Sections 1 to 7: sequence limits, series and partial sums, geometric and telescoping series, the integral test, p-series, comparison tests, alternating series, ratio and root testsCALC II第 1 至 7 节：数列极限、级数与部分和、等比级数与裂项级数、积分判别法、p-级数、比较判别法、交错级数、比值判别法与根式判别法微积分II
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PART I · CORE TECHNIQUES第一部分 · 核心技巧Computational fluency · 28 marks计算能力 · 28分
Sequence Limits, Series Evaluation, and the Integral Test数列极限、级数求值与积分判别法
Show all working. For each sequence limit, state the rule or theorem you invoke. For each series, state whether it converges or diverges and give the exact sum when one exists. A verdict given without justification earns no credit.须写出完整解题过程。对于每个数列极限，请注明所用的法则或定理。对于每个级数，请说明其收敛或发散，并在存在精确和的情况下给出精确结果。未作任何说明的结论不得分。
Q1MEDIUMCOREsequence limits and the monotone convergence theorem数列极限与单调收敛定理[8 marks]
Determine whether each sequence converges or diverges. If it converges, find the exact limit.判断每个数列是否收敛或发散。若收敛，求其精确极限。
(a) $a_n = \dfrac{3n^{2}-2n+1}{5n^{2}+4}$ [2]
(b) $b_n = \dfrac{(-1)^{n}\,n}{n^{2}+1}$ [2]
(c)Define $c_1 = 1$ and $c_{n+1} = \sqrt{2 + c_n}$ for $n \ge 1$. Show that $\{c_n\}$ is increasing and bounded above by $3$, and hence converges. Then find $\displaystyle\lim_{n\to\infty} c_n$.设 $c_1 = 1$，$c_{n+1} = \sqrt{2 + c_n}$（$n \ge 1$）。证明 $\{c_n\}$ 单调递增且有上界 $3$，从而收敛，再求 $\displaystyle\lim_{n\to\infty} c_n$。[4]
Q2MEDIUMCOREgeometric series and the nth-term divergence test等比级数与第n项发散判别法[8 marks]
For each series, state whether it converges or diverges. If it converges, give the exact sum. Identify the test or formula you use.对每个级数，说明其收敛或发散。若收敛，给出精确和。注明所用判别法或公式。
Q3MEDIUMCOREtelescoping series via partial fractions利用部分分式求裂项级数[5 marks]
Consider the series $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+2)}$.考虑级数 $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+2)}$。
(a)Write $\dfrac{1}{n(n+2)}$ as a partial fraction $\dfrac{A}{n}+\dfrac{B}{n+2}$ and find $A$ and $B$.将 $\dfrac{1}{n(n+2)}$ 分解为部分分式 $\dfrac{A}{n}+\dfrac{B}{n+2}$，求 $A$ 和 $B$。[2]
(b)Write down the $N$th partial sum $S_N$ by listing which terms survive the telescoping cancellation, then evaluate $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+2)}$ exactly.列出裂项相消后保留的项，写出第 $N$ 个部分和 $S_N$，然后精确求 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+2)}$。[3]
Q4HARDCOREintegral test and p-series积分判别法与p-级数[7 marks]
For each series, state a convergence verdict with a complete justification. Do not assume the integral test applies without verifying its hypotheses.对每个级数，给出完整论证后的收敛结论。在验证积分判别法的前提条件之前，不得直接引用该判别法。
(b)$\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}$. Apply the integral test: verify its three hypotheses, compute the improper integral, and state the verdict.$\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}$。应用积分判别法：验证三个前提条件，计算广义积分，并给出结论。[4]
