Worked Solutions to the AP-Style Practice SetAP 风格练习题配套解析
Topics 10.1 - 10.15专题 10.1 至 10.15BC
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}=$$\displaystyle\sum_{n=1}^{\infty}\frac{1}{(2n-1)(2n+1)}=$
Decompose the general term: $\dfrac{1}{(2n-1)(2n+1)}=\dfrac{1}{2}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)$. (M1)分解通项：$\dfrac{1}{(2n-1)(2n+1)}=\dfrac{1}{2}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)$。(M1)
$$ S_N=\frac{1}{2}\sum_{n=1}^{N}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)=\frac{1}{2}\left(1-\frac{1}{2N+1}\right)\longrightarrow\frac{1}{2}. $$(A1)
$\displaystyle\sum_{n=1}^{\infty}\frac{5}{4^{n}}=$$\displaystyle\sum_{n=1}^{\infty}\frac{5}{4^{n}}=$
The series starts at $n=1$ with first term $a=5/4$ and common ratio $r=1/4$; since $|r|<1$, it converges. (M1)级数从 $n=1$ 开始，首项 $a=5/4$，公比 $r=1/4$；因 $|r|<1$，级数收敛。(M1)
$$ S=\frac{a}{1-r}=\frac{5/4}{1-1/4}=\frac{5/4}{3/4}=\frac{5}{3}. $$(A1)
Which of the following series diverges?下列哪个级数发散？
The ratios are $r=0.9,\ -0.5,\ 3/2,\ 1/3$ respectively; a geometric series converges precisely when $|r|<1$. (M1)各公比依次为 $r=0.9,\ -0.5,\ 3/2,\ 1/3$；等比级数当且仅当 $|r|<1$ 时收敛。(M1)
Only option (C) has $|r|=3/2>1$, so it is the only one that diverges. (A1)仅选项 (C) 满足 $|r|=3/2>1$，故只有它发散。(A1)
$\displaystyle\sum_{n=1}^{\infty}\frac{2n^{2}+1}{3n^{2}+n}$$\displaystyle\sum_{n=1}^{\infty}\frac{2n^{2}+1}{3n^{2}+n}$
Divide numerator and denominator by $n^{2}$: $\displaystyle\lim_{n\to\infty}\frac{2n^{2}+1}{3n^{2}+n}=\lim_{n\to\infty}\frac{2+1/n^{2}}{3+1/n}=\frac{2}{3}$. (M1)分子分母同除以 $n^{2}$：$\displaystyle\lim_{n\to\infty}\frac{2n^{2}+1}{3n^{2}+n}=\lim_{n\to\infty}\frac{2+1/n^{2}}{3+1/n}=\frac{2}{3}$。(M1)
Since $\tfrac{2}{3}\ne 0$, the nth Term Test gives divergence immediately. (A1)因 $\tfrac{2}{3}\ne 0$，由第 n 项判别法立即得知级数发散。(A1)
$\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}$$\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}$
Let $f(x)=\dfrac{1}{x\ln x}$: it is positive, continuous, and decreasing for $x\ge 2$, so the integral test applies. (M1)设 $f(x)=\dfrac{1}{x\ln x}$：在 $x\ge 2$ 上为正、连续且单调递减，故可用积分判别法。(M1)
$$ \int_2^{\infty}\frac{dx}{x\ln x}=\lim_{b\to\infty}\bigl[\ln(\ln x)\bigr]_2^{b}=\lim_{b\to\infty}\ln(\ln b)-\ln(\ln 2)=\infty. $$The improper integral diverges, so the series diverges by the integral test. (A1)该反常积分发散，故由积分判别法，级数发散。(A1)
Which series converges?下列哪个级数收敛？
