\(\qquad\)《階的估計基礎》\(^{\color{blue}{[1]}}\)(潘承洞、於秀源著)一書着重展現了常見估計方法的具體過程，是關於漸近分析的一本不錯的入門讀物。它是在第一版《階的估計》的精簡版本，想進一步了解估計技巧的讀者可閱讀《階的估計》。
\(\qquad\)書中的正文和題目由於印刷等因素出現了一些錯誤，加之估計漸近式常需要比較復雜的計算，所以初讀《階的估計基礎》者可能會遇到一定的困難。我將習題做了一些調整，選出\(41\)個題目給出解答。本系列文章所引用的定理、例題等對應着原書相應的內容，如\(1.3\)節的推論\(1\)，指的是《階的估計基礎》一書中第\(1\)章第\(3\)節的推論\(1\)。希望讀者先自己嘗試，當實在無法可循時再參考解答。
\(\qquad\)當然我給出的解答並非最佳的，如果有其他解法的朋友，歡迎一起交流。
\(\qquad\)為了更好的體驗，請在電腦端閱讀此文。
題1. 設\(f(x)\)為下列函數：
\[\begin{align*} (1)& \ \sqrt[\uproot{7}3]{x+\sqrt[\uproot{3}3]{x+\sqrt[\uproot{2}4]{x}}};\\ (2)& \ \sin \frac{1}{x^2}\log\Big(1+\frac{1}{x^a}\Big),a>0;\\ (3)& \ e^{\sin \frac{1}{x}+\cos \frac{1}{x}}\log \frac{1}{x+5}. \end{align*}\]
分別確定當\(x\to +\infty\)時\(f(x)\)的等價量.
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)\(\ f(x)=e^{\frac{1}{x}+o\left(x^{-2}\right)+1+O\left(x^{-2}\right)}\log \frac{1}{x+5}\thicksim e^{1+\frac{1}{x}}\log \frac{1}{x},\ x\to +\infty.\qquad \vartriangleleft\)
題2. 設\(f(x)\)為下列函數：
\[(1)\tan^3 x-3\tan x,\ x\to \frac{\pi}{3};\ \ (2)\big(1-2x\big)^{x^{-1/2}},\ x\to 0;\ \ (3)x^x-1,\ x\to 1. \]
請分別確定\(f(x)\)的等價量 .
解. (1)利用\(\text{Lagrange}\)中值定理，在\(x\)與\(\frac{\pi}{3}\)間存在\(\xi\)，使
\[\begin{align*} f(x)& =\tan x\Big(\tan x+\tan \frac{\pi}{3}\Big)\Big(\tan x-\tan \frac{\pi}{3}\Big)\\ & =\tan x\Big(\tan x+\tan \frac{\pi}{3}\Big)\Big(x-\frac{\pi}{3}\Big)\sec^2 \xi. \end{align*}\]
於是當\(x\to \frac{\pi}{3}\)時，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)\(\ f(x)=e^{x\log x}-1\thicksim x\log x=x\log(1+x-1)\thicksim x-1,\ x\to 1.\qquad \vartriangleleft\)
題3. 當\(x\to +\infty\)時，按階的大小將下列函數進行排列：
\[\left(\log x\right)^a;\ \ \left(\log\log x\right)^{\beta};\ \ x^a;\ \ x^{\left(\log x\right)^a};\ \ x^{\left(\log x\right)^{x^a}};\ \ e^{ax};\ \ e^{x^a};\ \ e^{\left(\log x\right)^A};\ \ \left(1+x\right)^a. \]
其中\(0<a<1,0<\beta<1,A>2.\)
解. 將這些函數依次編號\(a_1,a_2,\cdots,a_9\)，則它們階的大小關系為
\[a_5>a_7>a_6>a_8>a_4>a_9>a_3>a_1>a_2. \]
鑒於本題和下題中的函數過多，遂略去比較的步驟，留給讀者自行補充.\(\qquad \vartriangleleft\)
題4. 設\(a>0\)，當\(x\to 0^+\)時，按階的大小將下列函數進行排列：
\[(x+1)\sin x^2;\ \ \cos x-1;\ \ \left(1+x\right)^x-1;\ \ x^x-1;\ \ e^{2x}-1-2x^2; \]
\[\sqrt[\uproot{3}3]{1-\sqrt{x}}-1;\ \ 2^{x^2}-1;\ \ 3^{-\frac{1}{x}};\ \ e^{-\frac{a}{x}};\ \ \left(\log x\right)^{-\frac{a}{x}};\ \ \left(\log x^{-1}\right)^{\log x}. \]
解. 將這些函數依次編號\(a_1,a_2,\cdots,a_{11}\)，當\(a>\log 3\)時，階的大小關系為
\[a_{10}>a_9>a_8>a_{11}>a_1=a_2=a_3=a_7>a_5>a_6>a_4. \]
當\(a\leqslant \log 3\)時，\(a_{10}>a_8\geqslant a_9>a_{11}>\cdots\)，其他大小關系不變.\(\qquad \vartriangleleft\)
題5. 證明：若\(0<k<1\)，則
\[\int_0^1{\frac{\text{d}x}{\sqrt{\left(1-x^2\right)\left(1-k^2 x^2\right)}}}\thicksim \frac{1}{2}\log\frac{1}{1-k},\ k\to 1^-. \]
證. 將積分記為\(I(k)\)，並將被積函數對\(k^2\)作冪級數展開，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，我們有
\[a_n:=\frac{\pi}{2}\left(\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}\right)^2\thicksim b_n:=\frac{1}{2n},\ n\to \infty. \]
依\(1.3\)節的推論\(1\)，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\(\frac{1}{2}\log\frac{1}{1+k}=O(1)\).\(\qquad \vartriangleleft\)
題6. 設\(\sigma>0\)，若級數\(\sum_{n=1}^{\infty}\frac{a_n}{n^{\sigma}}\)收斂，則
\[\sum_{n=1}^{\infty}{a_n x^n}=o\left(\frac{1}{\left(1-x\right)^{\sigma}}\right),\ x\to 1^-. \]
證. 先證明一個命題\(^{\color{blue}{[3]}}\)：
引理1. 設\(\{b_n\}\)單調趨於0，且\(\sum_{n=1}^{\infty}{a_n b_n}\)收斂，則\(\underset{n\to \infty}{\lim}\left(a_1+a_2+\cdots+a_n\right)b_n=0.\)
\(\color{blue}{引理的證明：}\)不妨設\(\{b_n\}\)遞減，則\(\forall \varepsilon>0,\ \exists N\in \mathbb{N},\ \forall n>N_1,\ \text{s.t.}\ \left|\sum_{k=N_1+1}^n{a_k b_k}\right|<\frac{\varepsilon}{6}\). 固定\(n\)，則\(\left\{\frac{b_n}{b_k}\right\}\)關於\(k\)單調增，由Abel引理，有
\[\left|\sum_{k=N_1+1}^n{a_k b_n}\right|=\left|\sum_{k=N_1+1}^n{a_k b_k\cdot\frac{b_n}{b_k}}\right|<\frac{\varepsilon}{6} \left(\frac{b_n}{b_{N_1+1}}+\frac{2b_n}{b_n}\right)\leqslant \frac{\varepsilon}{2}. \]
由於\(b_n\downarrow 0\)，對於上述給定的\(\varepsilon,N_1\)，\(\exists N_2\in \mathbb{N},\forall n>N_2,\ \text{s.t.}\ \left|(a_1+a_2+\cdots+a_{N_1})b_n\right|<\frac{\varepsilon}{2}\).
因此，當\(n>\max\{N_1,N_2\}\)時，
\[\left|\sum_{k=1}^n{a_k b_n}\right|\leqslant \left|\sum_{k=1}^{N_1}{a_k b_n}\right|+\left|\sum_{k=N_1+1}^n{a_k b_n}\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \]
引理得證！\(\qquad\square\)
接原證明：取\(b_n=\frac{1}{n^{\sigma}}\)，則\(\underset{n\to \infty}{\lim}\left(a_1+a_2+\cdots+a_n\right)n^{-\sigma}=0\).
