阶的估计基础—习题解答

\(\qquad\)《阶的估计基础》\(^{\color{blue}{[1]}}\)(潘承洞、于秀源著)一书着重展现了常见估计方法的具体过程，是关于渐近分析的一本不错的入门读物。它是在第一版《阶的估计》的精简版本，想进一步了解估计技巧的读者可阅读《阶的估计》。
\(\qquad\)书中的正文和题目由于印刷等因素出现了一些错误，加之估计渐近式常需要比较复杂的计算，所以初读《阶的估计基础》者可能会遇到一定的困难。我将习题做了一些调整，选出\(41\)个题目给出解答。本系列文章所引用的定理、例题等对应着原书相应的内容，如\(1.3\)节的推论\(1\)，指的是《阶的估计基础》一书中第\(1\)章第\(3\)节的推论\(1\)。希望读者先自己尝试，当实在无法可循时再参考解答。
\(\qquad\)当然我给出的解答并非最佳的，如果有其他解法的朋友，欢迎一起交流。
\(\qquad\)为了更好的体验，请在电脑端阅读此文。

题1. 设\(f(x)\)为下列函数：

\[\begin{align*} (1)& \ \sqrt[\uproot{7}3]{x+\sqrt[\uproot{3}3]{x+\sqrt[\uproot{2}4]{x}}};\\ (2)& \ \sin \frac{1}{x^2}\log\Big(1+\frac{1}{x^a}\Big),a>0;\\ (3)& \ e^{\sin \frac{1}{x}+\cos \frac{1}{x}}\log \frac{1}{x+5}. \end{align*}\]

分别确定当\(x\to +\infty\)时\(f(x)\)的等价量.

解. (1)\(\ f(x)=\sqrt[\uproot{2}3]{x}\cdot \sqrt[\uproot{7}3]{1+x^{-1}\sqrt[\uproot{3}3]{x+\sqrt[\uproot{2}4]{x}}}\thicksim \sqrt[\uproot{2}3]{x},\ x\to +\infty.\)

(2)\(\ f(x)=\Big(x^{-2}+o\big(x^{-4}\big)\Big)\Big(x^{-a}+o\big(x^{-a}\big)\Big)\thicksim x^{-a-2},\ x\to +\infty.\)

(3)\(\ f(x)=e^{\frac{1}{x}+o\left(x^{-2}\right)+1+O\left(x^{-2}\right)}\log \frac{1}{x+5}\thicksim e^{1+\frac{1}{x}}\log \frac{1}{x},\ x\to +\infty.\qquad \vartriangleleft\)

题2. 设\(f(x)\)为下列函数：

\[(1)\tan^3 x-3\tan x,\ x\to \frac{\pi}{3};\ \ (2)\big(1-2x\big)^{x^{-1/2}},\ x\to 0;\ \ (3)x^x-1,\ x\to 1. \]

请分别确定\(f(x)\)的等价量 .

解. (1)利用\(\text{Lagrange}\)中值定理，在\(x\)与\(\frac{\pi}{3}\)间存在\(\xi\)，使

\[\begin{align*} f(x)& =\tan x\Big(\tan x+\tan \frac{\pi}{3}\Big)\Big(\tan x-\tan \frac{\pi}{3}\Big)\\ & =\tan x\Big(\tan x+\tan \frac{\pi}{3}\Big)\Big(x-\frac{\pi}{3}\Big)\sec^2 \xi. \end{align*}\]

于是当\(x\to \frac{\pi}{3}\)时，有

\[f(x)\thicksim \sqrt{3}\cdot 2\sqrt{3}\Big(x-\frac{\pi}{3}\Big)\sec^2 \frac{\pi}{3}=24\Big(x-\frac{\pi}{3}\Big). \]

(2)\(\ \log f(x)=x^{-\frac{1}{2}}\log(1-2x)\thicksim -2\sqrt{x}.\) 故\(f(x)\thicksim e^{-2\sqrt{x}}\to 1,\ x\to 0\).

(3)\(\ f(x)=e^{x\log x}-1\thicksim x\log x=x\log(1+x-1)\thicksim x-1,\ x\to 1.\qquad \vartriangleleft\)

题3. 当\(x\to +\infty\)时，按阶的大小将下列函数进行排列：

\[\left(\log x\right)^a;\ \ \left(\log\log x\right)^{\beta};\ \ x^a;\ \ x^{\left(\log x\right)^a};\ \ x^{\left(\log x\right)^{x^a}};\ \ e^{ax};\ \ e^{x^a};\ \ e^{\left(\log x\right)^A};\ \ \left(1+x\right)^a. \]

其中\(0<a<1,0<\beta<1,A>2.\)

解. 将这些函数依次编号\(a_1,a_2,\cdots,a_9\)，则它们阶的大小关系为

\[a_5>a_7>a_6>a_8>a_4>a_9>a_3>a_1>a_2. \]

鉴于本题和下题中的函数过多，遂略去比较的步骤，留给读者自行补充.\(\qquad \vartriangleleft\)

题4. 设\(a>0\)，当\(x\to 0^+\)时，按阶的大小将下列函数进行排列：

\[(x+1)\sin x^2;\ \ \cos x-1;\ \ \left(1+x\right)^x-1;\ \ x^x-1;\ \ e^{2x}-1-2x^2; \]

\[\sqrt[\uproot{3}3]{1-\sqrt{x}}-1;\ \ 2^{x^2}-1;\ \ 3^{-\frac{1}{x}};\ \ e^{-\frac{a}{x}};\ \ \left(\log x\right)^{-\frac{a}{x}};\ \ \left(\log x^{-1}\right)^{\log x}. \]

解. 将这些函数依次编号\(a_1,a_2,\cdots,a_{11}\)，当\(a>\log 3\)时，阶的大小关系为

\[a_{10}>a_9>a_8>a_{11}>a_1=a_2=a_3=a_7>a_5>a_6>a_4. \]

当\(a\leqslant \log 3\)时，\(a_{10}>a_8\geqslant a_9>a_{11}>\cdots\)，其他大小关系不变.\(\qquad \vartriangleleft\)

题5. 证明：若\(0<k<1\)，则

\[\int_0^1{\frac{\text{d}x}{\sqrt{\left(1-x^2\right)\left(1-k^2 x^2\right)}}}\thicksim \frac{1}{2}\log\frac{1}{1-k},\ k\to 1^-. \]

证. 将积分记为\(I(k)\)，并将被积函数对\(k^2\)作幂级数展开，可得\(^{\color{blue}{[2]}}\)

\[\begin{align*} I(k)& =\int_0^1\frac{\text{d}x}{\sqrt{1-x^2}}\left(\sum_{n=1}^{\infty}\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}k^{2n}x^{2n}\right)\\ & =\sum_{n=1}^{\infty}\left({\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}k^{2n}\cdot \int_0^1{\frac{x^{2n}}{\sqrt{1-x^2}}\text{d}x}}\right)\\ & =\sum_{n=1}^{\infty}\left({\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}k^{2n}\cdot \int_0^{\frac{\pi}{2}}{\sin^{2n}x\text{d}x}}\right)\\ & =\frac{\pi}{2}\sum_{n=1}^{\infty}{\left(\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}\right)^2 k^{2n}}. \end{align*}\]

利用\(\text{Wallis}\)公式，我们有

\[a_n:=\frac{\pi}{2}\left(\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}\right)^2\thicksim b_n:=\frac{1}{2n},\ n\to \infty. \]

依\(1.3\)节的推论\(1\)，可以得到

\[I(k)\thicksim \sum_{n=1}^{\infty}{\frac{\left(k^2\right)^n}{2n}}=\frac{1}{2}\log\frac{1}{1-k^2}\thicksim \frac{1}{2}\log\frac{1}{1-k},\ k\to 1^-. \]

其中最后一步是因\(\frac{1}{2}\log\frac{1}{1+k}=O(1)\).\(\qquad \vartriangleleft\)

题6. 设\(\sigma>0\)，若级数\(\sum_{n=1}^{\infty}\frac{a_n}{n^{\sigma}}\)收敛，则

\[\sum_{n=1}^{\infty}{a_n x^n}=o\left(\frac{1}{\left(1-x\right)^{\sigma}}\right),\ x\to 1^-. \]

证. 先证明一个命题\(^{\color{blue}{[3]}}\)：

引理1. 设\(\{b_n\}\)单调趋于0，且\(\sum_{n=1}^{\infty}{a_n b_n}\)收敛，则\(\underset{n\to \infty}{\lim}\left(a_1+a_2+\cdots+a_n\right)b_n=0.\)

\(\color{blue}{引理的证明：}\)不妨设\(\{b_n\}\)递减，则\(\forall \varepsilon>0,\ \exists N\in \mathbb{N},\ \forall n>N_1,\ \text{s.t.}\ \left|\sum_{k=N_1+1}^n{a_k b_k}\right|<\frac{\varepsilon}{6}\). 固定\(n\)，则\(\left\{\frac{b_n}{b_k}\right\}\)关于\(k\)单调增，由Abel引理，有

\[\left|\sum_{k=N_1+1}^n{a_k b_n}\right|=\left|\sum_{k=N_1+1}^n{a_k b_k\cdot\frac{b_n}{b_k}}\right|<\frac{\varepsilon}{6} \left(\frac{b_n}{b_{N_1+1}}+\frac{2b_n}{b_n}\right)\leqslant \frac{\varepsilon}{2}. \]

由于\(b_n\downarrow 0\)，对于上述给定的\(\varepsilon,N_1\)，\(\exists N_2\in \mathbb{N},\forall n>N_2,\ \text{s.t.}\ \left|(a_1+a_2+\cdots+a_{N_1})b_n\right|<\frac{\varepsilon}{2}\).