The exponents are $p=1,\ 0.5,\ 1.5,\ 0.9$ respectively; a $p$-series converges exactly when $p>1$. (M1)各指数依次为 $p=1,\ 0.5,\ 1.5,\ 0.9$；$p$ 级数当且仅当 $p>1$ 时收敛。(M1)
Only $p=1.5>1$ in option (C) satisfies this, so it is the only convergent series. (A1)仅选项 (C) 的 $p=1.5>1$ 满足此条件，故只有它收敛。(A1)
Which of the following is a true statement about the harmonic series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$?下列关于调和级数 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ 的说法中，哪个正确？
(A) is false: it is the alternating harmonic series $\sum(-1)^{n+1}/n$ that converges to $\ln 2$, not $\sum 1/n$ itself. (B) is false: $\int_1^{\infty}\tfrac{1}{x}\,dx=\lim_{b\to\infty}\ln b=\infty$, so the integral test shows divergence, not convergence. (D) is false as a justification: since $\lim 1/n=0$, the nth term test is inconclusive here, not a valid reason for divergence. (M1)(A) 错误：收敛到 $\ln 2$ 的是交错调和级数 $\sum(-1)^{n+1}/n$，而非 $\sum 1/n$ 本身。(B) 错误：$\int_1^{\infty}\tfrac{1}{x}\,dx=\lim_{b\to\infty}\ln b=\infty$，积分判别法显示的是发散而非收敛。(D) 作为理由不成立：因 $\lim 1/n=0$，第 n 项判别法在此不能下结论，不能作为发散的正确依据。(M1)
(C) is the correct statement: the series does diverge (by the integral test above), even though the individual terms shrink to $0$. (A1)(C) 为正确说法：级数确实发散（由上面的积分判别法可知），尽管各项本身趋于 $0$。(A1)
Since $0\le\dfrac{1}{n^{3}+1}\le\dfrac{1}{n^{3}}$ for all $n\ge 1$, and $\displaystyle\sum\frac{1}{n^{3}}$ converges, what does the Direct Comparison Test conclude about $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{3}+1}$?因对所有 $n\ge 1$ 有 $0\le\dfrac{1}{n^{3}+1}\le\dfrac{1}{n^{3}}$，且 $\displaystyle\sum\frac{1}{n^{3}}$ 收敛，比较判别法对 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{3}+1}$ 得出什么结论？
$\displaystyle\sum\frac{1}{n^{3}}$ is a $p$-series with $p=3>1$, so it converges. (M1)$\displaystyle\sum\frac{1}{n^{3}}$ 是 $p$ 级数，$p=3>1$，故收敛。(M1)
Since the given series is squeezed between $0$ and a convergent series termwise, the Direct Comparison Test concludes $\displaystyle\sum\frac{1}{n^{3}+1}$ also converges. (A1)因给定级数逐项夹在 $0$ 与一个收敛级数之间，由比较判别法知 $\displaystyle\sum\frac{1}{n^{3}+1}$ 同样收敛。(A1)
$\displaystyle\sum_{n=1}^{\infty}\frac{n+2}{n^{3}-n+1}$$\displaystyle\sum_{n=1}^{\infty}\frac{n+2}{n^{3}-n+1}$
For large $n$, $a_n=\dfrac{n+2}{n^{3}-n+1}\approx\dfrac{n}{n^{3}}=\dfrac{1}{n^{2}}$, so compare with $b_n=1/n^{2}$ (a convergent $p$-series, $p=2$). (M1)当 $n$ 较大时，$a_n=\dfrac{n+2}{n^{3}-n+1}\approx\dfrac{n}{n^{3}}=\dfrac{1}{n^{2}}$，故与 $b_n=1/n^{2}$（收敛的 $p$ 级数，$p=2$）比较。(M1)
