{k!}+\sqrt[2(k+1)]{(k+1)!}\right)^{2}\right]=\frac{2}{\ue^{2\gamma+1}}.$$

W9. Let $\alpha>0$ be a real number. Compute the limit of the sequence $\left\{x_n\right\}_{n \geq 1}$ defined by

$$x_n=\left\{\begin{array}{ll} \sum_{k=1}^n \sinh \left(\frac{k}{n^2}\right), & n>\frac{1}{\alpha} ; \\ 0, & n \leq \frac{1}{\alpha} \end{array} .\right.$$

提示

用 Taylor 展开 $\sinh x=x+\frac{x^{3}}{6}+o\left(u^{3}\right)$, 结果为 $1/2$.

W10. If $\mathrm{si}(x)=-\int_x^{\infty}\left(\frac{\sin t}{t}\right) \ud t$, $x>0$, then:

$$\int_{\ue}^{\ue^{2}}\frac{\mathrm{si}(\ue^{4}x)-\mathrm{si}(\ue^{3}x)}{x}\ud x=\int_{\ue^{3}}^{\ue^{4}}\frac{\mathrm{si}(\ue^{2}x)-\mathrm{si}(\ue x)}{x}\ud x.$$

提示

这本质上是一道交换积分次序问题, 注意到

$$\mathrm{si}(x)=-\frac{\pi}{2}+\int_{0}^{x}\frac{\sin t}{t}\ud t,$$

所以等式的左边为

$$\begin{align*} \int_{\ue}^{\ue^{2}}\frac{\mathrm{si}(\ue^{4}x)-\mathrm{si}(\ue^{3}x)}{x}\ud x & =\int_{\ue}^{\ue^{2}}\int_{\ue^{3}x}^{\ue^{4}x}\frac{\sin t}{tx}\ud t\ud x\\ & =\int_{\ue^{4}}^{\ue^{6}}\left(\int_{\ue}^{t/\ue^{3}}\frac{\sin t}{tx}\ud x+\int_{t/\ue^{4}}^{\ue^{2}}\frac{\sin t}{tx}\ud x\right)\ud t\\ & =\int_{\ue^{4}}^{\ue^{6}}\frac{2\sin t}{t}\ud t; \end{align*}$$

同理, 等式的右边为

$$\begin{align*} \int_{\ue^{3}}^{\ue^{4}}\frac{\mathrm{si}(\ue^{2}x)-\mathrm{si}(\ue x)}{x}\ud x & =\int_{\ue^{3}}^{\ue^{4}}\int_{\ue x}^{\ue^{2}x}\frac{\sin t}{tx}\ud t\ud x\\ & =\int_{\ue^{4}}^{\ue^{6}}\left(\int_{\ue^{3}}^{t/\ue}\frac{\sin t}{tx}\ud x+\int_{t/\ue^{2}}^{\ue^{4}}\frac{\sin t}{tx}\ud x\right)\ud t\\ & =\int_{\ue^{4}}^{\ue^{6}}\frac{2\sin t}{t}\ud t. \end{align*}$$

W11. Let $\left(s_n\right)_{n \geq 1}$ be a sequence given by $s_n=-2 \sqrt{n}+\sum_{k=1}^n \frac{1}{\sqrt{k}}$ with $\lim\limits_{n \to \infty} s_n=s=$ Ioachimescu constant and $\left(a_n\right)_{n \geq 1}$, $\left(b_n\right)_{n \geq 1}$ be a positive real sequences such that

$$\lim_{n \to \infty} \frac{a_{n+1}}{n a_n}=a \in R_{+}^*, \quad\lim_{n \to \infty} \frac{b_{n+1}}{b_n \sqrt{n}}=b \in R_{+}^*$$

Compute

$$\lim_{n \to \infty}\left(1+e^{s_n}-e^{s_{n+1}}\right)^{\sqrt[n]{a_n b_n}}$$

提示

由

$$\lim_{n\to\infty}\frac{a_{n}}{na_{n-1}}=a,\quad\lim_{n\to\infty}\frac{b_{n}}{b_{n-1}\sqrt{n}}=b,$$