因此，当\(n>\max\{N_1,N_2\}\)时，

\[\left|\sum_{k=1}^n{a_k b_n}\right|\leqslant \left|\sum_{k=1}^{N_1}{a_k b_n}\right|+\left|\sum_{k=N_1+1}^n{a_k b_n}\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \]

引理得证！\(\qquad\square\)

接原证明：取\(b_n=\frac{1}{n^{\sigma}}\)，则\(\underset{n\to \infty}{\lim}\left(a_1+a_2+\cdots+a_n\right)n^{-\sigma}=0\).

设\(\{c_n\}\)满足\(\left(1-x\right)^{-\sigma}=\sum\limits_{n=0}^{\infty}{c_n x^n}\)，则

\[\left(1-x\right)^{-\sigma-1}=\frac{1}{1-x}\sum_{n=0}^{\infty}{c_n x^n}=\sum_{n=0}^{\infty}{\left(c_0+c_1+\cdots+c_n\right)x^n}. \]

因此

\[\sum_{j=0}^n{c_j}=\binom{\sigma+n}{n}=\frac{(\sigma+1)(\sigma+2)\cdots(\sigma+n)}{n!}\thicksim \frac{n^{\sigma}}{\sigma \Gamma(\sigma)},\ n\to \infty. \]

这蕴涵着\(a_1+a_2+\cdots+a_n=o\left(n^{\sigma}\right)=o(c_0+c_1+\cdots+c_n),\ n\to \infty.\)

利用\(1.3\)节的推论\(2\)，我们有

\[\sum_{n=1}^{\infty}{a_n x^n}=o\left(\sum_{n=0}^{\infty}{c_n x^n}\right)=o\left(\frac{1}{\left(1-x\right)^{\sigma}}\right),\ x\to 1^-. \]

证毕！\(\qquad \vartriangleleft\)

题7. 证明并确定\(c>0\)，使

\[\sum_{n=1}^{\infty}{\left(x^{n^2}-x^{2n^2}\right)}\thicksim \frac{c}{\sqrt{1-x}},\ x\to 1^-. \]

证. 易见

\[\int_0^{\infty}{x^{t^2}\text{d}t}\leqslant \sum_{n=0}^{\infty}{x^{n^2}}\leqslant 1+\int_0^{\infty}{x^{t^2}\text{d}t}. \]

当\(x\to 1^-\)时，

\[\int_0^{\infty}{x^{t^2}\text{d}t}=\int_0^{\infty}{e^{-t^2 \log\frac{1}{x}}\text{d}t}=\frac{1}{2}\sqrt{\frac{\pi}{\log x^{-1}}}\thicksim \frac{1}{2}\sqrt{\frac{\pi}{1-x}}. \]

因此，\(\sum_{n=0}^{\infty}{x^{n^2}}\thicksim \frac{1}{2}\sqrt{\frac{\pi}{1-x}},\ x\to 1^-\). 据此整理可得\(c=\frac{\left(2-\sqrt{2}\right)\sqrt{\pi}}{4}\).\(\qquad \vartriangleleft\)

\(\color{red}{注：}\) 本题亦可直接利用\(1.6\)节例\(6\)的结论.事实上，还可以展开任意多项\(^{\color{blue}{[4]}}\).

题8. 设\(b_n>0\)，级数\(\sum_{n=0}^{\infty}{b_n x^n}\)对一切\(x\)收敛，又设存在\(s<+\infty\)，使得

\[\underset{n\to \infty}{\lim}\frac{a_n}{b_n}=s, \]

则\(\sum_{n=0}^{\infty}{a_n x^n}\)也对一切\(x\)收敛，而且

\[\sum_{n=0}^{\infty}{a_n x^n}\thicksim s\sum_{n=0}^{\infty}{b_n x^n},\ x\to +\infty. \]

证. 下面对\(\text{Toeplitz}\)定理的一个推论述而不证：

引理2. 设函数\(\varphi_n(x)\)在\((0,1)\)内非负，且\(\sum_{n=0}^{\infty}{\varphi_n(x)}=1\)，则对固定的\(m\in \mathbb{N}\)，\(\underset{x\to +\infty}{\lim}\varphi_m(x)=0\)的充要条件是对每一个收敛于\(s\)的序列\(\{s_n\}\)，总有

\[\underset{x\to +\infty}{\lim}\left(s_0 \varphi_0(x)+s_1 \varphi_1(x)+\cdots+s_n \varphi_n(x)+\cdots\right)=s. \]

首先记

\[s_n:=\frac{a_n}{b_n},\ \varphi_n(x):=\frac{b_n x^n}{b_0+b_1 x+b_2 x^2+\cdots}. \]

给定\(\varepsilon>0\)及\(m\in \mathbb{N}\)，对充分大的\(x\)，选择\(n\)使得

\[b_0+b_1 x+b_2 x^2+\cdots+b_n x^n>\frac{b_m x^m}{\varepsilon}. \]

于是

\[\varphi_m(x)<\frac{b_m x^m}{b_0+b_1 x+\cdots+b_n x^n}<\varepsilon. \]

因此\(\underset{x\to +\infty}{\lim}\varphi_m(x)=0\). 利用引理\(2\)，我们有

\[\underset{x\to +\infty}{\lim}\sum_{n=0}^{\infty}{s_n \varphi_n(x)}=\underset{x\to \infty}{\lim}{\frac{a_0+a_1 x+a_2 x^2+\cdots}{b_0+b_1 x+b_2 x^2+\cdots}}=s. \]

故\(\sum_{n=0}^{\infty}{a_n x^n}\thicksim s\sum_{n=0}^{\infty}{b_n x^n},\ x\to +\infty.\) \(\qquad \vartriangleleft\)

题9. 设\(\underset{n\to \infty}{\lim}a_n=A\)存在且有限，则

\[\sum_{n=0}^{\infty}{a_n\frac{x^n}{n!}}=Ae^x\left(1+o(1)\right),\ x\to +\infty. \]

证. 利用引理\(1.2\)，并注意到\(\sum_{n=0}^{\infty}{\frac{A}{n!}x^n}=Ae^x\).\(\qquad \vartriangleleft\)

题10. 设\(\pi(x)\)是不超过\(x>0\)的素数个数，且\(\pi(x)\thicksim \frac{x}{\log x},x\to +\infty\)，则第\(n\)个素数\(p_n\thicksim ?\)

解. 注意到

\[\underset{x\to +\infty}{\lim}\frac{\pi(x)\log \pi(x)}{x}=\underset{x\to +\infty}{\lim}x^{-1}\frac{x}{\log x}\log \frac{x}{\log x}=1. \]

令\(x=p_n\)，则\(\pi(x)\log \pi(x)=n\log n\).利用上式，即有\(p_n\thicksim n\log n\).\(\qquad \vartriangleleft\)

\(\color{red}{注：}\) 更一般地\(^{\color{blue}{[5]}}\)，\(\forall n\geqslant 20,\ n\left(\log n+\log\log n-\frac{3}{2}\right)<p_n<n\left(\log n+\log\log n-\frac{1}{2}\right)\).

题11. 证明：\(\left(x+1+O\left(\frac{1}{x}\right)\right)^x=ex^x+O\left(x^{x-1}\right),\ x\to +\infty\).

证. 由\(x\log\left(x+1+O\left(\frac{1}{x}\right)\right)=x\log x+x\log\left(1+\frac{1}{x}+O\left(\frac{1}{x^2}\right)\right)\)，有

\[\text{LHS}=x^x\exp\left(1+O\left(\frac{1}{x}\right)\right)=ex^x\left(1+O\left(\frac{1}{x}\right)\right)=ex^x+O\left(x^{x-1}\right),\ x\to +\infty. \]

证毕！\(\qquad \vartriangleleft\)

题12. 证明：对\(\forall n\in \mathbb{N}\)，总有\(\int_1^x{\log^n t\text{d}t}=O\left(x\log^n x\right).\)
证. 设\(f(x,t)=\frac{\log^n t}{x\log^n x}(1\leqslant t\leqslant x)\)，则

\[f(x,t)=\exp\left(n\log\frac{\log t}{\log x}-\log x\right)=\exp\left(-\log x+O(1)\right)=O\left(\frac{1}{x}\right). \]

从而有\(\int_1^x{f(x,t)\text{d}t}=O\left(\frac{1}{x}\right)\int_1^x{\text{d}t}=O(1)\). 整理之即证毕.\(\qquad \vartriangleleft\)

\(\color{red}{注：}\) 本题证明的是不等式关系，即\(|f|\leqslant Cg\)，而非\(x\to 1^+\)或\(x\to \infty\)时的渐近式.

题13. 设\(\{na_n\}\)递减趋于0，\(\sum_{n=1}^{\infty}{a_n}\)收敛，则\(a_n=o\big(\frac{1}{n\log n}\big),\ n\to \infty\).

证. \(\forall \varepsilon>0,\exists N\in \mathbb{N},\forall n>N,\ \text{s.t.}\ \sum_{k=\left[\sqrt{n}\right]}^n{ka_k}<\varepsilon\). 于是有

\[\sum_{k=\left[\sqrt{n}\right]}^n{ka_k\cdot \frac{1}{k}}=o(1)\geqslant na_n\cdot \sum_{k=\left[\sqrt{n}\right]}^n{\frac{1}{k}}=na_n\left(\log n-\log\left[\sqrt{n}\right]\right)(1+o(1))=a_n\cdot O(n\log n). \]

因此，\(a_n=o\left(\frac{1}{n\log n}\right),\ n\to \infty\).\(\qquad \vartriangleleft\)

\(\color{red}{注：}\) 本题亦可利用引理\(1.1\).