$$ \lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{n+2}{n^{3}-n+1}\cdot n^{2}=\lim_{n\to\infty}\frac{n^{3}+2n^{2}}{n^{3}-n+1}=1. $$Since $0<1<\infty$ and $\sum 1/n^{2}$ converges, the LCT gives that $\sum a_n$ converges too. (A1)因 $0<1<\infty$ 且 $\sum 1/n^{2}$ 收敛，由极限比较判别法知 $\sum a_n$ 也收敛。(A1)
$\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\frac{n}{n^{2}+1}$$\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\frac{n}{n^{2}+1}$
$b_n=\dfrac{n}{n^{2}+1}>0$ for all $n\ge 1$. (M1)对所有 $n\ge 1$，$b_n=\dfrac{n}{n^{2}+1}>0$。(M1)
Letting $f(x)=\dfrac{x}{x^{2}+1}$, $f'(x)=\dfrac{(x^{2}+1)-x(2x)}{(x^{2}+1)^{2}}=\dfrac{1-x^{2}}{(x^{2}+1)^{2}}<0$ for $x>1$, so $b_n$ is eventually decreasing; also $\displaystyle\lim_{n\to\infty}b_n=0$. (M1)令 $f(x)=\dfrac{x}{x^{2}+1}$，则 $f'(x)=\dfrac{(x^{2}+1)-x(2x)}{(x^{2}+1)^{2}}=\dfrac{1-x^{2}}{(x^{2}+1)^{2}}<0$（$x>1$ 时），故 $b_n$ 最终单调递减；且 $\displaystyle\lim_{n\to\infty}b_n=0$。(M1)
All three conditions hold, so the series converges by the Alternating Series Test. (A1)三个条件均满足，故由交错级数判别法，级数收敛。(A1)
$\displaystyle\sum_{n=1}^{\infty}\frac{2^{n}}{n!}$$\displaystyle\sum_{n=1}^{\infty}\frac{2^{n}}{n!}$
$\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{2^{n+1}/(n+1)!}{2^{n}/n!}=\dfrac{2\cdot n!}{(n+1)!}=\dfrac{2}{n+1}$. (M1)$\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{2^{n+1}/(n+1)!}{2^{n}/n!}=\dfrac{2\cdot n!}{(n+1)!}=\dfrac{2}{n+1}$。(M1)
$L=\displaystyle\lim_{n\to\infty}\frac{2}{n+1}=0<1$, so by the ratio test the series converges absolutely (its terms are already positive, so absolute convergence is just convergence here). (A1)$L=\displaystyle\lim_{n\to\infty}\frac{2}{n+1}=0<1$，故由比值判别法，级数绝对收敛（因各项已为正，此处绝对收敛即为收敛）。(A1)
$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{\sqrt{n}}$ is$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{\sqrt{n}}$ 是
$\displaystyle\sum\left|\frac{(-1)^{n+1}}{\sqrt{n}}\right|=\sum\frac{1}{n^{1/2}}$ is a $p$-series with $p=1/2\le 1$, so it diverges: the series is not absolutely convergent. (M1)$\displaystyle\sum\left|\frac{(-1)^{n+1}}{\sqrt{n}}\right|=\sum\frac{1}{n^{1/2}}$ 是 $p$ 级数，$p=1/2\le 1$，故发散：该级数不是绝对收敛。(M1)
But $b_n=1/\sqrt{n}$ is positive, decreasing, and $\to 0$, so the original series converges by the AST. Converging overall while $\sum|a_n|$ diverges is exactly conditional convergence. (A1)但 $b_n=1/\sqrt{n}$ 为正、递减且 $\to 0$，故由交错级数判别法，原级数收敛。整体收敛而 $\sum|a_n|$ 发散，正是条件收敛的定义。(A1)