設\(\{c_n\}\)滿足\(\left(1-x\right)^{-\sigma}=\sum\limits_{n=0}^{\infty}{c_n x^n}\)，則
\[\left(1-x\right)^{-\sigma-1}=\frac{1}{1-x}\sum_{n=0}^{\infty}{c_n x^n}=\sum_{n=0}^{\infty}{\left(c_0+c_1+\cdots+c_n\right)x^n}. \]
因此
\[\sum_{j=0}^n{c_j}=\binom{\sigma+n}{n}=\frac{(\sigma+1)(\sigma+2)\cdots(\sigma+n)}{n!}\thicksim \frac{n^{\sigma}}{\sigma \Gamma(\sigma)},\ n\to \infty. \]
這蘊涵着\(a_1+a_2+\cdots+a_n=o\left(n^{\sigma}\right)=o(c_0+c_1+\cdots+c_n),\ n\to \infty.\)
利用\(1.3\)節的推論\(2\)，我們有
\[\sum_{n=1}^{\infty}{a_n x^n}=o\left(\sum_{n=0}^{\infty}{c_n x^n}\right)=o\left(\frac{1}{\left(1-x\right)^{\sigma}}\right),\ x\to 1^-. \]
證畢！\(\qquad \vartriangleleft\)
題7. 證明並確定\(c>0\)，使
\[\sum_{n=1}^{\infty}{\left(x^{n^2}-x^{2n^2}\right)}\thicksim \frac{c}{\sqrt{1-x}},\ x\to 1^-. \]
證. 易見
\[\int_0^{\infty}{x^{t^2}\text{d}t}\leqslant \sum_{n=0}^{\infty}{x^{n^2}}\leqslant 1+\int_0^{\infty}{x^{t^2}\text{d}t}. \]
當\(x\to 1^-\)時，
\[\int_0^{\infty}{x^{t^2}\text{d}t}=\int_0^{\infty}{e^{-t^2 \log\frac{1}{x}}\text{d}t}=\frac{1}{2}\sqrt{\frac{\pi}{\log x^{-1}}}\thicksim \frac{1}{2}\sqrt{\frac{\pi}{1-x}}. \]
因此，\(\sum_{n=0}^{\infty}{x^{n^2}}\thicksim \frac{1}{2}\sqrt{\frac{\pi}{1-x}},\ x\to 1^-\). 據此整理可得\(c=\frac{\left(2-\sqrt{2}\right)\sqrt{\pi}}{4}\).\(\qquad \vartriangleleft\)
\(\color{red}{注：}\) 本題亦可直接利用\(1.6\)節例\(6\)的結論.事實上，還可以展開任意多項\(^{\color{blue}{[4]}}\).
題8. 設\(b_n>0\)，級數\(\sum_{n=0}^{\infty}{b_n x^n}\)對一切\(x\)收斂，又設存在\(s<+\infty\)，使得
\[\underset{n\to \infty}{\lim}\frac{a_n}{b_n}=s, \]
則\(\sum_{n=0}^{\infty}{a_n x^n}\)也對一切\(x\)收斂，而且
\[\sum_{n=0}^{\infty}{a_n x^n}\thicksim s\sum_{n=0}^{\infty}{b_n x^n},\ x\to +\infty. \]
證. 下面對\(\text{Toeplitz}\)定理的一個推論述而不證：
引理2. 設函數\(\varphi_n(x)\)在\((0,1)\)內非負，且\(\sum_{n=0}^{\infty}{\varphi_n(x)}=1\)，則對固定的\(m\in \mathbb{N}\)，\(\underset{x\to +\infty}{\lim}\varphi_m(x)=0\)的充要條件是對每一個收斂於\(s\)的序列\(\{s_n\}\)，總有
\[\underset{x\to +\infty}{\lim}\left(s_0 \varphi_0(x)+s_1 \varphi_1(x)+\cdots+s_n \varphi_n(x)+\cdots\right)=s. \]
首先記
\[s_n:=\frac{a_n}{b_n},\ \varphi_n(x):=\frac{b_n x^n}{b_0+b_1 x+b_2 x^2+\cdots}. \]
給定\(\varepsilon>0\)及\(m\in \mathbb{N}\)，對充分大的\(x\)，選擇\(n\)使得
\[b_0+b_1 x+b_2 x^2+\cdots+b_n x^n>\frac{b_m x^m}{\varepsilon}. \]
於是
\[\varphi_m(x)<\frac{b_m x^m}{b_0+b_1 x+\cdots+b_n x^n}<\varepsilon. \]
因此\(\underset{x\to +\infty}{\lim}\varphi_m(x)=0\). 利用引理\(2\)，我們有
\[\underset{x\to +\infty}{\lim}\sum_{n=0}^{\infty}{s_n \varphi_n(x)}=\underset{x\to \infty}{\lim}{\frac{a_0+a_1 x+a_2 x^2+\cdots}{b_0+b_1 x+b_2 x^2+\cdots}}=s. \]
故\(\sum_{n=0}^{\infty}{a_n x^n}\thicksim s\sum_{n=0}^{\infty}{b_n x^n},\ x\to +\infty.\) \(\qquad \vartriangleleft\)
題9. 設\(\underset{n\to \infty}{\lim}a_n=A\)存在且有限，則
\[\sum_{n=0}^{\infty}{a_n\frac{x^n}{n!}}=Ae^x\left(1+o(1)\right),\ x\to +\infty. \]
證. 利用引理\(1.2\)，並注意到\(\sum_{n=0}^{\infty}{\frac{A}{n!}x^n}=Ae^x\).\(\qquad \vartriangleleft\)
題10. 設\(\pi(x)\)是不超過\(x>0\)的素數個數，且\(\pi(x)\thicksim \frac{x}{\log x},x\to +\infty\)，則第\(n\)個素數\(p_n\thicksim ?\)
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\(x=p_n\)，則\(\pi(x)\log \pi(x)=n\log n\).利用上式，即有\(p_n\thicksim n\log n\).\(\qquad \vartriangleleft\)
\(\color{red}{注：}\) 更一般地\(^{\color{blue}{[5]}}\)，\(\forall n\geqslant 20,\ n\left(\log n+\log\log n-\frac{3}{2}\right)<p_n<n\left(\log n+\log\log n-\frac{1}{2}\right)\).
題11. 證明：\(\left(x+1+O\left(\frac{1}{x}\right)\right)^x=ex^x+O\left(x^{x-1}\right),\ x\to +\infty\).
證. 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
\[\text{LHS}=x^x\exp\left(1+O\left(\frac{1}{x}\right)\right)=ex^x\left(1+O\left(\frac{1}{x}\right)\right)=ex^x+O\left(x^{x-1}\right),\ x\to +\infty. \]
證畢！\(\qquad \vartriangleleft\)
題12. 證明：對\(\forall n\in \mathbb{N}\)，總有\(\int_1^x{\log^n t\text{d}t}=O\left(x\log^n x\right).\)
證. 設\(f(x,t)=\frac{\log^n t}{x\log^n x}(1\leqslant t\leqslant x)\)，則
\[f(x,t)=\exp\left(n\log\frac{\log t}{\log x}-\log x\right)=\exp\left(-\log x+O(1)\right)=O\left(\frac{1}{x}\right). \]
從而有\(\int_1^x{f(x,t)\text{d}t}=O\left(\frac{1}{x}\right)\int_1^x{\text{d}t}=O(1)\). 整理之即證畢.\(\qquad \vartriangleleft\)
\(\color{red}{注：}\) 本題證明的是不等式關系，即\(|f|\leqslant Cg\)，而非\(x\to 1^+\)或\(x\to \infty\)時的漸近式.
題13. 設\(\{na_n\}\)遞減趨於0，\(\sum_{n=1}^{\infty}{a_n}\)收斂，則\(a_n=o\big(\frac{1}{n\log n}\big),\ n\to \infty\).
證. \(\forall \varepsilon>0,\exists N\in \mathbb{N},\forall n>N,\ \text{s.t.}\ \sum_{k=\left[\sqrt{n}\right]}^n{ka_k}<\varepsilon\). 於是有
\[\sum_{k=\left[\sqrt{n}\right]}^n{ka_k\cdot \frac{1}{k}}=o(1)\geqslant na_n\cdot \sum_{k=\left[\sqrt{n}\right]}^n{\frac{1}{k}}=na_n\left(\log n-\log\left[\sqrt{n}\right]\right)(1+o(1))=a_n\cdot O(n\log n). \]
因此，\(a_n=o\left(\frac{1}{n\log n}\right),\ n\to \infty\).\(\qquad \vartriangleleft\)
\(\color{red}{注：}\) 本題亦可利用引理\(1.1\).
題14. 證明：對\(\forall x>0\)，有\(\left(1+\frac{x}{n}\right)^{n+x}\thicksim e^x\left(1+\frac{x^2}{2n}\right),\ n\to \infty\).