题14. 证明：对\(\forall x>0\)，有\(\left(1+\frac{x}{n}\right)^{n+x}\thicksim e^x\left(1+\frac{x^2}{2n}\right),\ n\to \infty\).

证. 由于

\[(n+x)\log\left(1+\frac{x}{n}\right)=(n+x)\left(\frac{x}{n}-\frac{x^2}{2n^2}+o\left(\frac{1}{n^3}\right)\right)=x+\frac{x^2}{2n}+o\left(\frac{1}{n}\right), \]

所以

\[\left(1+\frac{x}{n}\right)^{n+x}=\exp\left(x+\frac{x^2}{2n}+o\left(\frac{1}{n}\right)\right)\thicksim e^x\left(1+\frac{x^2}{2n}\right),\ n\to \infty. \]

证毕！\(\qquad \vartriangleleft\)

题15. 设\(a_n\thicksim \frac{1}{\log n},\ n\to\infty\)，则\(\sum_{n=0}^{\infty}{a_n x^n}\thicksim \frac{1}{(x-1)\log(1-x)},\ x\to 1^-\).

证. 设\(S(j,r):=\sum_{k=j}^r{\frac{1}{(n-k+1)\log(k+2)}}\)，当\(|x|<1\)时，我们有

\[f(x):=\log\frac{1}{1-x}\cdot \sum_{k=2}^{\infty}{\frac{x^k}{\log k}}=\sum_{n=1}^{\infty}{\frac{x^{n+1}}{n+1}}\cdot \sum_{k=2}^{\infty}{\frac{x^{k+2}}{\log (k+2)}}=\sum_{n=0}^{\infty}{S(0,n)x^n}. \]

固定\(\varepsilon>0\)，使\(\varepsilon n\in \mathbb{N}\)，有\(S(0,\varepsilon n)\leqslant \frac{(\varepsilon n+1)}{(1-\varepsilon)n\log 2}\leqslant \frac{2\varepsilon}{(1-\varepsilon)\log 2}\). 同时有

\[\begin{align*} & \varlimsup_{n\to \infty}S(\varepsilon n,n)\leqslant \varlimsup_{n\to \infty}\frac{1}{\log \varepsilon n}\cdot \sum_{k=\varepsilon n}^n{\frac{1}{n-k+1}}=\underset{n\to \infty}{\lim}\frac{\log(1-\varepsilon)n}{\log \varepsilon n}=1;\\ & \varliminf_{n\to \infty}S(\varepsilon n,n)\geqslant \varliminf_{n\to \infty}\frac{1}{\log(n+2)}\cdot \sum_{k=\varepsilon n}^n{\frac{1}{n-k+1}}=\underset{n\to \infty}{\lim}\frac{\log(1-\varepsilon)n}{\log(n+2)}=1. \end{align*}\]

令\(\varepsilon \to 0^+\)，有\(\underset{n\to \infty}{\lim}S(0,n)=1\). 这蕴涵着\(\sum\limits_{n=0}^{\infty}{a_n x^n}\thicksim \frac{-f(x)}{\log(1-x)}\thicksim \frac{1}{(x-1)\log(1-x)},\ x\to 1^-\).\(\qquad \vartriangleleft\)

\(\color{red}{注：}\) 与\(S(0,n)\to 1\)类似的极限有

\[\underset{n\to \infty}{\lim}\frac{1}{n}\sum_{k=0}^n{\frac{\log(k+2)}{\log(n-k+2)}}=1,\quad \underset{n\to \infty}{\lim}\frac{1}{\sqrt{n}}\sum_{k=1}^n{\frac{\sqrt[k]{k}}{\sqrt{n+1-k}}}=2. \]

题16. 设\(x\ne 0,-1,-2,\cdots\)，证明：级数

\[\sum_{n=1}^{\infty}{\frac{n!a_n}{x(x+1)(x+2)\cdots(x+n)}} \]

与级数\(\sum_{n=1}^{\infty}{\frac{a_n}{n^x}}\)有相同的收敛性.

证. 对\(x\ne 0,-1,-2,\cdots\)，定义

\[b_n:=\frac{n!n^x}{x(x+1)(x+2)\cdots(x+n)},\ n=1,2,\cdots. \]

接下来证明\(\{b_n\}\)对固定的\(x\)是单调有界的，再结合\(\text{Abel}\)判别法，即证得结论.

由\(\Gamma(x)\)的无穷乘积表示\(^{\color{blue}{[6]}}\)，有\(\underset{n\to \infty}{\lim}b_n=\Gamma(x)\)，这蕴涵着有界性. 同时，我们有\(^{\color{blue}{[7]}}\)

\[\frac{b_{n+1}}{b_n}=\left(1+\frac{1}{n}\right)^{x+1}\left(1+\frac{x+1}{n}\right)^{-1}=1+\frac{x(x+1)}{2n^2}+o\left(\frac{1}{n^2}\right). \]

\(x(x+1)>0\)时，比值最后将大于\(1\)；\(x(x+1)<0\)时，最后将小于\(1\). 即\(\{b_n\}\)单调有界.\(\qquad \vartriangleleft\)

题17. 求极限：

\[(1)\underset{x\to 0}{\lim}x\int_x^1{\frac{\cos t}{t^2}\text{d}t};\quad (2)\underset{x\to +\infty}{\lim}\frac{\int_0^x{\sqrt{1+t^4}\text{d}t}}{x^3};\quad (3)\underset{x\to 0^+}{\lim}\frac{\int_x^{\infty}{t^{-1}e^{-t}\text{d}t}}{\log x}. \]

证. (1)首先分部积分，有

\[\int_x^1{\frac{\cos t}{t^2}\text{d}t}=\frac{\cos x}{x}-\cos 1+\int_x^1{\frac{\sin t}{t}\text{d}t}=\frac{\cos x}{x}+O(1). \]

因此，极限为\(\underset{x\to 0}{\lim}\left(\cos x+O(x)\right)=1.\)

(2)使用一次\(\text{L'Hospital}\)法则，则得极限为\(\frac{1}{3}\).

(3)当\(0<t<1\)时，\(-1<\frac{1}{t}\left(e^{-t}-1\right)<e^{-1}-1.\)从而有

\[\int_x^{\infty}{\frac{\text{d}t}{te^t}}=\log \frac{1}{x}+\int_x^1{\frac{e^{-t}-1}{t}\text{d}t}+\int_1^{\infty}{\frac{\text{d}t}{te^t}}=\log \frac{1}{x}+O(1). \]

因此，极限为\(-1\).\(\qquad \vartriangleleft\)

\(\color{red}{注：}\)切勿滥用\(\text{L'Hospital}\)法则解题，只有在满足条件时才能使用.

题18. 设正项数列\(\{a_n\}\)满足条件

\[\frac{a_{n+1}}{a_n}=1-\frac{1}{n}-\frac{\alpha_n}{n\log n},\ n=2,3,\cdots, \]

则级数\(\sum_{n=2}^{\infty}{a_n}\)当\(\alpha_n \geqslant \alpha>1\)时收敛，而当\(\alpha_n \leqslant \alpha'<1\)时发散.

证. 不妨设\(n\geqslant 2\)时，\(\left|\frac{1}{n}+\frac{\alpha_n}{n\log n}\right|<\frac{1}{2}.\)同时有

\[\begin{align*} a_n=a_2\prod_{k=2}^{n-1}{\left(1-\frac{1}{k}-\frac{\alpha_k}{k\log k}\right)},\ n=3,4,\cdots.\tag{1.1} \end{align*}\]

我们只需证明\(\alpha_n=\alpha>1\)和\(\alpha_n=\alpha'<1\)时的结论均成立.

事实上，依\((1.1)\)式可知在这两种情况下，数列分别被缩小、扩大，再利用比较审敛法，便证得结论. 因此，不妨设\(\alpha_n \equiv a\).

由于

\[\begin{align*} \sum_{k=2}^{n-1}{\log \left(1-\frac{1}{k}-\frac{a}{k\log k}\right)}& =\sum_{k=2}^{n-1}{\left(-\frac{1}{k}-\frac{a}{k\log k}+O\left(\frac{1}{k^2}\right)\right)}\\ & =-\log n-a\log\log n+c'+O\left(n^{-1}\right). \end{align*}\]

所以当\(n\to \infty\)时，有

\[a_n=cn^{-1}\left(\log n\right)^{-a}e^{O(\frac{1}{n})}=\frac{c}{n\log^a n}\left(1+O\left(\frac{1}{n}\right)\right). \]

其中\(c,c'\)是常数且\(c\ne 0\). 易见，\(c<1\)时，\(\sum{a_n}\)收敛；\(c>1\)时，\(\sum{a_n}\)收敛.\(\qquad \vartriangleleft\)

题19. 设级数

\[\sum_{n=1}^{\infty}{a_n\left(x^2-1\right)\left(x^2-2^2\right)\cdots\left(x^2-n^2\right)} \]

在点\(x=x_0\not\in \mathbb{Z}\)收敛，则它在所有点\(x\in (-\infty,+\infty)\)收敛.