For $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n!}$, what is the smallest $n$ such that the partial sum $S_n$ approximates the sum with error less than $0.001$?对 $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n!}$，使部分和 $S_n$ 的误差小于 $0.001$ 的最小 $n$ 是多少？
The error bound requires $|a_{n+1}|=\dfrac{1}{(n+1)!}<0.001$, i.e. $(n+1)!>1000$. (M1)误差界要求 $|a_{n+1}|=\dfrac{1}{(n+1)!}<0.001$，即 $(n+1)!>1000$。(M1)
Check factorials: $6!=720<1000$, but $7!=5040>1000$. So the smallest $n+1$ that works is $7$, giving $n=6$. (A1)检验阶乘：$6!=720<1000$，而 $7!=5040>1000$。故满足条件的最小 $n+1$ 为 $7$，即 $n=6$。(A1)
In the 4th-degree Maclaurin polynomial for $f(x)=\cos x$, the coefficient of $x^{4}$ is在 $f(x)=\cos x$ 的 4 次麦克劳林多项式中，$x^{4}$ 的系数为
$f(x)=\cos x$, $f''''(x)=\cos x$ (the 4th derivative returns to $\cos x$), so $f''''(0)=\cos 0=1$. (M1)$f(x)=\cos x$，$f''''(x)=\cos x$（四阶导数回到 $\cos x$），故 $f''''(0)=\cos 0=1$。(M1)
The coefficient of $x^{4}$ in $P_4(x)$ is $\dfrac{f''''(0)}{4!}=\dfrac{1}{24}$. (A1)$P_4(x)$ 中 $x^{4}$ 的系数为 $\dfrac{f''''(0)}{4!}=\dfrac{1}{24}$。(A1)
Let $P_3(x)$ be the 3rd-degree Taylor polynomial for $f(x)=e^{x}$ centered at $a=0$. If $|f^{(4)}(c)|\le 3$ for $0\le c\le 1$, the Lagrange error bound for $|f(1)-P_3(1)|$ is设 $P_3(x)$ 为 $f(x)=e^{x}$ 在 $a=0$ 处的 3 次泰勒多项式。若对 $0\le c\le 1$ 有 $|f^{(4)}(c)|\le 3$，则 $|f(1)-P_3(1)|$ 的拉格朗日误差界为
Here $M=3$, $n=3$ (so $n+1=4$), and $|x-a|=|1-0|=1$. (M1)此处 $M=3$，$n=3$（故 $n+1=4$），$|x-a|=|1-0|=1$。(M1)
$$ |R_3(1)|\le\frac{M}{(n+1)!}|x-a|^{n+1}=\frac{3}{4!}\cdot 1^{4}=\frac{3}{24}=\frac{1}{8}. $$(A1)
The radius of convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{(x+1)^{n}}{n\cdot 3^{n}}$ is$\displaystyle\sum_{n=1}^{\infty}\frac{(x+1)^{n}}{n\cdot 3^{n}}$ 的收敛半径为
$\left|\dfrac{a_{n+1}}{a_n}\right|=\left|\dfrac{(x+1)^{n+1}}{(n+1)3^{n+1}}\cdot\dfrac{n\cdot 3^{n}}{(x+1)^{n}}\right|=\dfrac{|x+1|}{3}\cdot\dfrac{n}{n+1}\longrightarrow\dfrac{|x+1|}{3}$. (M1)$\left|\dfrac{a_{n+1}}{a_n}\right|=\left|\dfrac{(x+1)^{n+1}}{(n+1)3^{n+1}}\cdot\dfrac{n\cdot 3^{n}}{(x+1)^{n}}\right|=\dfrac{|x+1|}{3}\cdot\dfrac{n}{n+1}\longrightarrow\dfrac{|x+1|}{3}$。(M1)
The series converges when $\dfrac{|x+1|}{3}<1$, i.e. $|x+1|<3$, so $R=3$. (A1)当 $\dfrac{|x+1|}{3}<1$ 即 $|x+1|<3$ 时级数收敛，故 $R=3$。(A1)