證. 由於
\[(n+x)\log\left(1+\frac{x}{n}\right)=(n+x)\left(\frac{x}{n}-\frac{x^2}{2n^2}+o\left(\frac{1}{n^3}\right)\right)=x+\frac{x^2}{2n}+o\left(\frac{1}{n}\right), \]
所以
\[\left(1+\frac{x}{n}\right)^{n+x}=\exp\left(x+\frac{x^2}{2n}+o\left(\frac{1}{n}\right)\right)\thicksim e^x\left(1+\frac{x^2}{2n}\right),\ n\to \infty. \]
證畢！\(\qquad \vartriangleleft\)
題15. 設\(a_n\thicksim \frac{1}{\log n},\ n\to\infty\)，則\(\sum_{n=0}^{\infty}{a_n x^n}\thicksim \frac{1}{(x-1)\log(1-x)},\ x\to 1^-\).
證. 設\(S(j,r):=\sum_{k=j}^r{\frac{1}{(n-k+1)\log(k+2)}}\)，當\(|x|<1\)時，我們有
\[f(x):=\log\frac{1}{1-x}\cdot \sum_{k=2}^{\infty}{\frac{x^k}{\log k}}=\sum_{n=1}^{\infty}{\frac{x^{n+1}}{n+1}}\cdot \sum_{k=2}^{\infty}{\frac{x^{k+2}}{\log (k+2)}}=\sum_{n=0}^{\infty}{S(0,n)x^n}. \]
固定\(\varepsilon>0\)，使\(\varepsilon n\in \mathbb{N}\)，有\(S(0,\varepsilon n)\leqslant \frac{(\varepsilon n+1)}{(1-\varepsilon)n\log 2}\leqslant \frac{2\varepsilon}{(1-\varepsilon)\log 2}\). 同時有
\[\begin{align*} & \varlimsup_{n\to \infty}S(\varepsilon n,n)\leqslant \varlimsup_{n\to \infty}\frac{1}{\log \varepsilon n}\cdot \sum_{k=\varepsilon n}^n{\frac{1}{n-k+1}}=\underset{n\to \infty}{\lim}\frac{\log(1-\varepsilon)n}{\log \varepsilon n}=1;\\ & \varliminf_{n\to \infty}S(\varepsilon n,n)\geqslant \varliminf_{n\to \infty}\frac{1}{\log(n+2)}\cdot \sum_{k=\varepsilon n}^n{\frac{1}{n-k+1}}=\underset{n\to \infty}{\lim}\frac{\log(1-\varepsilon)n}{\log(n+2)}=1. \end{align*}\]
令\(\varepsilon \to 0^+\)，有\(\underset{n\to \infty}{\lim}S(0,n)=1\). 這蘊涵着\(\sum\limits_{n=0}^{\infty}{a_n x^n}\thicksim \frac{-f(x)}{\log(1-x)}\thicksim \frac{1}{(x-1)\log(1-x)},\ x\to 1^-\).\(\qquad \vartriangleleft\)
\(\color{red}{注：}\) 與\(S(0,n)\to 1\)類似的極限有
\[\underset{n\to \infty}{\lim}\frac{1}{n}\sum_{k=0}^n{\frac{\log(k+2)}{\log(n-k+2)}}=1,\quad \underset{n\to \infty}{\lim}\frac{1}{\sqrt{n}}\sum_{k=1}^n{\frac{\sqrt[k]{k}}{\sqrt{n+1-k}}}=2. \]
題16. 設\(x\ne 0,-1,-2,\cdots\)，證明：級數
\[\sum_{n=1}^{\infty}{\frac{n!a_n}{x(x+1)(x+2)\cdots(x+n)}} \]
與級數\(\sum_{n=1}^{\infty}{\frac{a_n}{n^x}}\)有相同的收斂性.
證. 對\(x\ne 0,-1,-2,\cdots\)，定義
\[b_n:=\frac{n!n^x}{x(x+1)(x+2)\cdots(x+n)},\ n=1,2,\cdots. \]
接下來證明\(\{b_n\}\)對固定的\(x\)是單調有界的，再結合\(\text{Abel}\)判別法，即證得結論.
由\(\Gamma(x)\)的無窮乘積表示\(^{\color{blue}{[6]}}\)，有\(\underset{n\to \infty}{\lim}b_n=\Gamma(x)\)，這蘊涵着有界性. 同時，我們有\(^{\color{blue}{[7]}}\)
\[\frac{b_{n+1}}{b_n}=\left(1+\frac{1}{n}\right)^{x+1}\left(1+\frac{x+1}{n}\right)^{-1}=1+\frac{x(x+1)}{2n^2}+o\left(\frac{1}{n^2}\right). \]
\(x(x+1)>0\)時，比值最后將大於\(1\)；\(x(x+1)<0\)時，最后將小於\(1\). 即\(\{b_n\}\)單調有界.\(\qquad \vartriangleleft\)
題17. 求極限：
\[(1)\underset{x\to 0}{\lim}x\int_x^1{\frac{\cos t}{t^2}\text{d}t};\quad (2)\underset{x\to +\infty}{\lim}\frac{\int_0^x{\sqrt{1+t^4}\text{d}t}}{x^3};\quad (3)\underset{x\to 0^+}{\lim}\frac{\int_x^{\infty}{t^{-1}e^{-t}\text{d}t}}{\log x}. \]
證. (1)首先分部積分，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，極限為\(\underset{x\to 0}{\lim}\left(\cos x+O(x)\right)=1.\)
(2)使用一次\(\text{L'Hospital}\)法則，則得極限為\(\frac{1}{3}\).
(3)當\(0<t<1\)時，\(-1<\frac{1}{t}\left(e^{-t}-1\right)<e^{-1}-1.\)從而有
\[\int_x^{\infty}{\frac{\text{d}t}{te^t}}=\log \frac{1}{x}+\int_x^1{\frac{e^{-t}-1}{t}\text{d}t}+\int_1^{\infty}{\frac{\text{d}t}{te^t}}=\log \frac{1}{x}+O(1). \]
因此，極限為\(-1\).\(\qquad \vartriangleleft\)
\(\color{red}{注：}\)切勿濫用\(\text{L'Hospital}\)法則解題，只有在滿足條件時才能使用.
題18. 設正項數列\(\{a_n\}\)滿足條件
\[\frac{a_{n+1}}{a_n}=1-\frac{1}{n}-\frac{\alpha_n}{n\log n},\ n=2,3,\cdots, \]
則級數\(\sum_{n=2}^{\infty}{a_n}\)當\(\alpha_n \geqslant \alpha>1\)時收斂，而當\(\alpha_n \leqslant \alpha'<1\)時發散.
證. 不妨設\(n\geqslant 2\)時，\(\left|\frac{1}{n}+\frac{\alpha_n}{n\log n}\right|<\frac{1}{2}.\)同時有
\[\begin{align*} a_n=a_2\prod_{k=2}^{n-1}{\left(1-\frac{1}{k}-\frac{\alpha_k}{k\log k}\right)},\ n=3,4,\cdots.\tag{1.1} \end{align*}\]
我們只需證明\(\alpha_n=\alpha>1\)和\(\alpha_n=\alpha'<1\)時的結論均成立.
事實上，依\((1.1)\)式可知在這兩種情況下，數列分別被縮小、擴大，再利用比較審斂法，便證得結論. 因此，不妨設\(\alpha_n \equiv a\).
由於
\[\begin{align*} \sum_{k=2}^{n-1}{\log \left(1-\frac{1}{k}-\frac{a}{k\log k}\right)}& =\sum_{k=2}^{n-1}{\left(-\frac{1}{k}-\frac{a}{k\log k}+O\left(\frac{1}{k^2}\right)\right)}\\ & =-\log n-a\log\log n+c'+O\left(n^{-1}\right). \end{align*}\]
所以當\(n\to \infty\)時，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\(c,c'\)是常數且\(c\ne 0\). 易見，\(c<1\)時，\(\sum{a_n}\)收斂；\(c>1\)時，\(\sum{a_n}\)收斂.\(\qquad \vartriangleleft\)
題19. 設級數
\[\sum_{n=1}^{\infty}{a_n\left(x^2-1\right)\left(x^2-2^2\right)\cdots\left(x^2-n^2\right)} \]
在點\(x=x_0\not\in \mathbb{Z}\)收斂，則它在所有點\(x\in (-\infty,+\infty)\)收斂.