证. 对\(x_0\not\in \mathbb{Z},x\in \mathbb{R}\)，定义

\[b_n:=\frac{\left(x^2-1\right)\left(x^2-2^2\right)\cdots\left(x^2-n^2\right)}{\left(x_0^2-1\right)\left(x_0^2-2^2\right)\cdots\left(x_0^2-n^2\right)}. \]

易见，它是单调变化的. 由于

\[\alpha_n:=\frac{x^2-n^2}{x_0^2-n^2}=1+\frac{x^2-x_0^2}{x_0^2-n^2}=1+O\left(\frac{1}{n^2}\right). \]

所以\(\underset{n\to \infty}{\lim}b_n=\prod{\alpha_n}\)收敛，故\(\{b_n\}\)有界. 利用\(\text{Abel}\)判别法，原级数在\(x\in \mathbb{R}\)点收敛.\(\qquad \vartriangleleft\)

题20. 判定无穷乘积\(\prod{a_n}\)的收敛性，其通项为：

\[\begin{align*} & (1)a_n=1+\frac{1}{x^n}\left(1+\frac{1}{n}\right)^{n^\alpha};\qquad\ (2)a_n=1+\frac{1}{n^{\alpha}}\left(1-\frac{x\log n}{n}\right)^n;\\ & (3)a_n=1+\frac{2n+1}{\left(n^2-1\right)\left(n+1\right)^{\alpha}};\quad (4)a_n=1+\frac{n^{\alpha}}{\left(1+\sin \frac{\pi}{n}\right)^{2n}}. \end{align*}\]

解. 下面仅将答案陈列出来，留给读者自证.

(1)\(\alpha>2\)时发散\(;\alpha=2\)时\(,|x|\leqslant e\)发散\(,|x|>e\)收敛\(;\alpha<2\)时\(,|x|\leqslant 1\)发散\(,|x|>1\)收敛.

(2)\(x+\alpha>1\)收敛\(,x+\alpha \leqslant 1\)发散.\(\quad\)(3)\(\alpha>0\)收敛\(,\alpha \leqslant 0\)发散.

(4)\(\alpha>1\)收敛\(,\alpha \leqslant 1\)发散.\(\qquad \vartriangleleft\)

题21. 证明：当\(\alpha<-1,\beta<-2\)或\(-1<\alpha<\frac{\beta}{2}-1\)时，\(\int_0^{\infty}{x^{\alpha}\left|\cos x\right|^{x^{\beta}}\text{d}x}\)收敛.

证. 将原积分写成两项，我们有

\[I:=\int_0^{\infty}{x^{\alpha}\left|\cos x\right|^{x^{\beta}}\text{d}x}=\left(\int_0^1+\int_1^{\infty}\right)x^{\alpha}\left|\cos x\right|^{x^{\beta}}\text{d}x=I_1+I_2. \]

(1)\(\alpha<-1,\beta<-2\)时，\(I_2\)显然收敛. 而\(x\to 0^+\)时\(x^{\alpha}\left|\cos x\right|^{x^{\beta}}<\exp \left(\frac{1}{2}x^{\beta}\log|x|\right)\)，故\(I_1\)收敛.

(2)\(-1<\alpha<\frac{\beta}{2}-1\)时，\(I_1\)显然收敛，且有

\[\begin{align*} I_2& =\int_1^{\frac{\pi}{2}}{x^{\alpha}\left|\cos x\right|^{x^{\beta}}\text{d}x}+\sum_{k=1}^{\infty}{\int_{k\pi+\frac{\pi}{2}}^{k\pi+\frac{\pi}{2}}{x^{\alpha}\left|\cos x\right|^{x^{\beta}}\text{d}x}}\\ & =c_1+\sum_{k=1}^{\infty}{\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\left(x+k\pi\right)^{\alpha}\left|\cos x\right|^{\left(x+k\pi\right)^{\beta}}\text{d}x}}\\ & <c_1+\sum_{k=1}^{\infty}{\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\left(k\pi+\frac{\pi}{2}\right)^{\alpha}\left|\cos x\right|^{\left(k\pi-\frac{\pi}{2}\right)^{\beta}}\text{d}x}}\\ & \overset{(a)}{<}c_1+c_{\beta}\sum_{k=1}^{\infty}{\left(k\pi+\frac{\pi}{2}\right)^{\alpha-\frac{\beta}{2}}}<+\infty. \end{align*}\]

其中\((a)\)处利用了\(\text{Stirling}\)公式，\(c_{\beta}\)是只依赖于\(\beta\)的常数. 综合(1)(2)情况，命题成立.\(\qquad \vartriangleleft\)

题22. 证明：当且仅当\(x\geqslant \frac{1}{2}\)时，数列

\[a_n=\left(1+\frac{1}{n}\right)^n\left(1+\frac{x}{n}\right),\ n=1,2,3,\cdots \]

单调减少.

证. (1)充分性：将\(a_n\)表示为

\[a_n=\left(1+\frac{1}{n}\right)^{n+1/2}\left(1+\frac{x}{n}\right)\left(1+\frac{1}{n}\right)^{-1/2}. \]

其中第一个因子递减，第二个因子的平方\(1+\frac{2x-1}{n+1}+\frac{x^2}{n(n+1)}\)在\(x\geqslant \frac{1}{2}\)时关于\(n\)递减.

(2)必要性：当\(|x|<1\)时，我们有

\[\log \frac{1+x}{1-x}=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right). \]

分别以\(\frac{1}{2n+1},\frac{x}{2n+x}\)代替\(x\)，得到

\[\begin{align*} \log a_n& =n\log\left(1+\frac{1}{n}\right)+\log\left(1+\frac{x}{n}\right)\\ & =\frac{2n}{2n+1}+\frac{2x}{2n+x}+\frac{1}{12n^2}+O\left(\frac{1}{n^3}\right). \end{align*}\]

于是有

\[\log a_n-\log a_{n+1}=\frac{4x-2}{4n^2}+O\left(\frac{1}{n^3}\right). \]

这蕴涵着必要性成立.\(\qquad \vartriangleleft\)

题23. 设\(r>0\)，数列\(\{a_n\},\{b_n\}(n\in \mathbb{N})\)满足条件：

\[(1)b_n\ne 0;\quad (2)f(x)=\sum_{n=0}^{\infty}{a_n x^n}(|x|<r)\text{收敛};\quad (3)\underset{n\to \infty}{\lim}\frac{b_n}{b_{n+1}}=q,|q|<r. \]

令\(c_n=a_0 b_n+a_1 b_{n-1}+\cdots+a_n b_0(n=1,2,\cdots)\)，则\(\underset{n\to \infty}{\lim}\frac{c_n}{b_n}=f(q).\)

证. 取充分小的\(\varepsilon>0\)，使\(|q|+\varepsilon<r\)，则存在不依赖于\(n\)和\(k\)的常数\(A\)，使

\[\left|\frac{b_{n-k}}{b_n}\right|<A\left(|q|+\varepsilon\right)^k,\ k=0,1,\cdots,n;n=0,1,\cdots. \]

当\(n>m\)时，得到

\[\begin{align*} \frac{c_n}{b_n}-f(q)=\sum_{k=0}^m{a_k\left(\frac{b_{n-k}}{b_n}-q^k\right)}+\sum_{k=m+1}^n{a_k\left(\frac{b_{n-k}}{b_n}-q^k\right)}.\tag{1.2} \end{align*}\]

第二项的绝对值小于

\[g_1(m):=A\sum_{k=m+1}^{\infty}{|a_k|\left(|q|+\varepsilon\right)^k}+\sum_{k=m+1}^{\infty}{|a_k|\cdot\left|q\right|^k}. \]

易见，对\(|t|\leqslant |q|+\varepsilon\)，级数\(\sum{a_n t^n}\)绝对收敛. 事实上，我们有

\[\sum{\left|a_n t^n\right|}=\sum{|a_n|\left(|q|+\varepsilon\right)^n\cdot \left(\frac{|t|}{|q|+\varepsilon}\right)^n}\leqslant M\sum{\left(\frac{|t|}{|q|+\varepsilon}\right)^n}<+\infty. \]

固定\(m\)，选取\(n\)，使得

\[g_2(m)=\left|\sum_{k=0}^m{a_k\left(\frac{b_{n-k}}{b_n}-q^k\right)}\right|<\frac{\varepsilon}{2}. \]

选取充分大的\(m\)，使\(2g_1(m)<\varepsilon\)，从而有\(g_1(m)+g_2(m)<\varepsilon\). 所以\((1.2)\)式左边即为无穷小.\(\qquad \vartriangleleft\)

题24. 证明下面的二重积分可以交换积分次序：

\[\int_0^{\infty}{\text{d}x\int_0^{\infty}{y(\cos2mx)e^{-y^2\left(1+x^2\right)}\text{d}y}}. \]

证. 设\(f(x,y):=e^{-y^2\left(1+x^2\right)}\cos(2mx)\)，则

\[g(x):=\int_0^{\infty}{ye^{-y^2\left(1+x^2\right)}\text{d}y},\ a\leqslant x\leqslant A \]

是一致收敛的，其中\((A>a>0)\). 于是有

\[\begin{align*} I& :=\int_a^A{g(x)\cos(2mx)\text{d}x}=\int_0^{\infty}{y\text{d}y\int_a^A{f(x,y)\text{d}x}}\\ & =\int_0^{\infty}{y\text{d}y\int_0^{\infty}{f(x,y)\text{d}x}}-\int_0^{\infty}{y\text{d}y\int_0^a{f(x,y)\text{d}x}}-\int_0^{\infty}{y\text{d}y\int_A^{\infty}{f(x,y)\text{d}x}}\\ & =\int_0^{\infty}{y\text{d}y\int_0^{\infty}{f(x,y)\text{d}x}}-I_1-I_2. \end{align*}\]

只需证明：\(\underset{a\to 0^+}{\lim}I_1=\underset{A\to \infty}{\lim}I_2=0\). 首先，我们有

\[|I_1|\leqslant \int_0^{\infty}{y\text{d}y\int_0^a{e^{-y^2\left(1+x^2\right)}\text{d}x}}\leqslant a\int_0^{\infty}{ye^{-y^2}\text{d}y}=\frac{a}{2}. \]