Which series represents $\cos(x^{2})$?下列哪个级数表示 $\cos(x^{2})$？
Start from $\cos u=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^{n}u^{2n}}{(2n)!}=1-\frac{u^{2}}{2!}+\frac{u^{4}}{4!}-\cdots$, and let $u=x^{2}$. (M1)从 $\cos u=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^{n}u^{2n}}{(2n)!}=1-\frac{u^{2}}{2!}+\frac{u^{4}}{4!}-\cdots$ 出发，令 $u=x^{2}$。(M1)
$$ \cos(x^{2})=1-\frac{(x^{2})^{2}}{2!}+\frac{(x^{2})^{4}}{4!}-\cdots=1-\frac{x^{4}}{2!}+\frac{x^{8}}{4!}-\cdots, $$which is option (A). (A1)即选项 (A)。(A1)
A power series representation of $\dfrac{x}{1-x^{3}}$ is$\dfrac{x}{1-x^{3}}$ 的幂级数表示为
Start from $\dfrac{1}{1-u}=\displaystyle\sum_{n=0}^{\infty}u^{n}$ for $|u|<1$, and let $u=x^{3}$: $\dfrac{1}{1-x^{3}}=\displaystyle\sum_{n=0}^{\infty}x^{3n}$. (M1)从 $\dfrac{1}{1-u}=\displaystyle\sum_{n=0}^{\infty}u^{n}$（$|u|<1$）出发，令 $u=x^{3}$：$\dfrac{1}{1-x^{3}}=\displaystyle\sum_{n=0}^{\infty}x^{3n}$。(M1)
Multiply every term by $x$: $\dfrac{x}{1-x^{3}}=\displaystyle\sum_{n=0}^{\infty}x\cdot x^{3n}=\displaystyle\sum_{n=0}^{\infty}x^{3n+1}=x+x^{4}+x^{7}+\cdots$. (A1)每项乘以 $x$：$\dfrac{x}{1-x^{3}}=\displaystyle\sum_{n=0}^{\infty}x\cdot x^{3n}=\displaystyle\sum_{n=0}^{\infty}x^{3n+1}=x+x^{4}+x^{7}+\cdots$。(A1)
For each series below, state whether it converges or diverges, name the test used, and (if it converges by a direct computation) give the sum: (a) $\displaystyle\sum_{n=1}^{\infty}\frac{n}{n+5}$; (b) $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+2)}$; (c) $\displaystyle\sum_{n=0}^{\infty}\frac{3^{n}}{n!}$.对下列各级数，判断其是收敛还是发散，写出所用判别法，并（若可直接求和）给出和：(a) $\displaystyle\sum_{n=1}^{\infty}\frac{n}{n+5}$；(b) $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+2)}$；(c) $\displaystyle\sum_{n=0}^{\infty}\frac{3^{n}}{n!}$。
$\displaystyle\lim_{n\to\infty}\frac{n}{n+5}=1\ne 0$. (M1)$\displaystyle\lim_{n\to\infty}\frac{n}{n+5}=1\ne 0$。(M1)
By the nth Term Test, the series diverges. (A1)由第 n 项判别法，级数发散。(A1)
$\dfrac{1}{n(n+2)}=\dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+2}\right)$, so $S_N=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{N+1}-\dfrac{1}{N+2}\right)$. (M1)$\dfrac{1}{n(n+2)}=\dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+2}\right)$，故 $S_N=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{N+1}-\dfrac{1}{N+2}\right)$。(M1)