所以有

$$\lim_{n\to\infty}\frac{a_{n}b_{n}}{a_{n-1}b_{n-1}\cdot n^{3/2}}=ab,$$

取对数, 做 Cesaro 平均的极限有

$$\lim_{n\to\infty}\left(\frac{\ln(a_{n}b_{n})-\frac{3}{2}\ln(n!)}{n}\right)=\ln(ab),$$

所以

$$\ln\sqrt[n]{a_{n}b_{n}}=\frac{3}{2}\ln\left(\frac{n}{\ue}\right)+\ln(ab)+o(1)\Longrightarrow\sqrt[n]{a_{n}b_{n}}\sim\left(\frac{n}{\ue}\right)^{3/2}ab.$$

由于 $\lim s_{n}=s$ 存在, 所以 $\lim\left(\ue^{s_{n}}-\ue^{s_{n+1}}\right)=0$, 有等价无穷小代换 $\ln\left(1+\ue^{s_{n}}-\ue^{s_{n+1}}\right)\sim\ue^{s_{n}}\left(1-\ue^{s_{n+1}-s_{n}}\right)$, 对后一项做幂级数展开可得:

$$1-\ue^{s_{n+1}-s_{n}}=1-\ue^{-\frac{2}{\sqrt{n+1}+\sqrt{n}}+\frac{1}{\sqrt{n+1}}}=1-\ue^{-\frac{1}{4n\sqrt{n}}+o\left(\frac{1}{n\sqrt{n}}\right)}=\frac{1}{4n\sqrt{n}}+o\left(\frac{1}{n\sqrt{n}}\right).$$

因 $\ue^{s_{n}}\to\ue^{s}$, 所以原极限等于

$$\lim_{n\to\infty}\ue^{\ln\left(1+\ue^{s_{n}}-\ue^{s_{n+1}}\right)\cdot\sqrt[n]{a_{n}b_{n}}}=\lim_{n\to\infty}\ue^{\ue^{s}\left(\frac{1}{4n\sqrt{n}}+o\left(\frac{1}{n\sqrt{n}}\right)\right)\cdot ab\left(\frac{n}{\ue}\right)^{3/2}}=\ue^{\frac{ab\ue^{s-3/2}}{4}}.$$

W12. If $0<a<b$ then:

$$\frac{\int_a^{\frac{a+b}{2}}\left(\tan ^{-1} t\right) d t}{\int_a^b\left(\tan ^{-1} t\right) d t}<\frac{1}{2}$$

提示

根据定积分的几何意义, 比较两个曲边梯形的面积.

W13. Let $a, b$ and $c$ be complex numbers such that $abc=1$. Find the value of the cubic root of

$$\begin{vmatrix} b+n^3 c & n(c-b) & n^2(b-c) \\ n^2(c-a) & c+n^3 a & n(a-c) \\ n(b-a) & n^2(a-b) & a+n^3 b \end{vmatrix}.$$

提示

结果为 $(n+1)(n^2-n+1)$.

W14. If $a, b, c>0$, $a b+b c+c a=3$ then:

$$4\left(\tan ^{-1} 2\right)\left(\tan ^{-1}(\sqrt[3]{a b c})\right) \leq \pi \tan ^{-1}(1+\sqrt[3]{a b c}).$$

提示

先证明: $1\ge abc\eqqcolon x$, 再用一阶导数与 $\left(1+x^{2}\right)\arctan x$ 的单调性, 证明函数 $\frac{\arctan(1+x)}{\arctan x}$ 单调递减.

W15. It is possible to partition the set $\{100,101, \cdots, 1000\}$ into two subsets so that for any two distinct elements $x$ and $y$ belonging to the same subset $\sqrt[3]{x+y}$ is irrational?