现将\(I_2\)写成两部分，可得

\[I_2=\int_0^1{y\text{d}y\int_A^{\infty}{f(x,y)\text{d}x}}+\int_1^{\infty}{y\text{d}y\int_A^{\infty}{f(x,y)\text{d}x}}=I_2^1+I_2^2. \]

当\(A\to \infty\)时，\(\int_A^{\infty}{f(x,y)\text{d}x}\to 0\)，因此

\[I_2^1=\int_0^1{y\cdot o(1)\text{d}y}=o(1). \]

同时有

\[\begin{align*} \left|I_2^2\right|\leqslant \int_1^{\infty}{y\text{d}y\int_A^{\infty}{e^{-y^2\left(1+x^2\right)}\text{d}x}}=\int_1^{\infty}{ye^{-y^2}\text{d}y\int_A^{\infty}{e^{-x^2 y^2}\text{d}x}}.\tag{1.3} \end{align*}\]

注意到

\[\int_A^{\infty}{e^{-x^2 y^2}\text{d}x}=\int_A^{\infty}{\frac{-1}{2xy^2}\text{d}e^{-x^2 y^2}}=\frac{1}{2Ay^2}e^{-A^2 y^2}-\frac{1}{2y^2}\int_A^{\infty}{x^{-1}e^{-x^2 y^2}\text{d}x}<\frac{1}{2Ay^2}e^{-A^2 y^2}. \]

因此\((1.3)\)式化为

\[\left|I_2^2\right|<\frac{1}{2A}\int_1^{\infty}{y^{-1}e^{-y^2\left(1+A^2\right)}\text{d}y}=o(1). \]

这便证明了\(\underset{a\to 0^+}{\lim}I_1=\underset{A\to \infty}{\lim}I_2=0\).\(\qquad \vartriangleleft\)

题25. 设\(c>b>0,c>a+b\)，证明：

\[\sum_{n=0}^{\infty}{\frac{a(a+1)\cdots(a+n-1)}{n!}\int_0^1{t^{b+n-1}\left(1-t\right)^{c-b-1}\text{d}t}}=\int_0^1{t^{b-1}\left(1-t\right)^{c-a-b-1}\text{d}t}. \]

证. 将\(\left(1-t\right)^{-a}\)展开为幂级数，并仿照\(2.3\)节的例\(3\)来处理.\(\qquad \vartriangleleft\)

题26. 设\(\varphi(x)\)在\(x>0\)定义，数列\(\{a_n\}_{n=0}^{\infty}\subset \mathbb{R}\)，且\(x\geqslant 1\)时有

\[\varphi(x)=a_0+\frac{a_1}{x}+\frac{a_2}{x^2}+\cdots+\frac{a_n}{x^n}+\cdots. \]

证明无穷乘积\(\varphi(1)\varphi(2)\cdots \varphi(n)\cdots\)收敛的充要条件是\(a_0=1,a_1=0\).

证. 无穷乘积收敛的必要条件是\(\underset{n\to \infty}{\lim}\varphi(n)=a_0=1\)，则

\[\psi(x):=\log \varphi(n)=\frac{a_1}{x}+\frac{a_2-\frac{1}{2}a_1^2}{x^2}+\cdots. \]

应用\(2.4\)节的例\(2\)，可知\(\sum{\psi(n)}\)收敛的充要条件是\(a_1=0\).证毕.\(\qquad \vartriangleleft\)

题27. 设\(\frac{1}{2}\leqslant \sigma<1\)，证明：

\[\sum_{1\leqslant m<n}{\frac{e^{-(m+n)\delta}}{m^{\sigma}n^{\sigma}\log \frac{n}{m}}}=O\left({\delta}^{2\sigma-2}\log \frac{1}{\delta}\right) \]

当\(\delta>0\)时成立.

证. 令\(T=\delta^{-1}\)，则有

\[S^*(T)=\sum_{T\leqslant m<n}{\frac{e^{-(m+n)\delta}}{m^{\sigma}n^{\sigma}\log \frac{n}{m}}}=\sum_{T\leqslant m<n}{\frac{e^{-(m+n)\frac{\delta}{2}}}{m^{\frac{\sigma}{2}}n^{\frac{\sigma}{2}}}\cdot O(1)}=O\left(\delta^{\sigma}\right). \]

结合\(3.5\)节的例\(3\)，我们有

\[S(T)=\sum_{m<n<T}{\frac{e^{-(m+n)\delta}}{m^{\sigma}n^{\sigma}\log \frac{n}{m}}}\leqslant \sum_{m<n<T}{\frac{1}{m^{\sigma}n^{\sigma}\log \frac{n}{m}}}=O\left(T^{2-2\sigma}\log T\right). \]

因此，\(S(T)+S^*(T)=O\left({\delta}^{2\sigma-2}\log \frac{1}{\delta}\right)\).\(\qquad \vartriangleleft\)

题28. 证明：

\[\begin{align*} & (1)\sum_{n=0}^{\infty}{\left(-1\right)^n\left(\frac{1}{1\cdot (2n+1)}+\frac{1}{3\cdot (2n-1)}+\cdots+\frac{1}{(2n+1)\cdot 1}\right)}=\frac{\pi^2}{16};\\ & (2)\sum_{n=1}^{\infty}{\frac{\left(-1\right)^n}{n+1}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)}=-\frac{1}{2}\log^2 2. \end{align*}\]

证. (1)逆用级数的\(\text{Cauchy}\)乘积，有

\[\mathrm{LHS}=\sum_{n=1}^{\infty}{\sum_{i+j=n+1}{\frac{\left(-1\right)^{i+j-2}}{(2i-1)(2j-1)}}}=\left(\sum_{n=1}^{\infty}{\frac{\left(-1\right)^{k-1}}{2k-1}}\right)^2=\frac{\pi^2}{16}. \]

(2)仿照(1)，有

\[\begin{align*} \mathrm{LHS}& =\sum_{n=1}^{\infty}{\frac{\left(-1\right)^n}{n+1}\sum_{k=1}^n{\frac{1}{k}}}=\frac{1}{2}\sum_{n=1}^{\infty}{\frac{\left(-1\right)^n}{n+1}\sum_{i+j=n+1}{\left(\frac{1}{i}+\frac{1}{j}\right)}}\\ & =\frac{1}{2}\sum_{n=1}^{\infty}{\sum_{i+j=n+1}{\frac{\left(-1\right)^{i+j-1}}{i+j}\left(\frac{1}{i}+\frac{1}{j}\right)}}=-\frac{1}{2}\sum_{n=1}^{\infty}{\sum_{i+j=n+1}{\frac{\left(-1\right)^{i+j}}{ij}}}\\ & =-\frac{1}{2}\left(\sum_{n=1}^{\infty}{\frac{\left(-1\right)^n}{n}}\right)^2=-\frac{1}{2}\log^2 2. \end{align*}\]

\(\color{red}{注：}\) 事实上，本题解法不唯一.

题29. 证明：当\(t\to 0^+\)时，

\[\sum_{n=1}^{\infty}{\frac{1}{n}e^{-n^2 t}}=-\frac{1}{2}\log t-\frac{1}{2}\gamma+O\left(\sqrt{t}\right). \]

其中\(\gamma\)是\(\text{Euler-Mascheroni}\) 常数.

证. 利用\(\text{Euler--Maclaurin}\)公式，有

\[f(t):=\sum_{n=1}^{\infty}{\frac{1}{n}e^{-n^2 t}}=\frac{1}{2}e^{-t}+\int_1^{\infty}{\frac{1}{x}e^{-x^2 t}\text{d}x}-\int_1^{\infty}{\left(2t+\frac{1}{x^2}\right)e^{-x^2 t}\left(\{x\}-\frac{1}{2}\right)\text{d}x}. \]

由于\(x^{-1}\text{d}x=\text{d}(\log x)\)，再分部积分，可得

\[f(t)=\frac{1}{2}e^{-t}+2t\int_1^{\infty}{xe^{-x^2 t}\log x\text{d}x}-\int_1^{\infty}{\left(2t+\frac{1}{x^2}\right)e^{-x^2 t}\left(\{x\}-\frac{1}{2}\right)\text{d}x}. \]

由于

\[\int_1^{\infty}{2te^{-x^2 t}\{x\}\text{d}x}=2t\int_1^{\infty}{e^{-x^2 t}O(1)\text{d}x}\xlongequal{u=x^2 t} O\left(\sqrt{t}\right)\int_t^{\infty}{u^{-\frac{1}{2}}e^{-u}\text{d}u}=O\left(\sqrt{t}\right), \]

所以

\[\begin{align*} f(t)\xlongequal{u=x^2 t} e^{-t}+O\left(\sqrt{t}\right)+\frac{1}{2}\int_t^{\infty}{e^{-u}\log u\text{d}u}-\frac{1}{2}e^{-t}\log t-\int_1^{\infty}{x^{-2}e^{-x^2 t}\{x\}\text{d}x}.\tag{1.4} \end{align*}\]

通过\(\text{L'Hospital}\)法则，不难验证

\[\int_t^{\infty}{e^{-u}\log u\text{d}u}=\int_0^{\infty}{e^{-u}\log u\text{d}u}-\int_0^t{e^{-u}\log u\text{d}u}=-\gamma+o\left(\sqrt{t}\right). \]

考虑到

\[\begin{align*} \int_1^{\infty}{\frac{\{x\}}{x^2}e^{-x^2 t}\text{d}x} & =\int_1^{\frac{1}{\sqrt{t}}}+\int_{\frac{1}{\sqrt{t}}}^{\infty}=\int_1^{\frac{1}{\sqrt{t}}}{\frac{\{x\}}{x^2}\left(1+O\left(\sqrt{t}\right)\right)\text{d}x}+\int_{\frac{1}{\sqrt{t}}}^{\infty}{\frac{O(1)}{x^2}\text{d}x}\\ & =O\left(\sqrt{t}\right)+\left(1+O\left(\sqrt{t}\right)\right)\int_1^{\infty}{\frac{\{x\}}{x^2}\text{d}x}=1-\gamma+O\left(\sqrt{t}\right). \end{align*}\]

结合上式与\((1.4)\)式，即得\(f(t)=-\frac{1}{2}\log t-\frac{1}{2}\gamma+O\left(\sqrt{t}\right).\qquad \vartriangleleft\)

题30. 证明：当\(n\to \infty\)时，有

\[\frac{1}{\log n}-\frac{1}{\log(n+1)}+\frac{1}{\log(n+2)}-\cdots=\frac{1}{2}\log n+O\left(\frac{1}{n\log^2 n}\right). \]

证. 设\(f(x)=\frac{1}{\log(n+x)}(x\geqslant 0)\)，则\(f'(x)=-\frac{1}{(x+n)\log^2(x+n)},f''(x)\geqslant 0\).