As $N\to\infty$, the sum $\to\dfrac{1}{2}\left(\dfrac{3}{2}\right)=\dfrac{3}{4}$. (A1)当 $N\to\infty$ 时，和 $\to\dfrac{1}{2}\left(\dfrac{3}{2}\right)=\dfrac{3}{4}$。(A1)
$\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{3}{n+1}\to 0<1$. (M1)$\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{3}{n+1}\to 0<1$。(M1)
By the ratio test, the series converges (in fact, to $e^{3}$, though the sum is not required here). (A1)由比值判别法，级数收敛（实际上收敛到 $e^{3}$，但此处不要求求和）。(A1)
Let $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n+1}$.设 $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n+1}$。
$b_n=\dfrac{1}{2n+1}>0$ for all $n\ge 1$. (M1)对所有 $n\ge 1$，$b_n=\dfrac{1}{2n+1}>0$。(M1)
$b_{n+1}=\dfrac{1}{2n+3}<\dfrac{1}{2n+1}=b_n$ since the denominator is strictly increasing, so $\{b_n\}$ is decreasing. (M1)因分母严格递增，$b_{n+1}=\dfrac{1}{2n+3}<\dfrac{1}{2n+1}=b_n$，故 $\{b_n\}$ 递减。(M1)
$\displaystyle\lim_{n\to\infty}b_n=\lim_{n\to\infty}\frac{1}{2n+1}=0$. (A1)$\displaystyle\lim_{n\to\infty}b_n=\lim_{n\to\infty}\frac{1}{2n+1}=0$。(A1)
All three conditions hold, so the series converges by the Alternating Series Test. (R1)三个条件均满足，故由交错级数判别法，级数收敛。(R1)
$\displaystyle\sum|a_n|=\sum\frac{1}{2n+1}$; compare with $b_n=1/n$: $\displaystyle\lim_{n\to\infty}\frac{1/(2n+1)}{1/n}=\lim_{n\to\infty}\frac{n}{2n+1}=\frac{1}{2}$. (M1)$\displaystyle\sum|a_n|=\sum\frac{1}{2n+1}$；与 $b_n=1/n$ 比较：$\displaystyle\lim_{n\to\infty}\frac{1/(2n+1)}{1/n}=\lim_{n\to\infty}\frac{n}{2n+1}=\frac{1}{2}$。(M1)
Since $0<\tfrac12<\infty$ and $\sum 1/n$ (harmonic) diverges, the LCT gives $\sum|a_n|$ diverges: the series is not absolutely convergent. (A1)因 $0<\tfrac12<\infty$ 且 $\sum 1/n$（调和级数）发散，由极限比较判别法知 $\sum|a_n|$ 发散：该级数不是绝对收敛。(A1)
Since the original series converges (part (a)) but $\sum|a_n|$ diverges, the series is conditionally convergent. (R1)因原级数收敛（(a) 部分）而 $\sum|a_n|$ 发散，该级数为条件收敛。(R1)
Need $|a_{n+1}|=\dfrac{1}{2(n+1)+1}=\dfrac{1}{2n+3}<0.05$, i.e. $2n+3>20$, so $n>8.5$. (M1)需要 $|a_{n+1}|=\dfrac{1}{2(n+1)+1}=\dfrac{1}{2n+3}<0.05$，即 $2n+3>20$，故 $n>8.5$。(M1)
The smallest integer satisfying this is $n=9$ (check: $\tfrac{1}{2(9)+3}=\tfrac{1}{21}\approx 0.048<0.05$, while $n=8$ gives $\tfrac{1}{19}\approx 0.053>0.05$). (A1)满足条件的最小整数为 $n=9$（检验：$\tfrac{1}{2(9)+3}=\tfrac{1}{21}\approx 0.048<0.05$，而 $n=8$ 时 $\tfrac{1}{19}\approx 0.053>0.05$）。(A1)
Let $f(x)=\ln x$, and let $P_3(x)$ be the 3rd-degree Taylor polynomial for $f$ centered at $a=1$.设 $f(x)=\ln x$，$P_3(x)$ 为 $f$ 在 $a=1$ 处的 3 次泰勒多项式。