(c)For the $p$-series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{p}}$, state the exact values of $p$ for which it converges and for which it diverges, identifying the boundary case $p=1$.对于 $p$-级数 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{p}}$，说明使其收敛和发散的 $p$ 的精确范围，并指明边界情形 $p=1$。[1]
PART II · DEFINITIONS AND PROOF第二部分 · 定义与证明Rigorous arguments · 26 marks严格论证 · 26分
Proofs: Geometric Series, the Divergence Test, and Comparison证明：等比级数、发散判别法与比较判别法
These items are graded on the logic of the argument. State every theorem you invoke by name before applying it. In a convergence proof, verify all hypotheses before citing the conclusion.本部分按论证逻辑评分。每次引用定理前须先点名定理。在收敛性证明中，需先验证所有前提条件，再得出结论。
Q5HARDPROOFderiving the geometric series sum formula推导等比级数求和公式[8 marks]
Let $r$ be a real number with $|r|<1$ and let $S_N = \displaystyle\sum_{n=0}^{N} r^n$.设 $r$ 为实数且 $|r|<1$，令 $S_N = \displaystyle\sum_{n=0}^{N} r^n$。
(a)Compute $S_N - r\,S_N$ and hence prove that $S_N = \dfrac{1-r^{N+1}}{1-r}$ for any $N \ge 0$.计算 $S_N - r\,S_N$，从而证明对任意 $N \ge 0$ 均有 $S_N = \dfrac{1-r^{N+1}}{1-r}$。[4]
(b)Using the result of (a), prove that $\displaystyle\sum_{n=0}^{\infty} r^{n} = \dfrac{1}{1-r}$ when $|r|<1$. You must show that $r^{N+1}\to 0$ as $N\to\infty$, and explain why the argument fails when $|r|\ge 1$.利用 (a) 的结果，证明当 $|r|<1$ 时 $\displaystyle\sum_{n=0}^{\infty} r^{n} = \dfrac{1}{1-r}$。须证明 $r^{N+1}\to 0$（$N\to\infty$），并解释当 $|r|\ge 1$ 时论证失效的原因。[4]
Q6HARDPROOFproving the nth-term divergence test and the divergence of the harmonic series证明第n项发散判别法与调和级数的发散性[9 marks]
This question establishes two foundational results.本题建立两个基础结论。
(a)Suppose $\displaystyle\sum_{n=1}^{\infty} a_n$ converges to $S$. Let $S_N = \displaystyle\sum_{n=1}^{N} a_n$ be the $N$th partial sum. Express $a_N$ in terms of $S_N$ and $S_{N-1}$, and use the convergence of $\{S_N\}$ to prove that $a_N \to 0$ as $N\to\infty$. State clearly why this implies: if $a_n \not\to 0$, then $\displaystyle\sum a_n$ diverges.设 $\displaystyle\sum_{n=1}^{\infty} a_n$ 收敛于 $S$，令 $S_N = \displaystyle\sum_{n=1}^{N} a_n$ 为第 $N$ 个部分和。用 $S_N$ 和 $S_{N-1}$ 表示 $a_N$，并利用 $\{S_N\}$ 的收敛性证明 $a_N \to 0$（$N\to\infty$）。清楚说明由此可得：若 $a_n \not\to 0$，则 $\displaystyle\sum a_n$ 发散。[5]
(b)Use the integral test to prove that $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ diverges. Verify the three hypotheses of the integral test for $f(x)=\dfrac{1}{x}$ on $[1,\infty)$, evaluate $\displaystyle\int_{1}^{\infty}\frac{1}{x}\,dx$, and state the conclusion.用积分判别法证明 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ 发散。对 $[1,\infty)$ 上的 $f(x)=\dfrac{1}{x}$ 验证积分判别法的三个前提条件，计算 $\displaystyle\int_{1}^{\infty}\frac{1}{x}\,dx$，并给出结论。[4]
Q7HARDPROOFcomparison and limit-comparison tests比较判别法与极限比较判别法[9 marks]
Use the stated test in each part. For each, name the comparison series, justify the comparison or limit, and give the verdict.在每小问中使用题目指定的判别法。对每小问，写明比较级数，论证比较关系或极限，并给出结论。