根据\(3.3\)节的定理\(1\)，有

\[\sum_{k=0}^{2N+1}{\left(-1\right)^k f(k)}=\frac{1}{2\log n}-\frac{1}{2\log(n+2N+2)}+O\left(\int_0^{2N+2}{f''(x)\text{d}x}\right). \]

其中\(\int_0^{2N+2}{f''(x)\text{d}x}=f'(2N+2)-f'(0)\). 令\(N\to \infty\)，有

\[\sum_{k=0}^{\infty}{\left(-1\right)^k f(k)}=\frac{1}{2}\log n+O(f'(0))=\frac{1}{2}\log n+O\left(\frac{1}{n\log^2 n}\right). \]

命题得证.\(\qquad \vartriangleleft\)

题31. 以\(p\)表示素数，已知对任意的\(m\in \mathbb{Z}_+\)，有

\[\sum_{p\leqslant x}{\log p}=x+O\left(x\left(\log x\right)^{-m}\right). \]

证明：当\(t\to 0^+\)时，有

\[\sum_p{e^{-pt}}=-\frac{1}{t\log t}(1+o(1)). \]

证. 取\(a_n=\log p_n,b(x)=\frac{e^{-tx}}{\log x}(x>1)\)，其中\(p_n\)是第\(n\)个素数，则

\[\begin{align*} \sum_{p\leqslant x}{e^{-pt}}& =\sum_{p_n\leqslant x}{a_n b(p_n)}=\frac{S(x)}{e^{tx}\log x}+\int_2^x{S(y)\cdot\frac{1+ty\log y}{y\log^2 y}\cdot \frac{\text{d}y}{e^{ty}}}\\ & =\frac{x}{e^{tx}\log x}\left(1+O\left(\frac{1}{\log^m x}\right)\right)+\int_2^x{\frac{1+ty\log y}{\log^2 y}\left(1+O\left(\frac{1}{\log^m y}\right)\right)\frac{\text{d}y}{e^{ty}}}. \end{align*}\]

令\(x\to \infty\)，有

\[\sum_{p\leqslant x}{e^{-pt}}=t\int_2^{\infty}{\frac{y}{\log y}e^{-ty}\text{d}y}(1+o(1))+O(1)\int_2^{\infty}{\frac{e^{-ty}}{\log^2 y}\text{d}y}. \]

现在我们来估计上式的第一项，第二项则可按照完全相同的步骤得到，即

\[O(1)\int_2^{\infty}{\frac{e^{-ty}}{\log^2 y}\text{d}y}=O\left(\frac{1}{t\log^2 t}\right),\ t\to 0^+. \]

将第一项写成三部分，并记\(\phi :=\frac{1}{t}\left(1+\log\frac{1}{t}\right)\)，有

\[I=\int_2^{\frac{1}{t}}+\int_{\frac{1}{t}}^{\phi}+\int_{\phi}^{\infty}{\frac{y}{\log y}e^{-ty}\text{d}y}\overset{\bigtriangleup}{=}I_1+I_2+I_3. \]

今取充分小的\(t>0\)，则当\(x\in [2t,1]\)时，有

\[0\leqslant \frac{\log x}{\log t}\leqslant 1+\frac{\log 2}{\log t}<1. \]

这蕴涵着

\[0\leqslant \sum_{n=2}^{\infty}{\left(\frac{\log x}{\log t}\right)^{n-2}}\leqslant \sum_{n=0}^{\infty}{\left(1+\frac{\log x}{\log t}\right)^n}=-\frac{\log t}{\log 2}. \]

因此，

\[\sum_{n=0}^{\infty}{\left(\frac{\log x}{\log t}\right)^n}=1+\frac{\log x}{\log t}+\left(\frac{\log x}{\log t}\right)^2\sum_{n=2}^{\infty}{\left(\frac{\log x}{\log t}\right)^{n-2}}=1+\frac{\log x}{\log t}+O\left(\frac{\log^2 x}{\log\frac{1}{t}}\right). \]

从而可得

\[\begin{align*} I_1& \overset{x=ty}{=}-\frac{1}{t^2\log t}\int_{2t}^1{xe^{-x}\left(1-\frac{\log x}{\log t}\right)^{-1}\text{d}x}=-\frac{1}{t^2\log t}\int_{2t}^1{xe^{-x}\sum_{n=0}^{\infty}{\left(\frac{\log x}{\log t}\right)^n}\text{d}x}\\ & =-\frac{1}{t^2\log t}\int_{2t}^1{xe^{-x}\text{d}x}(1+o(1))=-\frac{1}{t^2\log t}\int_0^1{xe^{-x}\text{d}x}(1+o(1)). \end{align*}\]

当\(x\in \left[1,1+\log\frac{1}{t}\right]\)时，\(1\leqslant 1-\frac{\log x}{\log t}\leqslant 1+\frac{\log\left(1+\log\frac{1}{t}\right)}{\log\frac{1}{t}}\to 1.\) 于是有

\[\begin{align*} I_2& =-\frac{1}{t^2\log t}\int_1^{1+\log\frac{1}{t}}{xe^{-x}\left(1-\frac{\log x}{\log t}\right)^{-1}\text{d}x}=-\frac{1+o(1)}{t^2\log t}\int_1^{1+\log\frac{1}{t}}{xe^{-x}\text{d}x}\\ & =\frac{1+o(1)}{t^2\log t}\int_{1+\log\frac{1}{t}}^{\infty}{xe^{-x}\text{d}x}-\frac{1+o(1)}{t^2\log t}\int_1^{\infty}{xe^{-x}\text{d}x}=-\frac{1+o(1)}{t^2\log t}\int_1^{\infty}{xe^{-x}\text{d}x}. \end{align*}\]

当\(x>1+\log\frac{1}{t}>1\)时，\(1-\frac{\log x}{\log t}>1\)，因此

\[I_3=-\frac{1}{t^2\log t}\int_{1+\log\frac{1}{t}}^{\infty}{xe^{-x}\left(1-\frac{\log x}{\log t}\right)^{-1}\text{d}x}\leqslant -\frac{1}{t^2\log t}\int_{1+\log\frac{1}{t}}^{\infty}{xe^{-x}\text{d}x}=O\left(\frac{1}{t^2\log\frac{1}{t}}\right). \]

于是，我们得到了\(\sum_p{e^{-pt}}=-\frac{1}{t\log t}(1+o(1)).\qquad \vartriangleleft\)

\(\color{red}{注：}\) 一般地，对任意的\(N\in \mathbb{N}\)，当\(t\to 0^+\)时，

\[\sum_p{e^{-pt}}=-\sum_{n=0}^N{\frac{1}{t\left(\log t\right)^{n+1}}}\cdot\left.\frac{\text{d}^n\Gamma \left( x+1 \right)}{\text{d}x^n} \right|_{x=0}+o\left(\frac{1}{t\left(\log t\right)^{N+1}}\right). \]

题32. 令\(\Delta^1 b_n=b_{n+1}-b_n,\Delta^m b_n=\Delta^{m-1}b_{n+1}-\Delta^{m-1}b_n(m\geqslant 2)\)，并设

\[b_n=0(n>N),\quad S_k^{(1)}=\sum_{n=1}^k{a_n},\quad S_k^{(m)}=\sum_{n=1}^k{S_n^{(m-1)}}. \]

证明：

\[\sum_{n=1}^N{a_n b_n}=\left(-1\right)^m\sum_{n=1}^N{S_n^{(m)}\Delta^m b_n}. \]

证. 当\(m=1\)时，由分部求和公式及\(b_{N+1}=0\)，结论成立.