$f(1)=0$; $f'(x)=1/x\Rightarrow f'(1)=1$; $f''(x)=-1/x^{2}\Rightarrow f''(1)=-1$; $f'''(x)=2/x^{3}\Rightarrow f'''(1)=2$. (M1)$f(1)=0$；$f'(x)=1/x\Rightarrow f'(1)=1$；$f''(x)=-1/x^{2}\Rightarrow f''(1)=-1$；$f'''(x)=2/x^{3}\Rightarrow f'''(1)=2$。(M1)
Substitute into $P_3(x)=f(1)+f'(1)(x-1)+\dfrac{f''(1)}{2!}(x-1)^{2}+\dfrac{f'''(1)}{3!}(x-1)^{3}$: (M1)代入 $P_3(x)=f(1)+f'(1)(x-1)+\dfrac{f''(1)}{2!}(x-1)^{2}+\dfrac{f'''(1)}{3!}(x-1)^{3}$：(M1)
$$ P_3(x)=(x-1)-\frac{(x-1)^{2}}{2}+\frac{(x-1)^{3}}{3}. $$(A1)
With $x-1=0.2$: (M1)取 $x-1=0.2$：(M1)
$$ P_3(1.2)=0.2-\frac{(0.2)^{2}}{2}+\frac{(0.2)^{3}}{3}=0.2-0.02+0.002\overline{6}\approx 0.18267. $$(A1)
Here $M=6$, $n=3$ (so $n+1=4$), $|x-a|=0.2$. (M1)此处 $M=6$，$n=3$（故 $n+1=4$），$|x-a|=0.2$。(M1)
$$ |R_3(1.2)|\le\frac{M}{4!}|x-a|^{4}=\frac{6}{24}(0.2)^{4}=0.25\times 0.0016=0.0004. $$(M1) (A1)
The actual error is $|0.18267-0.18232|\approx 0.00035$, which is indeed $\le 0.0004$, confirming the bound is valid. (R1)实际误差为 $|0.18267-0.18232|\approx 0.00035$，确实 $\le 0.0004$，验证该误差界成立。(R1)
Let $\displaystyle\sum_{n=1}^{\infty}\frac{(x-2)^{n}}{n\cdot 5^{n}}$.设 $\displaystyle\sum_{n=1}^{\infty}\frac{(x-2)^{n}}{n\cdot 5^{n}}$。
$\left|\dfrac{a_{n+1}}{a_n}\right|=\left|\dfrac{(x-2)^{n+1}}{(n+1)5^{n+1}}\cdot\dfrac{n\cdot 5^{n}}{(x-2)^{n}}\right|=\dfrac{|x-2|}{5}\cdot\dfrac{n}{n+1}$. (M1)$\left|\dfrac{a_{n+1}}{a_n}\right|=\left|\dfrac{(x-2)^{n+1}}{(n+1)5^{n+1}}\cdot\dfrac{n\cdot 5^{n}}{(x-2)^{n}}\right|=\dfrac{|x-2|}{5}\cdot\dfrac{n}{n+1}$。(M1)
Taking $n\to\infty$: $L=\dfrac{|x-2|}{5}$; convergence requires $L<1$. (M1)取 $n\to\infty$：$L=\dfrac{|x-2|}{5}$；收敛要求 $L<1$。(M1)
$|x-2|<5$, so $R=5$. (A1)$|x-2|<5$，故 $R=5$。(A1)
At $x=-3$: $(x-2)=-5$, so the term is $\dfrac{(-5)^{n}}{n\cdot 5^{n}}=\dfrac{(-1)^{n}}{n}$, the (negative of the) alternating harmonic series, which converges by the AST. (M1)(A1)在 $x=-3$ 处：$(x-2)=-5$，通项为 $\dfrac{(-5)^{n}}{n\cdot 5^{n}}=\dfrac{(-1)^{n}}{n}$，即交错调和级数（取相反数），由交错级数判别法收敛。(M1)(A1)
At $x=7$: $(x-2)=5$, so the term is $\dfrac{5^{n}}{n\cdot 5^{n}}=\dfrac{1}{n}$, the harmonic series, which diverges. (M1)(A1)在 $x=7$ 处：$(x-2)=5$，通项为 $\dfrac{5^{n}}{n\cdot 5^{n}}=\dfrac{1}{n}$，即调和级数，发散。(M1)(A1)
Include $x=-3$ (converges), exclude $x=7$ (diverges): the interval of convergence is $[-3,7)$. (A1)包含 $x=-3$（收敛），排除 $x=7$（发散）：收敛区间为 $[-3,7)$。(A1)