提示

这是一道线性代数问题. 考虑集合中任意两数和为立方数的所有可能结果为 $216, 343, 512, 729, 1000, 1331$, 现在需要划分两个集合, 使得同一集合中的任两个数之和不能是立方数. 此问题可以转化为和为立方数的两个整数应位于不同的集合, 求出这两个和为立方数的所有可能整数, 比如 $801+530=1331=11^3$, 所以 $801$ 与 $530$ 不在同一个集合内.

要确定集合划分为两个子集的方式是否存在, 这等价于验证是否有三个数 $x,y,z\in\{100,101, \cdots, 1000\}$, 使得它们任何两个数之和同时为立方数. 因为如果 $x,y,z$ 中的任何两个数之和为立方数, 则它们两两不能在一个集合内.

求解方程 $x+y=729$, $y+z=1000$, $z+x=1331$, 得到 $x=530$, $y=199$, $z=801$, 这三个数两两不能同时位于一个集合内, 因此集合的二划分不存在.

W16. If $f:[a, b] \to (0, \infty)$. $0<a \le b$. $f$ derivable, $f’$ continuous, then:

$$\int_a^b \frac{f^{\prime}(x) \sqrt{f(x)}}{f^3(x)+1} d x \leq \tan ^{-1}\left(\frac{f(b)-f(a)}{1+f(a) f(b)}\right).$$

提示

用凑微分法以及三角恒等式, 不等式等价于

$$\frac{2}{3}\left(\arctan\sqrt{f^{3}(b)}-\arctan\sqrt{f^{3}(a)}\right)\le\arctan f(b)-\arctan f(a),$$

然后用 Cauchy 中值定理证明

$$\frac{\arctan\sqrt{f^{3}(b)}-\arctan\sqrt{f^{3}(a)}}{\arctan f(b)-\arctan f(a)}\le\frac{3}{2}.$$

W17. Let $f_n=\left(1+\frac{1}{n}\right)^n\left((2 n-1)!!F_n\right)^{1 / n}$. Find $\lim\limits_{n \to \infty}\left(f_{n+1}-f_n\right)$, where $F_n$ denotes the $n$th Fibonacci number (given by $F_0=0$, $F_1=1$, and by $F_{n+1}=F_n+F_{n-1}$ for all $n \ge 1$).

提示

用幂级数展开, 给出每一个因子的较为精确的估计:

$$\left(1+\frac{1}{n}\right)^{n}=\ue^{n\ln\left(1+\frac{1}{n}\right)}=\ue\left(1-\frac{1}{2n}+O\left(\frac{1}{n^{2}}\right)\right),$$

和

$$\begin{align*} \left((2n-1)!!F_{n}\right)^{1/n} & =\ue^{\frac{1}{n}\left(\ln(2n-1)!!+\ln F_{n}\right)}\\ & =\ue^{\ln(2\phi n)-1+\frac{\ln(2/5)}{2n}+O\left(\frac{1}{n^{2}}\right)}\\ & =\frac{2\phi n}{\ue}+\frac{\phi}{\ue}\ln\frac{2}{5}+O\left(\frac{1}{n}\right) \end{align*}$$

所以

$$\begin{align*} f_{n} & =\ue\left(1-\frac{1}{2n}+O\left(\frac{1}{n^{2}}\right)\right)\cdot\left(\frac{2\phi n}{\ue}+\frac{\phi}{\ue}\ln\frac{2}{5}+O\left(\frac{1}{n}\right)\right)\\ & =2\phi n+\phi\ln\frac{2}{5}-\phi+O\left(\frac{1}{n}\right), \end{align*}$$

代入原极限, 便可得

$$\lim_{n\to\infty}\left(f_{n+1}-f_{n}\right)=\lim_{n\to\infty}\left(2\phi+O\left(\frac{1}{n}\right)\right)=2\phi=1+\sqrt{5}.$$