假定命题对\(m\)成立，以\(S_n^{(m)},\Delta^m b_n\)分别代替\(a_n,b_n\)，利用分部求和公式，有

\[\sum_{n=1}^N{S_n^{(m)}\Delta^m b_n}=(-1)\sum_{n=1}^N{S_n^{(m+1)}\Delta^{m+1}b_n}. \]

因此，命题对\(m+1\)成立.\(\qquad \vartriangleleft\)

题33. 设\(f\in \text{C}^2(a,\infty)\)，且\(\int_a^{\infty}\left|f''(x)\right|\text{d}x<\infty\)，并记

\[\rho(x)=\frac{1}{2}-x+[x],\ \ \sigma(x)=\int_0^x{\rho(x)\text{d}x}. \]

证明：对任意的\(b\geqslant a\)，存在常数\(C\)，使

\[\sum_{a<n\leqslant b}{f(n)}=C+\int_a^b{f(x)\text{d}x}+\rho(b)f(b)-\rho(a)f(a)-\int_b^{\infty}{\sigma(x)f''(x)\text{d}x}. \]

证. 根据\(3.2\)节的定理\(3\)，我们有

\[\sum_{a<n\leqslant b}{f(n)}=\int_a^b{f(x)\text{d}x}+\rho(b)f(b)-\rho(a)f(a)-\int_a^b{\rho(x)f'(x)\text{d}x}. \]

同时由

\[\begin{align*} \int_a^b{\rho(x)f'(x)\text{d}x} & =\sigma(b)f'(b)-\sigma(a)f'(a)-\int_a^b{\sigma(x)f''(x)\text{d}x}\\ & =\sigma(b)f'(b)-\sigma(a)f'(a)-\int_a^{\infty}{\sigma(x)f''(x)\text{d}x}+\int_b^{\infty}{\sigma(x)f''(x)\text{d}x}, \end{align*}\]

取

\[C=\sigma(a)f'(a)-\sigma(b)f'(b)+\int_a^{\infty}{\sigma(x)f''(x)\text{d}x}, \]

即得所欲证.\(\qquad \vartriangleleft\)

题34. 设\(\alpha>0,\beta,\tau\in \mathbb{R}\)，估计下列和式当\(N\to \infty\)时的阶：

\[(1)\sum_{n=1}^N{n^{\alpha}\left(\log n\right)^{\beta}};\quad (2)\sum_{n=1}^N{e^{n\alpha}\left(\log n\right)^{\beta}};\quad (3)\sum_{n=1}^N{\left(\log n\right)^{\tau}}. \]

证. 仿照\(3.2\)节的例\(1\)，结果分别为

\[\frac{1}{\alpha+1}N^{\alpha+1}\left(\log N\right)^{\beta},\ \frac{1}{\alpha}e^{N\alpha}\left(\log N\right)^{\beta},\ N\left(\log N\right)^{\tau}. \]

过程从略.\(\qquad \vartriangleleft\)

题35. 设\(\alpha>-1\)，以\(p\)表示素数，证明：存在常数\(A,B\)，使得

\[(1)\prod_{p\leqslant x}\left(1-\frac{1}{p}\right)^{-1}=e^{\gamma}\log x+O(1);\ (2)A\frac{x^{1+\alpha}}{\log x}<\sum_{p\leqslant x}{p^{\alpha}}<B\frac{x^{1+\alpha}}{\log x},\ x\geqslant 2. \]

其中\(\gamma\)是\(\text{Euler--Mascheroni}\)常数.

证. (1)利用\(\text{Mertens}\)第二定理\(^{\color{blue}{[8]}}\)即可.

(2)取\(a_n=1,b(x)=x^{\alpha}(x\geqslant 2)\)，记\(p_n\)是第\(n\)个素数，则

\[\sum_{p\leqslant x}{p^{\alpha}}=\sum_{p_n\leqslant x}{a_n b(p_n)}=\pi(x)x^{\alpha}-\alpha \int_2^x{\pi(t)t^{\alpha-1}\text{d}t}. \]

注意到\(\pi(x)=\frac{x}{\log x}\left(1+O\left(\frac{1}{\log x}\right)\right)\)，所以

\[\sum_{p\leqslant x}{p^{\alpha}}=\frac{x^{\alpha+1}}{\log x}\left(1+O\left(\frac{1}{\log x}\right)\right)-\alpha\int_2^x{\frac{t^{\alpha}}{\log t}\left(1+O\left(\frac{1}{\log t}\right)\right)\text{d}t}. \]

对上式的积分分部积分，有\(\sum_{p\leqslant x}{p^{\alpha}}=\frac{x^{\alpha+1}}{(\alpha+1)\log x}+O\left(\frac{x^{\alpha+1}}{\log^2 x}\right)\). 因此，存在常数\(C\)，使

\[\left|\sum_{p\leqslant x}{p^{\alpha}}-\frac{x^{\alpha+1}}{(\alpha+1)\log x}\right|\leqslant C\frac{x^{\alpha+1}}{\log^2 x},\ x\geqslant 2. \]

取\(A=\frac{1}{\alpha+1}-\frac{2C}{\log 2},B=\frac{1}{\alpha+1}+\frac{2C}{\log 2}\)，即得所欲证.\(\qquad \vartriangleleft\)

题36. 当\(x\to 0^+\)时，估计\(\sum_{n=1}^{\infty}{ne^{-n^2 x}}\)的阶.

证. 利用\(3.5\)节的例\(11\)，有

\[\sqrt{x}\sum_{n=1}^{\infty}(n\sqrt{x})e^{-\left(n\sqrt{x}\right)^2}\thicksim \int_0^{\infty}te^{-t^2}\text{d}t=\frac{1}{2},\ x\to 0^+. \]

故\(x\to 0^+\)时，\(\sum_{n=1}^{\infty}{ne^{-n^2 x}}=\frac{1}{2x}(1+o(1)).\qquad \vartriangleleft\)

题37. 设\(f(x)\)在0处连续，且\(x\to 0\)时\(f(x)-f(\sin x)\thicksim Ax^3\)(\(A\)是常数)，则\(f'(0)=6A\).

证. 设\(g(x)=f(x)-6Ax\)，则有\(g(x)-g(\sin x)=o\left(x^3\right),\ x\to 0\)，且\(g(0)=0\).

由于

\[\underset{x\to 0}{\lim}\frac{\sin 2x-\frac{2x}{\sqrt{1+x^2}}}{x^3}=-\frac{1}{3}, \]

所以，\(\exists \eta \in(0,1)\)使\(0<|x|<\eta\)时，\(x^{-3}\left(\sin 2x-\frac{2x}{\sqrt{1+x^2}}\right)<0\). 故\(\frac{1}{\sqrt{n}}<\eta\)时，成立

\[\begin{align*} \sin \frac{2}{\sqrt{n}}<\frac{\frac{2}{\sqrt{n}}}{1+\frac{1}{n}}=\frac{2}{\sqrt{n+1}}.\tag{1.5} \end{align*}\]

由连续性，对\(\forall \varepsilon>0\)，存在\(\delta\in \left(0,\frac{\pi}{2}\right)\)，使得当\(0<|x|<\delta\)时，\(\left|g(x)-g(\sin x)\right|\leqslant \varepsilon \left|x\right|^3\).

记\(a_0(x)=x,a_{n+1}(x)=\sin\left(a_n(x)\right)\)，并取\(k\in \mathbb{Z}\)使\(1+\frac{1}{x^2}\leqslant k\leqslant \frac{4}{x^2}\)，则利用\((1.5)\)式归纳得

\[|a_n(x)|=a_n(|x|)\leqslant \frac{2}{\sqrt{k+n}},\quad n=0,1,2,\cdots. \]

于是，当\(0<|x|<\delta\)时，

\[\begin{align*} |g(x)|& \leqslant \sum_{n=0}^{\infty}{\left|g(a_n(x))-g(a_{n+1}(x))\right|}\leqslant \varepsilon \sum_{n=0}^{\infty}{\left|a_n(x)\right|^3}\\ & \leqslant \varepsilon \sum_{n=0}^{\infty}{\frac{8}{\left(k+n\right)^{\frac{3}{2}}}}\leqslant \varepsilon \int_{k-1}^{\infty}{\frac{8}{t^{\frac{3}{2}}}\text{d}t}=\frac{16\varepsilon}{\sqrt{k-1}}\leqslant 16\varepsilon|x|. \end{align*}\]

依\(g'(0)\)的定义，\(g'(0)=0\)，即\(f'(0)=g'(0)+6A=6A\).\(\qquad \vartriangleleft\)

题38. 设\(x=f(t)\)满足方程\(xe^x=\frac{1}{t}\).证明：当\(t\to +\infty\)时，对任意的\(m\in \mathbb{N}\)，有

\[x=\sum_{k=1}^m\frac{\left(-1\right)^{k-1}k^{k-1}}{k!}t^{-k}+O\left(\frac{1}{t^{m+1}}\right). \]

证. 令\(t^{-1}=\omega,f(z)=ze^{z},z_0=\omega_0=0\)，则\(\frac{\zeta-z_0}{f(\zeta)-\omega_0}=e^{-\zeta}\).利用\(\text{Lagrange}\)逆定理\(^{\color{blue}{[9]}}\)，有

\[z=\phi(\omega)=\sum_{n=1}^{\infty}\frac{1}{n!}\left(\frac{\text{d}^{n-1}}{\text{d}\zeta^{n-1}}e^{-n\zeta}\right)_{\zeta=0}\omega^n=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}n^{n-1}}{n!}t^{-n}. \]

将\(z=x\)代入，即证毕.\(\qquad \vartriangleleft\)

题39. 设\(x=x(t)\)是方程\(t=x\left(1+x\right)^m\)满足\(t=0,x=0\)的根，求其在\(t\to 0\)时的渐近表示.

解. 取\(f(z)=z\left(1+z\right)^m,z_0=\omega_0=0\)，则\(\frac{\zeta-z_0}{f(\zeta)-\omega_0}=\frac{1}{\left(1+\zeta\right)^m}\). 由\(\left(1+\zeta\right)^{-mk}\)在0处的展开

\[\left(1+\zeta\right)^{-mk}=\sum_{r=0}^{\infty}\frac{\left(-1\right)^r}{r!}\cdot mk(mk+1)\cdots(mk+r-1)\zeta^r, \]

我们有

\[z=\phi(\omega)=\sum_{k=1}^{\infty}\frac{1}{k!}\left(\frac{\text{d}^{k-1}}{\text{d}\zeta^{k-1}}\left(1+\zeta\right)^{-mk}\right)_{\zeta=0}\omega^k=\sum_{k=1}^{\infty}\left(-1\right)^{k-1}\frac{(mk+k-2)!}{k!(mk-1)!}\omega^k. \]

将\(z=x,\omega=t\)代入，即得

\[x=\sum_{k=1}^{\infty}\left(-1\right)^{k-1}\frac{(mk+k-2)!}{k!(mk-1)!}t^k,\ t\to 0. \]

这便是渐近表示.\(\qquad \vartriangleleft\)

题40. 设\(\{a_n\}\)满足\(a_{n+1}=a_n+O\left(\frac{1}{n}\right)+O\left(\frac{a_n^2}{n^2}\right)\)，且\(n\to \infty\)时\(a_n=o(n)\)，则\(a_n=O(\log n)\).