At $x=-3$, the series of absolute values is $\displaystyle\sum\left|\frac{(-1)^{n}}{n}\right|=\sum\frac{1}{n}$, the harmonic series, which diverges. (M1)在 $x=-3$ 处，绝对值级数为 $\displaystyle\sum\left|\frac{(-1)^{n}}{n}\right|=\sum\frac{1}{n}$，即调和级数，发散。(M1)
Since the series converges at $x=-3$ (part (b)) but the absolute series diverges, the convergence there is conditional. (A1)因该级数在 $x=-3$ 处收敛（(b) 部分）而绝对值级数发散，故该处为条件收敛。(A1)
Let $g(x)=\ln(1+x)$.设 $g(x)=\ln(1+x)$。
From $\dfrac{1}{1-u}=\sum u^{n}$, let $u=-x$: $\dfrac{1}{1+x}=\displaystyle\sum_{n=0}^{\infty}(-1)^{n}x^{n}$. (M1)由 $\dfrac{1}{1-u}=\sum u^{n}$，令 $u=-x$：$\dfrac{1}{1+x}=\displaystyle\sum_{n=0}^{\infty}(-1)^{n}x^{n}$。(M1)
Integrate term by term: $\displaystyle\int\frac{dx}{1+x}=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+1}}{n+1}+C$. (M1)逐项积分：$\displaystyle\int\frac{dx}{1+x}=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+1}}{n+1}+C$。(M1)
Since $g(0)=\ln 1=0$, evaluating the series at $x=0$ gives $C=0$, so $\ln(1+x)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+1}}{n+1}=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots$. (A1)因 $g(0)=\ln 1=0$，将级数在 $x=0$ 处求值得 $C=0$，故 $\ln(1+x)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+1}}{n+1}=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots$。(A1)
(M1)(M1)
$$ 0.1-\frac{(0.1)^{2}}{2}+\frac{(0.1)^{3}}{3}-\frac{(0.1)^{4}}{4}=0.1-0.005+0.0003\overline{3}-0.000025\approx 0.09531. $$(A1)
The first omitted term is the $n=4$ term: $\dfrac{x^{5}}{5}=\dfrac{(0.1)^{5}}{5}$. (M1)第一个被舍弃的项为 $n=4$ 项：$\dfrac{x^{5}}{5}=\dfrac{(0.1)^{5}}{5}$。(M1)
$\dfrac{0.00001}{5}=0.000002=2\times10^{-6}$, so the error in part (b) is at most $2\times10^{-6}$. (A1)$\dfrac{0.00001}{5}=0.000002=2\times10^{-6}$，故 (b) 中的误差至多为 $2\times10^{-6}$。(A1)
The geometric series for $1/(1+x)$ converges for $|x|<1$, and term-by-term integration preserves the radius, so $R=1$. (M1)$1/(1+x)$ 的等比级数在 $|x|<1$ 时收敛，逐项积分保持半径不变，故 $R=1$。(M1)
At $x=1$: $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+1}=1-\frac12+\frac13-\cdots$; $b_n=1/(n+1)$ is positive, decreasing, and $\to 0$, so this converges by the AST. (M1)在 $x=1$ 处：$\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n+1}=1-\frac12+\frac13-\cdots$；$b_n=1/(n+1)$ 为正、递减且 $\to 0$，由交错级数判别法收敛。(M1)
So $x=1$ is included, and the interval of convergence for the series in (a) is $(-1,1]$. (A1)故 $x=1$ 属于收敛区间，(a) 中级数的收敛区间为 $(-1,1]$。(A1)