以下题目是QQ群472466041的入群题，总的来说这个群还是非常不错的，里面有很多数学大佬，欢迎各位的加入。下面附上个人解析。

---

已知 \(\displaystyle a_n=\int_{0}^{1}x(1-x^3)^n{\rm d} x\),求 \(\displaystyle\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}\).

Solution:

\[\begin{align*} a_{n+1}&=\int_0^1x(1-x^3)^{n+1}{\rm d}x=\frac{1}{2}\int_0^1(1-x^3)^{n+1}{\rm d}x^2\\ &=\frac{1}{2}x^2(1-x^3)^{n+1}\bigg|_0^1+\frac{3(n+1)}{2}\int_0^1 x^4(1-x^3)^{n}{\rm d}x\\ &=\frac{3(n+1)}{2}\int_0^1 [x-x(1-x^3)](1-x^3)^{n}{\rm d}x\\ &=\frac{3(n+1)}{2}a_{n}-\frac{3(n+1)}{2}a_{n+1}\\ &=\frac{3n+3}{3n+5}\cdot a_n \end{align*} \]

所以

\[\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1 \]

也可以利用贝塔函数快速求解

\[\begin{align*} a_n&=\underbrace{\int_0^1x\left(1-x^3\right)^n\mathrm{~d}x}_{x^3\rightarrow x}\\ &=\frac{1}{3}\int_0^1x^{-\frac{1}{3}}(1-x)^n\mathrm{~d}x\\ &=\frac{1}{3}\mathrm{B}\left(\frac{2}{3},n+1\right) \end{align*} \]

\[\begin{align*} \therefore \lim_{n\to\infty}\frac{a_{n+1}}{a_n}&=\lim_{n\to\infty}\frac{\mathrm{B}\left(\frac{2}{3},n+2\right)}{\mathrm{B}\left(\frac{2}{3},n+1\right)}\\ &=\lim_{n\to\infty}\frac{\Gamma\left(n+2\right)\Gamma\left(n+\frac{5}{3}\right)}{\Gamma\left(n+1\right)\Gamma\left(n+\frac{8}{3}\right)}\\ &=\lim_{n\to\infty}\frac{n+1}{n+\frac{5}{3}}=1 \end{align*} \]

\(\square\)

---

已知 \(\{a_{n}\}\) 满足 \(\displaystyle a_n>0,\lim\limits_{n\to\infty}\left(a_{n}^2\sum_{k=1}^{n}a_k\right)=\frac{3}{2}\). 求极限 \(\lim\limits_{n\to\infty}a_n\sqrt[3]{n}\).

Solution:

设 \(\displaystyle S{n}=\sum_{k=1}^{n}a_k\).易知 \(\{S_{n}\}\) 严格单调递增且 \(\displaystyle \lim\limits_{n\to\infty}S_n=+\infty\),\(\lim\limits_{n\to\infty}a_{n}=0\).

由拉格朗日中值定理知

\[S_{n+1}^{\frac{3}{2}}-S_{n}^{\frac{3}{2}}=\frac{3}{2}\left(S_{n+1}-S_{n}\right) \sqrt{\xi}=\frac{3}{2} a_{n+1} \sqrt{\xi}, \xi \in\left(S_{n}, S_{n+1}\right) \]

又

\[a_{n+1} \sqrt{S_{n}}<a_{n+1} \sqrt{\xi}<a_{n+1} \sqrt{S_{n+1}} \]

且

\[\lim _{n \rightarrow \infty} a_{n+1} \sqrt{S_{n+1}}=\lim _{n \rightarrow \infty} \sqrt{a_{n}^{2} \sum_{k=1}^{n} a_{k}}=\sqrt{\frac{3}{2}} \]

\[\lim _{n \rightarrow \infty} a_{n+1} \sqrt{S_{n}}=\sqrt{\lim _{n \rightarrow \infty}\left(a_{n+1}^{2} S_{n+1}-a_{n+1}^{3}\right)}=\sqrt{\lim _{n \rightarrow \infty}\left(a_{n+1}^{2} S_{n+1}\right)}=\sqrt{\frac{3}{2}} \]

故由夹逼定理知

\[\lim _{n \rightarrow \infty} a_{n+1} \sqrt{\xi}=\sqrt{\frac{3}{2}} \Rightarrow \lim _{n \rightarrow \infty} S_{n+1}^{\frac{3}{2}}-S_{n}^{\frac{3}{2}}=\frac{3}{2} \sqrt{\frac{3}{2}} \]

由 Stolz 定理知

\[\begin{align*} \lim _{n \rightarrow \infty} a_{n} \sqrt[3]{n} &=\lim _{n \rightarrow \infty} \frac{a_{n} \sqrt{S_{n}} \sqrt[3]{n}}{\sqrt{S_{n}}} =\sqrt{\frac{3}{2}} \lim _{n \rightarrow \infty} \frac{\sqrt[3]{n}}{\sqrt{S_{n}}}\\ &=\sqrt{\frac{3}{2}} \sqrt[3]{\lim _{n \rightarrow \infty} \frac{n}{S_{n}^{\frac{3}{2}}}}=\sqrt{\frac{3}{2}} \sqrt[3]{\lim _{n \rightarrow \infty} \frac{(n+1)-n}{S_{n+1}^{\frac{3}{2}}-S_{n}^{\frac{3}{2}}}} \\ &=1 \end{align*} \]