证. 存在常数\(A,B>0\)，使\(|a_{n+1}|\leqslant |a_n|+\frac{A}{n}+B\frac{a_n^2}{n^2}\). 不妨取\(A=B=1\)，于是

\[|a_{n+1}|-\log(n+1)\leqslant (|a_n|-\log n)+\frac{a_n^2}{n^2}\leqslant (|a_n|-\log n)+\frac{\left(|a_n|-\log n\right)^2}{n^2}+\frac{2\log^2 n}{n^2}. \]

进而有

\[\begin{align*} |a_{n+1}|-2\log(n+1)& \leqslant (|a_n|-2\log n)+\frac{3\left(|a_n|-2\log n\right)^2}{n^2}+\frac{8\log^2 n}{n^2}-\frac{1}{n}\\ & \leqslant (|a_n|-2\log n)+\frac{3}{n^2}\left(|a_n|-2\log n\right)^2,\ n\geqslant 241. \end{align*}\]

记\(b_n:=3|a_n|-6\log n\)，则当\(n\)充分大时，\(b_n=o(n)\)，且

\[\begin{align*} |b_{n+1}|\leqslant |b_n|+\frac{b_n^2}{n^2}.\tag{1.6} \end{align*}\]

因此，\(\exists N\in \mathbb{N},\forall n\geqslant N\)，有\(|b_n|\leqslant \frac{n-1}{2}\)，故\(\frac{|b_{n+1}|}{n+1}\leqslant \left(1-\frac{1}{2n}\right)\frac{|b_n|}{n}\). 从而有

\[|b_n|\leqslant \frac{|b_N|}{N}\cdot n\prod_{k=N}^{n-1}\left(1-\frac{1}{2k}\right)=O\left(\sqrt{n}\right). \]

即\(b_n=O\left(\sqrt{n}\right)\).将之代入\((1.6)\)式，有\(|b_{n+1}|\leqslant |b_n|+O\left(\frac{1}{n}\right)\)，所以

\[|b_n|\leqslant |b_1|+\sum O\left(\frac{1}{k}\right)=O(\log n). \]

这蕴涵着\(a_n=O(\log n)\).\(\qquad \vartriangleleft\)

\(\color{red}{注：}\) \((1.6)\)式即为\(4.4\)节的例\(2\)的情形，本题给出了更为简洁的解答.

题41. 证明以下结论：

(1)若\(x=ye^{\left(\log y\right)^{5/8}}\)，则

\[y=xe^{-\left(\log x\right)^{\frac{5}{8}}+\frac{5}{8}\left(\log x\right)^{\frac{1}{4}}}(1+o(1)),\ x\to +\infty. \]

(2)若\(x=\frac{y}{\log y}\)，则当\(x\to +\infty\)时，

\[y=x\left(\log x+\log\log x+\frac{\log\log x}{\log x}\right)+O\left(\frac{x\left(\log\log x\right)^2}{\log^2 x}\right). \]

证. (1)设\(a=\frac{5}{8}\)，则\(3a-2=-\frac{1}{8}<0\)，且

\[\begin{align*} \log y=\log x-\log^a y.\tag{1.7} \end{align*}\]

当\(y>1\)时\(y<x\)，且易见\(x\to +\infty\)时\(y\to +\infty\)，于是\(\log y=O(\log x)\). 进一步地\(^{\color{blue}{[10]}}\)，

\[\begin{align*} \log\log y& =\log\left(\log x-O\left(\log^a x\right)\right)\\ & =\log\log x+\log\left(1-O\left(\log^{a-1}x\right)\right)\\ & =\log\log x+O\left(\log^{a-1}x\right). \end{align*}\]

取指数，有

\[\log y=\log x\cdot \left(1+O\left(\log^{a-1}x\right)\right). \]

将上式代入\((1.7)\)式，有

\[\log y=\log x-\log^a x\cdot \left(1+O\left(\log^{a-1}x\right)\right)^a=\log x-\log^a x+O\left(\log^{2a-1}x\right). \]

将上式代入\((1.7)\)式，有

\[\begin{align*} \log y& =\log x-\left(\log x-\log^a x+O\left(\log^{2a-1}x\right)\right)^a\\ & =\log x-\log^a x\cdot \left(1-\log^{a-1}x+O\left(\log^{2a-2}x\right)\right)^a\\ & =\log x-\log^a x+a\log^{2a-1}x+O\left(\log^{3a-2}x\right). \end{align*}\]

取指数，将\(a=\frac{5}{8}\)代入，整理得

\[y=xe^{-\left(\log x\right)^{\frac{5}{8}}+\frac{5}{8}\left(\log x\right)^{\frac{1}{4}}}(1+o(1)),\ x\to +\infty. \]

(2)由\(y=x\log y\)，有

\[\begin{align*} \log y=\log x+\log\log y.\tag{1.8} \end{align*}\]

易见，\(y=O(x)\)，所以\(\log y=\log x+O(\log\log x)\). 取对数，有

\[\begin{align*} \log\log y& =\log(\log x+O(\log\log x))\\ & =\log\log x+\log\left(1+O\left(\frac{\log\log x}{\log x}\right)\right)\\ & =\log\log x+O\left(\frac{\log\log x}{\log x}\right). \end{align*}\]

将之代入\((1.8)\)式，有

\[\log y=\log x+\log\log x+O\left(\frac{\log\log x}{\log x}\right). \]

取对数，可得

\[\begin{align*} \log\log y& =\log\left(\log x+\log\log x+O\left(\frac{\log\log x}{\log x}\right)\right)\\ & =\log\log x+\log\left(1+\frac{\log\log x}{\log x}+O\left(\frac{\log\log x}{\log^2 x}\right)\right)\\ & =\log\log x+\frac{\log\log x}{\log x}+O\left(\frac{\left(\log\log x\right)^2}{\log^2 x}\right). \end{align*}\]

将上式代入\((1.8)\)式，有

\[\log y=\log x+\log\log x+\frac{\log\log x}{\log x}+O\left(\frac{\left(\log\log x\right)^2}{\log^2 x}\right). \]

所以

\[y=x\log y=x\left(\log x+\log\log x+\frac{\log\log x}{\log x}\right)+O\left(\frac{x\left(\log\log x\right)^2}{\log^2 x}\right). \]

命题得证.\(\qquad \vartriangleleft\)

题42. 设\(a\geqslant 0,c>0,f'(t)>-ct^{a-1}(t>1)\)，且

\[\int_0^x f(t)\text{d}t=\frac{1}{a+1}x^{a+1}(1+o(1)),\ x\to +\infty. \]

证明：当\(x\to +\infty\)时，\(f(x)\thicksim x^a\).

证. 设\(g(x):=f(x)+\frac{c}{a}x^a(x\geqslant 0)\)，不妨令\(g(1)=0\)，则当\(x>1\)时，

\[g'(x)=f'(x)+cx^{a-1}>0\ \Rightarrow \ (xg(x))'=g(x)+xg'(x)>g(0)+xg'(x)>0. \]

所以\(x>1\)时，\(xg(x)\)是正的连续递增函数，且

\[\int_0^x g(t)\text{d}t=\frac{a+c}{a^2+a}x^{a+1}(1+o(1)),\ x\to +\infty. \]

利用\(4.3\)节的定理\(1\)，可得\(x\to +\infty\)时，\(g(x)\thicksim \frac{a+c}{a}x^a\)，即\(f(x)\thicksim x^a\).\(\qquad \vartriangleleft\)

参考文献：

\(\color{blue}{[1]}\) 潘承洞,于秀源. 阶的估计基础[M]. 北京: 高等教育出版社, 2015.

\(\color{blue}{[2]}\) G. Polya, G. Szego, Problems and Theorems in Analysis I, Springer-Verlag, Berlin, Germeny (1998).

\(\color{blue}{[3]}\) 薛春华,徐森林. 数学分析精选习题全解: 下册[M]. 北京: 清华大学出版社, 2010.

\(\color{blue}{[4]}\) I.S. Gradshteyn and I.M. Ryzhik, eds., Table of Integrals, Series, and Products, 5th ed., Academic Press, San Diego, 1994.

\(\color{blue}{[5]}\) L. Schoenfeld. Sharper bounds for the Chebyshev functions \(\vartheta(x)\) and \(\psi(x)\): II[J]. Math. Comp. , 1976(30): 337-360.

\(\color{blue}{[6]}\) Andrews, G., Askey, R., & Roy, R. (1999). Special Functions (Encyclopedia of Mathematics and Its Applications). Cambridge: Cambridge University Press.

\(\color{blue}{[7]}\) 菲赫金哥尔茨. 微积分学教程：第2卷, 第8版[M]. 徐献瑜, 冷生明, 梁文骐, 译. 北京: 高等教育出版社, 2006.

\(\color{blue}{[8]}\) 特伦鲍姆. 解析与概率数论导引[M]. 陈华一,译. 北京: 高等教育出版社, 2011.

\(\color{blue}{[9]}\) 沙巴特. 复分析导论：第4版, 第1卷[M]. 胥鸣伟, 李振宇译. 北京: 高等教育出版社, 2011.

\(\color{blue}{[10]}\) Olver, F. W. J., Asymptotics and Special Functions, Academic Press, New York, 1974.