證. 對\(x_0\not\in \mathbb{Z},x\in \mathbb{R}\)，定義
\[b_n:=\frac{\left(x^2-1\right)\left(x^2-2^2\right)\cdots\left(x^2-n^2\right)}{\left(x_0^2-1\right)\left(x_0^2-2^2\right)\cdots\left(x_0^2-n^2\right)}. \]
易見，它是單調變化的. 由於
\[\alpha_n:=\frac{x^2-n^2}{x_0^2-n^2}=1+\frac{x^2-x_0^2}{x_0^2-n^2}=1+O\left(\frac{1}{n^2}\right). \]
所以\(\underset{n\to \infty}{\lim}b_n=\prod{\alpha_n}\)收斂，故\(\{b_n\}\)有界. 利用\(\text{Abel}\)判別法，原級數在\(x\in \mathbb{R}\)點收斂.\(\qquad \vartriangleleft\)
題20. 判定無窮乘積\(\prod{a_n}\)的收斂性，其通項為：
\[\begin{align*} & (1)a_n=1+\frac{1}{x^n}\left(1+\frac{1}{n}\right)^{n^\alpha};\qquad\ (2)a_n=1+\frac{1}{n^{\alpha}}\left(1-\frac{x\log n}{n}\right)^n;\\ & (3)a_n=1+\frac{2n+1}{\left(n^2-1\right)\left(n+1\right)^{\alpha}};\quad (4)a_n=1+\frac{n^{\alpha}}{\left(1+\sin \frac{\pi}{n}\right)^{2n}}. \end{align*}\]
解. 下面僅將答案陳列出來，留給讀者自證.
(1)\(\alpha>2\)時發散\(;\alpha=2\)時\(,|x|\leqslant e\)發散\(,|x|>e\)收斂\(;\alpha<2\)時\(,|x|\leqslant 1\)發散\(,|x|>1\)收斂.
(2)\(x+\alpha>1\)收斂\(,x+\alpha \leqslant 1\)發散.\(\quad\)(3)\(\alpha>0\)收斂\(,\alpha \leqslant 0\)發散.
(4)\(\alpha>1\)收斂\(,\alpha \leqslant 1\)發散.\(\qquad \vartriangleleft\)
題21. 證明：當\(\alpha<-1,\beta<-2\)或\(-1<\alpha<\frac{\beta}{2}-1\)時，\(\int_0^{\infty}{x^{\alpha}\left|\cos x\right|^{x^{\beta}}\text{d}x}\)收斂.
證. 將原積分寫成兩項，我們有
\[I:=\int_0^{\infty}{x^{\alpha}\left|\cos x\right|^{x^{\beta}}\text{d}x}=\left(\int_0^1+\int_1^{\infty}\right)x^{\alpha}\left|\cos x\right|^{x^{\beta}}\text{d}x=I_1+I_2. \]
(1)\(\alpha<-1,\beta<-2\)時，\(I_2\)顯然收斂. 而\(x\to 0^+\)時\(x^{\alpha}\left|\cos x\right|^{x^{\beta}}<\exp \left(\frac{1}{2}x^{\beta}\log|x|\right)\)，故\(I_1\)收斂.
(2)\(-1<\alpha<\frac{\beta}{2}-1\)時，\(I_1\)顯然收斂，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\((a)\)處利用了\(\text{Stirling}\)公式，\(c_{\beta}\)是只依賴於\(\beta\)的常數. 綜合(1)(2)情況，命題成立.\(\qquad \vartriangleleft\)
題22. 證明：當且僅當\(x\geqslant \frac{1}{2}\)時，數列
\[a_n=\left(1+\frac{1}{n}\right)^n\left(1+\frac{x}{n}\right),\ n=1,2,3,\cdots \]
單調減少.
證. (1)充分性：將\(a_n\)表示為
\[a_n=\left(1+\frac{1}{n}\right)^{n+1/2}\left(1+\frac{x}{n}\right)\left(1+\frac{1}{n}\right)^{-1/2}. \]
其中第一個因子遞減，第二個因子的平方\(1+\frac{2x-1}{n+1}+\frac{x^2}{n(n+1)}\)在\(x\geqslant \frac{1}{2}\)時關於\(n\)遞減.
(2)必要性：當\(|x|<1\)時，我們有
\[\log \frac{1+x}{1-x}=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right). \]
分別以\(\frac{1}{2n+1},\frac{x}{2n+x}\)代替\(x\)，得到
\[\begin{align*} \log a_n& =n\log\left(1+\frac{1}{n}\right)+\log\left(1+\frac{x}{n}\right)\\ & =\frac{2n}{2n+1}+\frac{2x}{2n+x}+\frac{1}{12n^2}+O\left(\frac{1}{n^3}\right). \end{align*}\]
於是有
\[\log a_n-\log a_{n+1}=\frac{4x-2}{4n^2}+O\left(\frac{1}{n^3}\right). \]
這蘊涵着必要性成立.\(\qquad \vartriangleleft\)
題23. 設\(r>0\)，數列\(\{a_n\},\{b_n\}(n\in \mathbb{N})\)滿足條件：
\[(1)b_n\ne 0;\quad (2)f(x)=\sum_{n=0}^{\infty}{a_n x^n}(|x|<r)\text{收斂};\quad (3)\underset{n\to \infty}{\lim}\frac{b_n}{b_{n+1}}=q,|q|<r. \]
令\(c_n=a_0 b_n+a_1 b_{n-1}+\cdots+a_n b_0(n=1,2,\cdots)\)，則\(\underset{n\to \infty}{\lim}\frac{c_n}{b_n}=f(q).\)
證. 取充分小的\(\varepsilon>0\)，使\(|q|+\varepsilon<r\)，則存在不依賴於\(n\)和\(k\)的常數\(A\)，使
\[\left|\frac{b_{n-k}}{b_n}\right|<A\left(|q|+\varepsilon\right)^k,\ k=0,1,\cdots,n;n=0,1,\cdots. \]
當\(n>m\)時，得到
\[\begin{align*} \frac{c_n}{b_n}-f(q)=\sum_{k=0}^m{a_k\left(\frac{b_{n-k}}{b_n}-q^k\right)}+\sum_{k=m+1}^n{a_k\left(\frac{b_{n-k}}{b_n}-q^k\right)}.\tag{1.2} \end{align*}\]
第二項的絕對值小於
\[g_1(m):=A\sum_{k=m+1}^{\infty}{|a_k|\left(|q|+\varepsilon\right)^k}+\sum_{k=m+1}^{\infty}{|a_k|\cdot\left|q\right|^k}. \]
易見，對\(|t|\leqslant |q|+\varepsilon\)，級數\(\sum{a_n t^n}\)絕對收斂. 事實上，我們有
\[\sum{\left|a_n t^n\right|}=\sum{|a_n|\left(|q|+\varepsilon\right)^n\cdot \left(\frac{|t|}{|q|+\varepsilon}\right)^n}\leqslant M\sum{\left(\frac{|t|}{|q|+\varepsilon}\right)^n}<+\infty. \]
固定\(m\)，選取\(n\)，使得
\[g_2(m)=\left|\sum_{k=0}^m{a_k\left(\frac{b_{n-k}}{b_n}-q^k\right)}\right|<\frac{\varepsilon}{2}. \]
選取充分大的\(m\)，使\(2g_1(m)<\varepsilon\)，從而有\(g_1(m)+g_2(m)<\varepsilon\). 所以\((1.2)\)式左邊即為無窮小.\(\qquad \vartriangleleft\)
題24. 證明下面的二重積分可以交換積分次序：
\[\int_0^{\infty}{\text{d}x\int_0^{\infty}{y(\cos2mx)e^{-y^2\left(1+x^2\right)}\text{d}y}}. \]
證. 設\(f(x,y):=e^{-y^2\left(1+x^2\right)}\cos(2mx)\)，則