\(\square\)

---

\[\frac{1}{\pi}\sum_{n=0}^{\infty} \frac{(n !)^{2} 2^{n+1}}{(2 n+1) !} \]

Solution:

\[\begin{align*} &\sum_{n=0}^{\infty} \frac{(n !)^{2} 2^{n+1}}{(2 n+1) !}\\ =&2 \sum_{n=0}^{\infty} 2^{n} \frac{\Gamma(n+1) \Gamma(n+1)}{\Gamma(2 n+2)} \\ =&2 \sum_{n=0}^{\infty} 2^{n} \int_{0}^{1}\left(x-x^{2}\right)^{n} \mathrm{~d} x \\ =&2 \int_{0}^{1} \sum_{n=0}^{\infty}\left(2 x-2 x^{2}\right)^{n} \mathrm{~d} x \\ =&2 \int_{0}^{1} \frac{\mathrm{d} x}{1-2 x+2 x^{2}} \\ =&2 \int_{0}^{1} \frac{\mathrm{d}(2 x-1)}{1+(2 x-1)^{2}} \\ =&\left.2 \arctan (2 x-1)\right|_{0} ^{1} \\ =&\pi \end{align*} \]

于是

\[\frac{1}{\pi}\sum_{n=0}^{\infty} \frac{(n !)^{2} 2^{n+1}}{(2 n+1) !}=1 \]

\(\square\)

---

\[2\int_0^{\pi/4}\frac{\tan^{2e}x-2\sin^2x}{\sin(2x)\cdot\ln(\tan x)}{\rm d}x-\int_{0}^{\infty}\frac{e^x-1}{xe^x(e^x+1)}{\rm d}x \]

Solution: 令

\[I=\int_0^{\pi/4}\frac{\tan^{2e}x-2\sin^2x}{\sin(2x)\cdot\ln(\tan x)}{\rm d}x \]

考虑到

\[\sin^2 x=\frac{\tan^2 x}{1+\tan^2x}\qquad \sin 2x=\frac{2\tan x}{1+\tan^2 x} \]

则

\[\begin{aligned} I&=\frac12\int_{0}^{\pi/4}\frac{\tan^{2e}x(1+\tan^2x)-2\tan^2x}{\tan x\cdot \ln\tan x\cdot(1+\tan^2x)}{\rm d}(\tan x)\\ &=\frac{1}{2} \int_{0}^{1} \frac{u^{2 e-1} \cdot\left(1+u^{2}\right)-2 u}{\ln u} \cdot \frac{1}{1+u^{2}} {\rm d} u=\frac{1}{2} \sum_{n=0}^{\infty}(-1)^{n} \int_{0}^{1} \frac{u^{2 e-1} \cdot\left(1+u^{2}\right)-2 u}{\ln u} \cdot u^{2 n} {\rm d} u \\ &=\frac{1}{2} \sum_{n=0}^{\infty}(-1)^{n} \cdot\left(\int_{0}^{1} \frac{u^{2(n+e)-1}-u^{2 n+1}}{\ln u} {\rm d} u+\int_{0}^{1} \frac{u^{2(n+e)+1}-u^{2 n+1}}{\ln u} {\rm d}u\right) \\ &=\frac{1}{2} \sum_{n=0}^{\infty}(-1)^{n} \cdot\left(\ln \frac{n+e}{n+1}+\ln \frac{n+e+1}{n+1}\right)\qquad(\rm Wallis) \\ &=\frac{1}{2}\left(1+\ln \frac{\pi}{2}\right) \end{aligned} \]

令

\[J=\int_{0}^{\infty}\frac{e^x-1}{xe^x(e^x+1)}{\rm d}x \]

则

\[\begin{align*} J&=\int_{0}^{\infty}\frac{1-e^{-x}}{xe^x(1+e^{-x})}{\rm d}x=\sum_{n=0}^{\infty}(-1)^n\int_{0}^{\infty}\frac{1-e^{-x}}{xe^x}e^{-nx}{\rm d}x\\ &=\sum_{n=0}^{\infty}(-1)^n\int_{0}^{\infty}\frac{e^{-(n+1)x}-e^{-(n+2)x}}{x}{\rm d}x\\ &=\sum_{n=0}^{\infty}(-1)^n\ln\left(\frac{n+2}{n+1}\right)\qquad(\rm Wallis)\\ &=\ln\frac{\pi}{2} \end{align*} \]

于是

\[2\int_0^{\pi/4}\frac{\tan^{2e}x-2\sin^2x}{\sin(2x)\cdot\ln(\tan x)}{\rm d}x-\int_{0}^{\infty}\frac{e^x-1}{xe^x(e^x+1)}{\rm d}x=1 \]

\(\square\)