(a)Use the direct comparison test to determine whether $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}+n+1}$ converges or diverges. State clearly which series you compare to and why the inequality holds.用直接比较判别法判断 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}+n+1}$ 的收敛性。清楚说明选取的比较级数及不等式成立的理由。[4]
(b)Use the limit comparison test to determine whether $\displaystyle\sum_{n=1}^{\infty}\frac{3n^{2}-n+2}{n^{4}+5n}$ converges or diverges. Identify the comparison series, compute the limit $L = \displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}$, verify $0 < L < \infty$, and state the conclusion.用极限比较判别法判断 $\displaystyle\sum_{n=1}^{\infty}\frac{3n^{2}-n+2}{n^{4}+5n}$ 的收敛性。写明比较级数，计算极限 $L = \displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}$，验证 $0 < L < \infty$，并给出结论。[5]
PART III · APPLICATIONS AND SYNTHESIS第三部分 · 应用与综合Extended problems · 28 marks综合题 · 28分
Ratio and Root Tests, Absolute vs. Conditional Convergence, and the Alternating Series Remainder比值判别法与根式判别法、绝对收敛与条件收敛、交错级数余项估计
Set up each test cleanly and state its hypothesis. When a test is inconclusive, you must say so and switch to a different test. Carry exact values through intermediate steps.清晰建立每个判别法并陈述其假设条件。若判别法无法得出结论，须明确说明并改用其他判别法。中间步骤保留精确值。
Q8HARDAPPLIEDratio and root tests: test selection and full justification比值判别法与根式判别法：判别法选择与完整论证[9 marks]
For each series, choose the ratio test or root test, carry out the computation, and give the verdict. If the test returns $L=1$, say so and use a different method.对每个级数，选择比值判别法或根式判别法，完成计算并给出结论。若判别结果为 $L=1$，须明确说明并改用其他方法。
Q9HARDAPPLIEDabsolute versus conditional convergence and the Leibniz test绝对收敛与条件收敛及莱布尼茨判别法[9 marks]
For each series, determine whether it converges absolutely, converges conditionally, or diverges. Show all reasoning; a bare verdict earns no credit.对每个级数，判断其绝对收敛、条件收敛还是发散。须写出完整推理过程，仅给出结论不得分。
Q10HARDAPPLIEDalternating series remainder bound交错级数余项估计[10 marks]
Consider the alternating series $\displaystyle S = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{3}}= 1 - \frac{1}{8} + \frac{1}{27} - \frac{1}{64} + \cdots$考虑交错级数 $\displaystyle S = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{3}}= 1 - \frac{1}{8} + \frac{1}{27} - \frac{1}{64} + \cdots$
(a)Verify that the Leibniz Alternating Series Test applies to this series. You must check both hypotheses: that the terms $b_n = \dfrac{1}{n^{3}}$ are eventually decreasing, and that $b_n \to 0$.验证莱布尼茨交错级数判别法适用于本级数。须检验两个条件：各项 $b_n = \dfrac{1}{n^{3}}$ 最终单调递减，以及 $b_n \to 0$。[2]
(b)Write down the 4th partial sum $S_4 = \displaystyle\sum_{n=1}^{4}\frac{(-1)^{n+1}}{n^{3}}$ as an exact fraction.将第4个部分和 $S_4 = \displaystyle\sum_{n=1}^{4}\frac{(-1)^{n+1}}{n^{3}}$ 写成精确分数。[3]
(c)State the Alternating Series Remainder theorem and use it to find the smallest $N$ such that $S_N$ approximates $S$ to within $0.005$. Justify why the first omitted term $b_{N+1}$ bounds the error, and verify your $N$ satisfies the bound.陈述交错级数余项定理，并用它求最小的 $N$，使得 $S_N$ 对 $S$ 的近似误差不超过 $0.005$。解释第一个被省略的项 $b_{N+1}$ 何以控制误差，并验证所求 $N$ 满足该误差界。[5]