\[g(x):=\int_0^{\infty}{ye^{-y^2\left(1+x^2\right)}\text{d}y},\ a\leqslant x\leqslant A \]
是一致收斂的，其中\((A>a>0)\). 於是有
\[\begin{align*} I& :=\int_a^A{g(x)\cos(2mx)\text{d}x}=\int_0^{\infty}{y\text{d}y\int_a^A{f(x,y)\text{d}x}}\\ & =\int_0^{\infty}{y\text{d}y\int_0^{\infty}{f(x,y)\text{d}x}}-\int_0^{\infty}{y\text{d}y\int_0^a{f(x,y)\text{d}x}}-\int_0^{\infty}{y\text{d}y\int_A^{\infty}{f(x,y)\text{d}x}}\\ & =\int_0^{\infty}{y\text{d}y\int_0^{\infty}{f(x,y)\text{d}x}}-I_1-I_2. \end{align*}\]
只需證明：\(\underset{a\to 0^+}{\lim}I_1=\underset{A\to \infty}{\lim}I_2=0\). 首先，我們有
\[|I_1|\leqslant \int_0^{\infty}{y\text{d}y\int_0^a{e^{-y^2\left(1+x^2\right)}\text{d}x}}\leqslant a\int_0^{\infty}{ye^{-y^2}\text{d}y}=\frac{a}{2}. \]
現將\(I_2\)寫成兩部分，可得
\[I_2=\int_0^1{y\text{d}y\int_A^{\infty}{f(x,y)\text{d}x}}+\int_1^{\infty}{y\text{d}y\int_A^{\infty}{f(x,y)\text{d}x}}=I_2^1+I_2^2. \]
當\(A\to \infty\)時，\(\int_A^{\infty}{f(x,y)\text{d}x}\to 0\)，因此
\[I_2^1=\int_0^1{y\cdot o(1)\text{d}y}=o(1). \]
同時有
\[\begin{align*} \left|I_2^2\right|\leqslant \int_1^{\infty}{y\text{d}y\int_A^{\infty}{e^{-y^2\left(1+x^2\right)}\text{d}x}}=\int_1^{\infty}{ye^{-y^2}\text{d}y\int_A^{\infty}{e^{-x^2 y^2}\text{d}x}}.\tag{1.3} \end{align*}\]
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\((1.3)\)式化為
\[\left|I_2^2\right|<\frac{1}{2A}\int_1^{\infty}{y^{-1}e^{-y^2\left(1+A^2\right)}\text{d}y}=o(1). \]
這便證明了\(\underset{a\to 0^+}{\lim}I_1=\underset{A\to \infty}{\lim}I_2=0\).\(\qquad \vartriangleleft\)
題25. 設\(c>b>0,c>a+b\)，證明：
\[\sum_{n=0}^{\infty}{\frac{a(a+1)\cdots(a+n-1)}{n!}\int_0^1{t^{b+n-1}\left(1-t\right)^{c-b-1}\text{d}t}}=\int_0^1{t^{b-1}\left(1-t\right)^{c-a-b-1}\text{d}t}. \]
證. 將\(\left(1-t\right)^{-a}\)展開為冪級數，並仿照\(2.3\)節的例\(3\)來處理.\(\qquad \vartriangleleft\)
題26. 設\(\varphi(x)\)在\(x>0\)定義，數列\(\{a_n\}_{n=0}^{\infty}\subset \mathbb{R}\)，且\(x\geqslant 1\)時有
\[\varphi(x)=a_0+\frac{a_1}{x}+\frac{a_2}{x^2}+\cdots+\frac{a_n}{x^n}+\cdots. \]
證明無窮乘積\(\varphi(1)\varphi(2)\cdots \varphi(n)\cdots\)收斂的充要條件是\(a_0=1,a_1=0\).
證. 無窮乘積收斂的必要條件是\(\underset{n\to \infty}{\lim}\varphi(n)=a_0=1\)，則
\[\psi(x):=\log \varphi(n)=\frac{a_1}{x}+\frac{a_2-\frac{1}{2}a_1^2}{x^2}+\cdots. \]
應用\(2.4\)節的例\(2\)，可知\(\sum{\psi(n)}\)收斂的充要條件是\(a_1=0\).證畢.\(\qquad \vartriangleleft\)
題27. 設\(\frac{1}{2}\leqslant \sigma<1\)，證明：
\[\sum_{1\leqslant m<n}{\frac{e^{-(m+n)\delta}}{m^{\sigma}n^{\sigma}\log \frac{n}{m}}}=O\left({\delta}^{2\sigma-2}\log \frac{1}{\delta}\right) \]
當\(\delta>0\)時成立.
證. 令\(T=\delta^{-1}\)，則有
\[S^*(T)=\sum_{T\leqslant m<n}{\frac{e^{-(m+n)\delta}}{m^{\sigma}n^{\sigma}\log \frac{n}{m}}}=\sum_{T\leqslant m<n}{\frac{e^{-(m+n)\frac{\delta}{2}}}{m^{\frac{\sigma}{2}}n^{\frac{\sigma}{2}}}\cdot O(1)}=O\left(\delta^{\sigma}\right). \]
結合\(3.5\)節的例\(3\)，我們有
\[S(T)=\sum_{m<n<T}{\frac{e^{-(m+n)\delta}}{m^{\sigma}n^{\sigma}\log \frac{n}{m}}}\leqslant \sum_{m<n<T}{\frac{1}{m^{\sigma}n^{\sigma}\log \frac{n}{m}}}=O\left(T^{2-2\sigma}\log T\right). \]
因此，\(S(T)+S^*(T)=O\left({\delta}^{2\sigma-2}\log \frac{1}{\delta}\right)\).\(\qquad \vartriangleleft\)
題28. 證明：
\[\begin{align*} & (1)\sum_{n=0}^{\infty}{\left(-1\right)^n\left(\frac{1}{1\cdot (2n+1)}+\frac{1}{3\cdot (2n-1)}+\cdots+\frac{1}{(2n+1)\cdot 1}\right)}=\frac{\pi^2}{16};\\ & (2)\sum_{n=1}^{\infty}{\frac{\left(-1\right)^n}{n+1}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)}=-\frac{1}{2}\log^2 2. \end{align*}\]
證. (1)逆用級數的\(\text{Cauchy}\)乘積，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
\[\begin{align*} \mathrm{LHS}& =\sum_{n=1}^{\infty}{\frac{\left(-1\right)^n}{n+1}\sum_{k=1}^n{\frac{1}{k}}}=\frac{1}{2}\sum_{n=1}^{\infty}{\frac{\left(-1\right)^n}{n+1}\sum_{i+j=n+1}{\left(\frac{1}{i}+\frac{1}{j}\right)}}\\ & =\frac{1}{2}\sum_{n=1}^{\infty}{\sum_{i+j=n+1}{\frac{\left(-1\right)^{i+j-1}}{i+j}\left(\frac{1}{i}+\frac{1}{j}\right)}}=-\frac{1}{2}\sum_{n=1}^{\infty}{\sum_{i+j=n+1}{\frac{\left(-1\right)^{i+j}}{ij}}}\\ & =-\frac{1}{2}\left(\sum_{n=1}^{\infty}{\frac{\left(-1\right)^n}{n}}\right)^2=-\frac{1}{2}\log^2 2. \end{align*}\]
\(\color{red}{注：}\) 事實上，本題解法不唯一.
題29. 證明：當\(t\to 0^+\)時，
\[\sum_{n=1}^{\infty}{\frac{1}{n}e^{-n^2 t}}=-\frac{1}{2}\log t-\frac{1}{2}\gamma+O\left(\sqrt{t}\right). \]
其中\(\gamma\)是\(\text{Euler-Mascheroni}\) 常數.