W18. Let $\left\{c_k\right\}_{k \ge 1}$ be a sequence with $0 \le c_k \le 1$, $c_1 \ne 0$, $\alpha>1$. Let $C_n=c_1+\cdots+c_n$. Prove

$$\lim_{n \to \infty} \frac{C_1^\alpha+\cdots+C_n^\alpha}{\left(C_1+\cdots+C_n\right)^\alpha}=0.$$

提示

当 $\left\{ C_{n}\right\} $ 收敛于某个常数 $C$, 使用常见的不等式 $\left(x^{\alpha}+y^{\alpha}\right)^{1/\alpha}<x+y$, $x^{\alpha}+y^{\alpha}<\left(x+y\right)^{\alpha}$.

取 $\epsilon=\frac{C}{2}$, $\exists N$, s.t. $\forall n>N$, $\left|C_{n}-C\right|<\epsilon$,
也即 $\frac{3C}{2}>C_{n}>\frac{C}{2}$. 再用夹逼原理:

$$0<\frac{C_{1}^{\alpha}+\cdots+C_{N}^{\alpha}+\cdots+C_{n}^{\alpha}}{\left(C_{1}+\cdots+C_{N}+\cdots+C_{n}\right)^{\alpha}}<\frac{C(N)+(n-N)\left(\frac{3C}{2}\right)^{\alpha}}{\left(C(N)+(n-N)\frac{C}{2}\right)^{\alpha}}<\frac{C(N)+nC}{C(N)+n^{\alpha}C}\to0.$$

若 $\left\{ C_{n}\right\} $ 发散于无穷大, 用 Stolz 公式

$$\begin{align*} \lim_{n\to\infty}\frac{\left(C_{1}^{\alpha}+\cdots+C_{n}^{\alpha}\right)^{1/\alpha}}{C_{1}+\cdots+C_{n}} & =\lim_{n\to\infty}\frac{\left(1+\frac{C_{n}^{\alpha}}{C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}}\right)^{1/\alpha}-1}{C_{n}}\left(C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}\right)^{1/\alpha}\\ & =\lim_{n\to\infty}\left[\left(1+\frac{C_{n}^{\alpha}}{C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}}\right)^{1/\alpha}-1\right]\left(\frac{C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}}{C_{n}^{\alpha}}\right)^{1/\alpha} \end{align*}.$$

由于

$$\lim_{n\to\infty}\left(\frac{C_{n}^{\alpha}}{C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}}\right)=\lim_{n\to\infty}\frac{C_{n+1}^{\alpha}-C_{n}^{\alpha}}{C_{n}^{\alpha}}=\lim_{n\to\infty}\left(1+\frac{c_{n+1}}{C_{n}}\right)^{\alpha}-1=0,$$

所以

$$\left(1+\frac{C_{n}^{\alpha}}{C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}}\right)^{1/\alpha}-1\sim\frac{C_{n}^{\alpha}}{\alpha\left(C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}\right)}.$$

原极限化为

$$\begin{align*} \lim_{n\to\infty}\frac{\left(C_{1}^{\alpha}+\cdots+C_{n}^{\alpha}\right)^{1/\alpha}}{C_{1}+\cdots+C_{n}} & =\lim_{n\to\infty}\left[\left(1+\frac{C_{n}^{\alpha}}{C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}}\right)^{1/\alpha}-1\right]\left(\frac{C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}}{C_{n}^{\alpha}}\right)^{1/\alpha}\\ & =\lim_{n\to\infty}\left[\frac{C_{n}^{\alpha}}{\alpha\left(C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}\right)}\right]\left(\frac{C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}}{C_{n}^{\alpha}}\right)^{1/\alpha}\\ & =\lim_{n\to\infty}\left[\frac{C_{n}^{\alpha}}{\alpha\left(C_{1}^{\alpha}+\cdots+C_{n-1}^{\alpha}\right)}\right]^{1-\frac{1}{\alpha}}=\left(\frac{0}{\alpha}\right)^{1-\frac{1}{\alpha}}=0. \end{align*}$$