證. 利用\(\text{Euler--Maclaurin}\)公式，有
\[f(t):=\sum_{n=1}^{\infty}{\frac{1}{n}e^{-n^2 t}}=\frac{1}{2}e^{-t}+\int_1^{\infty}{\frac{1}{x}e^{-x^2 t}\text{d}x}-\int_1^{\infty}{\left(2t+\frac{1}{x^2}\right)e^{-x^2 t}\left(\{x\}-\frac{1}{2}\right)\text{d}x}. \]
由於\(x^{-1}\text{d}x=\text{d}(\log x)\)，再分部積分，可得
\[f(t)=\frac{1}{2}e^{-t}+2t\int_1^{\infty}{xe^{-x^2 t}\log x\text{d}x}-\int_1^{\infty}{\left(2t+\frac{1}{x^2}\right)e^{-x^2 t}\left(\{x\}-\frac{1}{2}\right)\text{d}x}. \]
由於
\[\int_1^{\infty}{2te^{-x^2 t}\{x\}\text{d}x}=2t\int_1^{\infty}{e^{-x^2 t}O(1)\text{d}x}\xlongequal{u=x^2 t} O\left(\sqrt{t}\right)\int_t^{\infty}{u^{-\frac{1}{2}}e^{-u}\text{d}u}=O\left(\sqrt{t}\right), \]
所以
\[\begin{align*} f(t)\xlongequal{u=x^2 t} e^{-t}+O\left(\sqrt{t}\right)+\frac{1}{2}\int_t^{\infty}{e^{-u}\log u\text{d}u}-\frac{1}{2}e^{-t}\log t-\int_1^{\infty}{x^{-2}e^{-x^2 t}\{x\}\text{d}x}.\tag{1.4} \end{align*}\]
通過\(\text{L'Hospital}\)法則，不難驗證
\[\int_t^{\infty}{e^{-u}\log u\text{d}u}=\int_0^{\infty}{e^{-u}\log u\text{d}u}-\int_0^t{e^{-u}\log u\text{d}u}=-\gamma+o\left(\sqrt{t}\right). \]
考慮到
\[\begin{align*} \int_1^{\infty}{\frac{\{x\}}{x^2}e^{-x^2 t}\text{d}x} & =\int_1^{\frac{1}{\sqrt{t}}}+\int_{\frac{1}{\sqrt{t}}}^{\infty}=\int_1^{\frac{1}{\sqrt{t}}}{\frac{\{x\}}{x^2}\left(1+O\left(\sqrt{t}\right)\right)\text{d}x}+\int_{\frac{1}{\sqrt{t}}}^{\infty}{\frac{O(1)}{x^2}\text{d}x}\\ & =O\left(\sqrt{t}\right)+\left(1+O\left(\sqrt{t}\right)\right)\int_1^{\infty}{\frac{\{x\}}{x^2}\text{d}x}=1-\gamma+O\left(\sqrt{t}\right). \end{align*}\]
結合上式與\((1.4)\)式，即得\(f(t)=-\frac{1}{2}\log t-\frac{1}{2}\gamma+O\left(\sqrt{t}\right).\qquad \vartriangleleft\)
題30. 證明：當\(n\to \infty\)時，有
\[\frac{1}{\log n}-\frac{1}{\log(n+1)}+\frac{1}{\log(n+2)}-\cdots=\frac{1}{2}\log n+O\left(\frac{1}{n\log^2 n}\right). \]
證. 設\(f(x)=\frac{1}{\log(n+x)}(x\geqslant 0)\)，則\(f'(x)=-\frac{1}{(x+n)\log^2(x+n)},f''(x)\geqslant 0\).
根據\(3.3\)節的定理\(1\)，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
\[\sum_{k=0}^{\infty}{\left(-1\right)^k f(k)}=\frac{1}{2}\log n+O(f'(0))=\frac{1}{2}\log n+O\left(\frac{1}{n\log^2 n}\right). \]
命題得證.\(\qquad \vartriangleleft\)
題31. 以\(p\)表示素數，已知對任意的\(m\in \mathbb{Z}_+\)，有
\[\sum_{p\leqslant x}{\log p}=x+O\left(x\left(\log x\right)^{-m}\right). \]
證明：當\(t\to 0^+\)時，有
\[\sum_p{e^{-pt}}=-\frac{1}{t\log t}(1+o(1)). \]
證. 取\(a_n=\log p_n,b(x)=\frac{e^{-tx}}{\log x}(x>1)\)，其中\(p_n\)是第\(n\)個素數，則
\[\begin{align*} \sum_{p\leqslant x}{e^{-pt}}& =\sum_{p_n\leqslant x}{a_n b(p_n)}=\frac{S(x)}{e^{tx}\log x}+\int_2^x{S(y)\cdot\frac{1+ty\log y}{y\log^2 y}\cdot \frac{\text{d}y}{e^{ty}}}\\ & =\frac{x}{e^{tx}\log x}\left(1+O\left(\frac{1}{\log^m x}\right)\right)+\int_2^x{\frac{1+ty\log y}{\log^2 y}\left(1+O\left(\frac{1}{\log^m y}\right)\right)\frac{\text{d}y}{e^{ty}}}. \end{align*}\]
令\(x\to \infty\)，有
\[\sum_{p\leqslant x}{e^{-pt}}=t\int_2^{\infty}{\frac{y}{\log y}e^{-ty}\text{d}y}(1+o(1))+O(1)\int_2^{\infty}{\frac{e^{-ty}}{\log^2 y}\text{d}y}. \]
現在我們來估計上式的第一項，第二項則可按照完全相同的步驟得到，即
\[O(1)\int_2^{\infty}{\frac{e^{-ty}}{\log^2 y}\text{d}y}=O\left(\frac{1}{t\log^2 t}\right),\ t\to 0^+. \]
將第一項寫成三部分，並記\(\phi :=\frac{1}{t}\left(1+\log\frac{1}{t}\right)\)，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\(t>0\)，則當\(x\in [2t,1]\)時，有
\[0\leqslant \frac{\log x}{\log t}\leqslant 1+\frac{\log 2}{\log t}<1. \]
這蘊涵着
\[0\leqslant \sum_{n=2}^{\infty}{\left(\frac{\log x}{\log t}\right)^{n-2}}\leqslant \sum_{n=0}^{\infty}{\left(1+\frac{\log x}{\log t}\right)^n}=-\frac{\log t}{\log 2}. \]
因此，
\[\sum_{n=0}^{\infty}{\left(\frac{\log x}{\log t}\right)^n}=1+\frac{\log x}{\log t}+\left(\frac{\log x}{\log t}\right)^2\sum_{n=2}^{\infty}{\left(\frac{\log x}{\log t}\right)^{n-2}}=1+\frac{\log x}{\log t}+O\left(\frac{\log^2 x}{\log\frac{1}{t}}\right). \]
從而可得
\[\begin{align*} I_1& \overset{x=ty}{=}-\frac{1}{t^2\log t}\int_{2t}^1{xe^{-x}\left(1-\frac{\log x}{\log t}\right)^{-1}\text{d}x}=-\frac{1}{t^2\log t}\int_{2t}^1{xe^{-x}\sum_{n=0}^{\infty}{\left(\frac{\log x}{\log t}\right)^n}\text{d}x}\\ & =-\frac{1}{t^2\log t}\int_{2t}^1{xe^{-x}\text{d}x}(1+o(1))=-\frac{1}{t^2\log t}\int_0^1{xe^{-x}\text{d}x}(1+o(1)). \end{align*}\]
當\(x\in \left[1,1+\log\frac{1}{t}\right]\)時，\(1\leqslant 1-\frac{\log x}{\log t}\leqslant 1+\frac{\log\left(1+\log\frac{1}{t}\right)}{\log\frac{1}{t}}\to 1.\) 於是有
\[\begin{align*} I_2& =-\frac{1}{t^2\log t}\int_1^{1+\log\frac{1}{t}}{xe^{-x}\left(1-\frac{\log x}{\log t}\right)^{-1}\text{d}x}=-\frac{1+o(1)}{t^2\log t}\int_1^{1+\log\frac{1}{t}}{xe^{-x}\text{d}x}\\ & =\frac{1+o(1)}{t^2\log t}\int_{1+\log\frac{1}{t}}^{\infty}{xe^{-x}\text{d}x}-\frac{1+o(1)}{t^2\log t}\int_1^{\infty}{xe^{-x}\text{d}x}=-\frac{1+o(1)}{t^2\log t}\int_1^{\infty}{xe^{-x}\text{d}x}. \end{align*}\]
當\(x>1+\log\frac{1}{t}>1\)時，\(1-\frac{\log x}{\log t}>1\)，因此
\[I_3=-\frac{1}{t^2\log t}\int_{1+\log\frac{1}{t}}^{\infty}{xe^{-x}\left(1-\frac{\log x}{\log t}\right)^{-1}\text{d}x}\leqslant -\frac{1}{t^2\log t}\int_{1+\log\frac{1}{t}}^{\infty}{xe^{-x}\text{d}x}=O\left(\frac{1}{t^2\log\frac{1}{t}}\right). \]
於是，我們得到了\(\sum_p{e^{-pt}}=-\frac{1}{t\log t}(1+o(1)).\qquad \vartriangleleft\)
\(\color{red}{注：}\) 一般地，對任意的\(N\in \mathbb{N}\)，當\(t\to 0^+\)時，
\[\sum_p{e^{-pt}}=-\sum_{n=0}^N{\frac{1}{t\left(\log t\right)^{n+1}}}\cdot\left.\frac{\text{d}^n\Gamma \left( x+1 \right)}{\text{d}x^n} \right|_{x=0}+o\left(\frac{1}{t\left(\log t\right)^{N+1}}\right). \]
題32. 令\(\Delta^1 b_n=b_{n+1}-b_n,\Delta^m b_n=\Delta^{m-1}b_{n+1}-\Delta^{m-1}b_n(m\geqslant 2)\)，並設
\[b_n=0(n>N),\quad S_k^{(1)}=\sum_{n=1}^k{a_n},\quad S_k^{(m)}=\sum_{n=1}^k{S_n^{(m-1)}}. \]
證明：
\[\sum_{n=1}^N{a_n b_n}=\left(-1\right)^m\sum_{n=1}^N{S_n^{(m)}\Delta^m b_n}. \]
證. 當\(m=1\)時，由分部求和公式及\(b_{N+1}=0\)，結論成立.
假定命題對\(m\)成立，以\(S_n^{(m)},\Delta^m b_n\)分別代替\(a_n,b_n\)，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，命題對\(m+1\)成立.\(\qquad \vartriangleleft\)
題33. 設\(f\in \text{C}^2(a,\infty)\)，且\(\int_a^{\infty}\left|f''(x)\right|\text{d}x<\infty\)，並記
\[\rho(x)=\frac{1}{2}-x+[x],\ \ \sigma(x)=\int_0^x{\rho(x)\text{d}x}. \]
證明：對任意的\(b\geqslant a\)，存在常數\(C\)，使
\[\sum_{a<n\leqslant b}{f(n)}=C+\int_a^b{f(x)\text{d}x}+\rho(b)f(b)-\rho(a)f(a)-\int_b^{\infty}{\sigma(x)f''(x)\text{d}x}. \]
證. 根據\(3.2\)節的定理\(3\)，我們有
\[\sum_{a<n\leqslant b}{f(n)}=\int_a^b{f(x)\text{d}x}+\rho(b)f(b)-\rho(a)f(a)-\int_a^b{\rho(x)f'(x)\text{d}x}. \]
同時由
\[\begin{align*} \int_a^b{\rho(x)f'(x)\text{d}x} & =\sigma(b)f'(b)-\sigma(a)f'(a)-\int_a^b{\sigma(x)f''(x)\text{d}x}\\ & =\sigma(b)f'(b)-\sigma(a)f'(a)-\int_a^{\infty}{\sigma(x)f''(x)\text{d}x}+\int_b^{\infty}{\sigma(x)f''(x)\text{d}x}, \end{align*}\]
取
\[C=\sigma(a)f'(a)-\sigma(b)f'(b)+\int_a^{\infty}{\sigma(x)f''(x)\text{d}x}, \]
即得所欲證.\(\qquad \vartriangleleft\)
題34. 設\(\alpha>0,\beta,\tau\in \mathbb{R}\)，估計下列和式當\(N\to \infty\)時的階：
\[(1)\sum_{n=1}^N{n^{\alpha}\left(\log n\right)^{\beta}};\quad (2)\sum_{n=1}^N{e^{n\alpha}\left(\log n\right)^{\beta}};\quad (3)\sum_{n=1}^N{\left(\log n\right)^{\tau}}. \]
證. 仿照\(3.2\)節的例\(1\)，結果分別為
\[\frac{1}{\alpha+1}N^{\alpha+1}\left(\log N\right)^{\beta},\ \frac{1}{\alpha}e^{N\alpha}\left(\log N\right)^{\beta},\ N\left(\log N\right)^{\tau}. \]
過程從略.\(\qquad \vartriangleleft\)
題35. 設\(\alpha>-1\)，以\(p\)表示素數，證明：存在常數\(A,B\)，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\(\text{Euler--Mascheroni}\)常數.
證. (1)利用\(\text{Mertens}\)第二定理\(^{\color{blue}{[8]}}\)即可.
(2)取\(a_n=1,b(x)=x^{\alpha}(x\geqslant 2)\)，記\(p_n\)是第\(n\)個素數，則
\[\sum_{p\leqslant x}{p^{\alpha}}=\sum_{p_n\leqslant x}{a_n b(p_n)}=\pi(x)x^{\alpha}-\alpha \int_2^x{\pi(t)t^{\alpha-1}\text{d}t}. \]
注意到\(\pi(x)=\frac{x}{\log x}\left(1+O\left(\frac{1}{\log x}\right)\right)\)，所以
\[\sum_{p\leqslant x}{p^{\alpha}}=\frac{x^{\alpha+1}}{\log x}\left(1+O\left(\frac{1}{\log x}\right)\right)-\alpha\int_2^x{\frac{t^{\alpha}}{\log t}\left(1+O\left(\frac{1}{\log t}\right)\right)\text{d}t}. \]
對上式的積分分部積分，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，存在常數\(C\)，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\(A=\frac{1}{\alpha+1}-\frac{2C}{\log 2},B=\frac{1}{\alpha+1}+\frac{2C}{\log 2}\)，即得所欲證.\(\qquad \vartriangleleft\)
題36. 當\(x\to 0^+\)時，估計\(\sum_{n=1}^{\infty}{ne^{-n^2 x}}\)的階.
證. 利用\(3.5\)節的例\(11\)，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\(x\to 0^+\)時，\(\sum_{n=1}^{\infty}{ne^{-n^2 x}}=\frac{1}{2x}(1+o(1)).\qquad \vartriangleleft\)
題37. 設\(f(x)\)在0處連續，且\(x\to 0\)時\(f(x)-f(\sin x)\thicksim Ax^3\)(\(A\)是常數)，則\(f'(0)=6A\).
證. 設\(g(x)=f(x)-6Ax\)，則有\(g(x)-g(\sin x)=o\left(x^3\right),\ x\to 0\)，且\(g(0)=0\).
由於
\[\underset{x\to 0}{\lim}\frac{\sin 2x-\frac{2x}{\sqrt{1+x^2}}}{x^3}=-\frac{1}{3}, \]
所以，\(\exists \eta \in(0,1)\)使\(0<|x|<\eta\)時，\(x^{-3}\left(\sin 2x-\frac{2x}{\sqrt{1+x^2}}\right)<0\). 故\(\frac{1}{\sqrt{n}}<\eta\)時，成立
\[\begin{align*} \sin \frac{2}{\sqrt{n}}<\frac{\frac{2}{\sqrt{n}}}{1+\frac{1}{n}}=\frac{2}{\sqrt{n+1}}.\tag{1.5} \end{align*}\]
由連續性，對\(\forall \varepsilon>0\)，存在\(\delta\in \left(0,\frac{\pi}{2}\right)\)，使得當\(0<|x|<\delta\)時，\(\left|g(x)-g(\sin x)\right|\leqslant \varepsilon \left|x\right|^3\).
記\(a_0(x)=x,a_{n+1}(x)=\sin\left(a_n(x)\right)\)，並取\(k\in \mathbb{Z}\)使\(1+\frac{1}{x^2}\leqslant k\leqslant \frac{4}{x^2}\)，則利用\((1.5)\)式歸納得
\[|a_n(x)|=a_n(|x|)\leqslant \frac{2}{\sqrt{k+n}},\quad n=0,1,2,\cdots. \]
於是，當\(0<|x|<\delta\)時，
\[\begin{align*} |g(x)|& \leqslant \sum_{n=0}^{\infty}{\left|g(a_n(x))-g(a_{n+1}(x))\right|}\leqslant \varepsilon \sum_{n=0}^{\infty}{\left|a_n(x)\right|^3}\\ & \leqslant \varepsilon \sum_{n=0}^{\infty}{\frac{8}{\left(k+n\right)^{\frac{3}{2}}}}\leqslant \varepsilon \int_{k-1}^{\infty}{\frac{8}{t^{\frac{3}{2}}}\text{d}t}=\frac{16\varepsilon}{\sqrt{k-1}}\leqslant 16\varepsilon|x|. \end{align*}\]
依\(g'(0)\)的定義，\(g'(0)=0\)，即\(f'(0)=g'(0)+6A=6A\).\(\qquad \vartriangleleft\)
題38. 設\(x=f(t)\)滿足方程\(xe^x=\frac{1}{t}\).證明：當\(t\to +\infty\)時，對任意的\(m\in \mathbb{N}\)，有
\[x=\sum_{k=1}^m\frac{\left(-1\right)^{k-1}k^{k-1}}{k!}t^{-k}+O\left(\frac{1}{t^{m+1}}\right). \]
證. 令\(t^{-1}=\omega,f(z)=ze^{z},z_0=\omega_0=0\)，則\(\frac{\zeta-z_0}{f(\zeta)-\omega_0}=e^{-\zeta}\).利用\(\text{Lagrange}\)逆定理\(^{\color{blue}{[9]}}\)，有
\[z=\phi(\omega)=\sum_{n=1}^{\infty}\frac{1}{n!}\left(\frac{\text{d}^{n-1}}{\text{d}\zeta^{n-1}}e^{-n\zeta}\right)_{\zeta=0}\omega^n=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}n^{n-1}}{n!}t^{-n}. \]
將\(z=x\)代入，即證畢.\(\qquad \vartriangleleft\)
題39. 設\(x=x(t)\)是方程\(t=x\left(1+x\right)^m\)滿足\(t=0,x=0\)的根，求其在\(t\to 0\)時的漸近表示.
解. 取\(f(z)=z\left(1+z\right)^m,z_0=\omega_0=0\)，則\(\frac{\zeta-z_0}{f(\zeta)-\omega_0}=\frac{1}{\left(1+\zeta\right)^m}\). 由\(\left(1+\zeta\right)^{-mk}\)在0處的展開
\[\left(1+\zeta\right)^{-mk}=\sum_{r=0}^{\infty}\frac{\left(-1\right)^r}{r!}\cdot mk(mk+1)\cdots(mk+r-1)\zeta^r, \]
我們有
\[z=\phi(\omega)=\sum_{k=1}^{\infty}\frac{1}{k!}\left(\frac{\text{d}^{k-1}}{\text{d}\zeta^{k-1}}\left(1+\zeta\right)^{-mk}\right)_{\zeta=0}\omega^k=\sum_{k=1}^{\infty}\left(-1\right)^{k-1}\frac{(mk+k-2)!}{k!(mk-1)!}\omega^k. \]
將\(z=x,\omega=t\)代入，即得
\[x=\sum_{k=1}^{\infty}\left(-1\right)^{k-1}\frac{(mk+k-2)!}{k!(mk-1)!}t^k,\ t\to 0. \]
這便是漸近表示.\(\qquad \vartriangleleft\)
題40. 設\(\{a_n\}\)滿足\(a_{n+1}=a_n+O\left(\frac{1}{n}\right)+O\left(\frac{a_n^2}{n^2}\right)\)，且\(n\to \infty\)時\(a_n=o(n)\)，則\(a_n=O(\log n)\).
證. 存在常數\(A,B>0\)，使\(|a_{n+1}|\leqslant |a_n|+\frac{A}{n}+B\frac{a_n^2}{n^2}\). 不妨取\(A=B=1\)，於是
\[|a_{n+1}|-\log(n+1)\leqslant (|a_n|-\log n)+\frac{a_n^2}{n^2}\leqslant (|a_n|-\log n)+\frac{\left(|a_n|-\log n\right)^2}{n^2}+\frac{2\log^2 n}{n^2}. \]
進而有
\[\begin{align*} |a_{n+1}|-2\log(n+1)& \leqslant (|a_n|-2\log n)+\frac{3\left(|a_n|-2\log n\right)^2}{n^2}+\frac{8\log^2 n}{n^2}-\frac{1}{n}\\ & \leqslant (|a_n|-2\log n)+\frac{3}{n^2}\left(|a_n|-2\log n\right)^2,\ n\geqslant 241. \end{align*}\]
記\(b_n:=3|a_n|-6\log n\)，則當\(n\)充分大時，\(b_n=o(n)\)，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\(\frac{|b_{n+1}|}{n+1}\leqslant \left(1-\frac{1}{2n}\right)\frac{|b_n|}{n}\). 從而有
\[|b_n|\leqslant \frac{|b_N|}{N}\cdot n\prod_{k=N}^{n-1}\left(1-\frac{1}{2k}\right)=O\left(\sqrt{n}\right). \]
即\(b_n=O\left(\sqrt{n}\right)\).將之代入\((1.6)\)式，有\(|b_{n+1}|\leqslant |b_n|+O\left(\frac{1}{n}\right)\)，所以
\[|b_n|\leqslant |b_1|+\sum O\left(\frac{1}{k}\right)=O(\log n). \]
這蘊涵着\(a_n=O(\log n)\).\(\qquad \vartriangleleft\)
\(\color{red}{注：}\) \((1.6)\)式即為\(4.4\)節的例\(2\)的情形，本題給出了更為簡潔的解答.
題41. 證明以下結論：
(1)若\(x=ye^{\left(\log y\right)^{5/8}}\)，則
\[y=xe^{-\left(\log x\right)^{\frac{5}{8}}+\frac{5}{8}\left(\log x\right)^{\frac{1}{4}}}(1+o(1)),\ x\to +\infty. \]
(2)若\(x=\frac{y}{\log y}\)，則當\(x\to +\infty\)時，
\[y=x\left(\log x+\log\log x+\frac{\log\log x}{\log x}\right)+O\left(\frac{x\left(\log\log x\right)^2}{\log^2 x}\right). \]
證. (1)設\(a=\frac{5}{8}\)，則\(3a-2=-\frac{1}{8}<0\)，且
\[\begin{align*} \log y=\log x-\log^a y.\tag{1.7} \end{align*}\]
當\(y>1\)時\(y<x\)，且易見\(x\to +\infty\)時\(y\to +\infty\)，於是\(\log y=O(\log x)\). 進一步地\(^{\color{blue}{[10]}}\)，
\[\begin{align*} \log\log y& =\log\left(\log x-O\left(\log^a x\right)\right)\\ & =\log\log x+\log\left(1-O\left(\log^{a-1}x\right)\right)\\ & =\log\log x+O\left(\log^{a-1}x\right). \end{align*}\]
取指數，有
\[\log y=\log x\cdot \left(1+O\left(\log^{a-1}x\right)\right). \]
將上式代入\((1.7)\)式，有
\[\log y=\log x-\log^a x\cdot \left(1+O\left(\log^{a-1}x\right)\right)^a=\log x-\log^a x+O\left(\log^{2a-1}x\right). \]
將上式代入\((1.7)\)式，有
\[\begin{align*} \log y& =\log x-\left(\log x-\log^a x+O\left(\log^{2a-1}x\right)\right)^a\\ & =\log x-\log^a x\cdot \left(1-\log^{a-1}x+O\left(\log^{2a-2}x\right)\right)^a\\ & =\log x-\log^a x+a\log^{2a-1}x+O\left(\log^{3a-2}x\right). \end{align*}\]
取指數，將\(a=\frac{5}{8}\)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
\[\begin{align*} \log y=\log x+\log\log y.\tag{1.8} \end{align*}\]
易見，\(y=O(x)\)，所以\(\log y=\log x+O(\log\log x)\). 取對數，有
\[\begin{align*} \log\log y& =\log(\log x+O(\log\log x))\\ & =\log\log x+\log\left(1+O\left(\frac{\log\log x}{\log x}\right)\right)\\ & =\log\log x+O\left(\frac{\log\log x}{\log x}\right). \end{align*}\]
將之代入\((1.8)\)式，有
\[\log y=\log x+\log\log x+O\left(\frac{\log\log x}{\log x}\right). \]
取對數，可得
\[\begin{align*} \log\log y& =\log\left(\log x+\log\log x+O\left(\frac{\log\log x}{\log x}\right)\right)\\ & =\log\log x+\log\left(1+\frac{\log\log x}{\log x}+O\left(\frac{\log\log x}{\log^2 x}\right)\right)\\ & =\log\log x+\frac{\log\log x}{\log x}+O\left(\frac{\left(\log\log x\right)^2}{\log^2 x}\right). \end{align*}\]
將上式代入\((1.8)\)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
\[y=x\log y=x\left(\log x+\log\log x+\frac{\log\log x}{\log x}\right)+O\left(\frac{x\left(\log\log x\right)^2}{\log^2 x}\right). \]
命題得證.\(\qquad \vartriangleleft\)
題42. 設\(a\geqslant 0,c>0,f'(t)>-ct^{a-1}(t>1)\)，且
\[\int_0^x f(t)\text{d}t=\frac{1}{a+1}x^{a+1}(1+o(1)),\ x\to +\infty. \]
證明：當\(x\to +\infty\)時，\(f(x)\thicksim x^a\).
證. 設\(g(x):=f(x)+\frac{c}{a}x^a(x\geqslant 0)\)，不妨令\(g(1)=0\)，則當\(x>1\)時，
\[g'(x)=f'(x)+cx^{a-1}>0\ \Rightarrow \ (xg(x))'=g(x)+xg'(x)>g(0)+xg'(x)>0. \]
所以\(x>1\)時，\(xg(x)\)是正的連續遞增函數，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\(4.3\)節的定理\(1\)，可得\(x\to +\infty\)時，\(g(x)\thicksim \frac{a+c}{a}x^a\)，即\(f(x)\thicksim x^a\).\(\qquad \vartriangleleft\)
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