Companion to the University-Style Practice Set配套大学风格练习题集
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
Determine convergence of: (a) $a_n=\dfrac{3n^{2}-2n+1}{5n^{2}+4}$; (b) $b_n=\dfrac{(-1)^{n}n}{n^{2}+1}$; (c) $c_1=1$, $c_{n+1}=\sqrt{2+c_n}$ (show monotone, bounded, then find the limit).判断以下数列的收敛性：(a) $a_n=\dfrac{3n^{2}-2n+1}{5n^{2}+4}$；(b) $b_n=\dfrac{(-1)^{n}n}{n^{2}+1}$；(c) $c_1=1$，$c_{n+1}=\sqrt{2+c_n}$（证明单调有界，再求极限）。
Divide every term in numerator and denominator by $n^{2}$: (M1)将分子与分母的每一项均除以 $n^{2}$：(M1)
$$ a_n = \frac{3 - 2/n + 1/n^{2}}{5 + 4/n^{2}} \longrightarrow \frac{3-0+0}{5+0} = \frac{3}{5}. $$The sequence converges to $\dfrac{3}{5}$. (A1)该数列收敛至 $\dfrac{3}{5}$。(A1)
For all $n \ge 1$, $n^{2}+1 > n^{2}$, so $|b_n| = \dfrac{n}{n^{2}+1} < \dfrac{n}{n^{2}} = \dfrac{1}{n}$. (M1)对所有 $n \ge 1$，$n^{2}+1 > n^{2}$，故 $|b_n| = \dfrac{n}{n^{2}+1} < \dfrac{n}{n^{2}} = \dfrac{1}{n}$。(M1)
Thus $-\dfrac{1}{n} < b_n < \dfrac{1}{n}$. Since $\pm 1/n \to 0$, the squeeze theorem gives $b_n \to 0$. The alternating sign does not prevent convergence; it is absorbed into the squeeze. (A1)因此 $-\dfrac{1}{n} < b_n < \dfrac{1}{n}$。由于 $\pm 1/n \to 0$，夹逼定理给出 $b_n \to 0$。交替符号不影响收敛性，已被夹逼吸收。(A1)
Increasing.单调递增。 Compute $c_2 = \sqrt{2+1} = \sqrt{3} \approx 1.73 > 1 = c_1$. Suppose $c_n > c_{n-1}$; then $c_{n+1} = \sqrt{2+c_n} > \sqrt{2+c_{n-1}} = c_n$. By induction $\{c_n\}$ is increasing. (M1)计算 $c_2 = \sqrt{2+1} = \sqrt{3} \approx 1.73 > 1 = c_1$。假设 $c_n > c_{n-1}$，则 $c_{n+1} = \sqrt{2+c_n} > \sqrt{2+c_{n-1}} = c_n$。由归纳法，$\{c_n\}$ 单调递增。(M1)
Bounded above by $3$.有上界 $3$。 Base: $c_1 = 1 < 3$. Inductive step: if $c_n < 3$ then $c_{n+1} = \sqrt{2+c_n} < \sqrt{2+3} = \sqrt{5} < 3$. (M1)基础情形：$c_1 = 1 < 3$。归纳步骤：若 $c_n < 3$，则 $c_{n+1} = \sqrt{2+c_n} < \sqrt{2+3} = \sqrt{5} < 3$。(M1)
By the Monotone Convergence Theorem, $L = \displaystyle\lim_{n\to\infty} c_n$ exists. (A1)由单调收敛定理，$L = \displaystyle\lim_{n\to\infty} c_n$ 存在。(A1)
Finding $L$.求 $L$。 Taking the limit on both sides of $c_{n+1} = \sqrt{2+c_n}$ gives $L = \sqrt{2+L}$, so $L^{2} = 2 + L$, i.e. $L^{2} - L - 2 = 0$, which factors as $(L-2)(L+1) = 0$. Since $c_n \ge 1 > 0$ for all $n$, we discard $L = -1$ and conclude $L = 2$. (A1)对 $c_{n+1} = \sqrt{2+c_n}$ 两边取极限得 $L = \sqrt{2+L}$，故 $L^{2} = 2 + L$，即 $L^{2} - L - 2 = 0$，因式分解得 $(L-2)(L+1) = 0$。由于对所有 $n$ 均有 $c_n \ge 1 > 0$，舍去 $L = -1$，得 $L = 2$。(A1)
Convergence of: (a) $\displaystyle\sum_{n=0}^{\infty}\tfrac{5}{3^{n}}$; (b) $\displaystyle\sum_{n=1}^{\infty}\tfrac{2^{n}+3^{n}}{6^{n}}$; (c) $\displaystyle\sum_{n=1}^{\infty}\tfrac{n^{2}}{n^{2}+1}$.判断以下级数的收敛性：(a) $\displaystyle\sum_{n=0}^{\infty}\tfrac{5}{3^{n}}$；(b) $\displaystyle\sum_{n=1}^{\infty}\tfrac{2^{n}+3^{n}}{6^{n}}$；(c) $\displaystyle\sum_{n=1}^{\infty}\tfrac{n^{2}}{n^{2}+1}$。
The series has first term $a = 5$ and common ratio $r = 1/3$, with $|r| = 1/3 < 1$. (M1) By the geometric series formula,该级数首项 $a = 5$，公比 $r = 1/3$，$|r| = 1/3 < 1$。(M1) 由等比级数公式，
$$ \sum_{n=0}^{\infty} \frac{5}{3^{n}} = \frac{5}{1 - 1/3} = \frac{5}{2/3} = \frac{15}{2}. \quad \text{(A1)} $$Write $\dfrac{2^{n}+3^{n}}{6^{n}} = \left(\dfrac{2}{6}\right)^{n} + \left(\dfrac{3}{6}\right)^{n} = \left(\dfrac{1}{3}\right)^{n} + \left(\dfrac{1}{2}\right)^{n}$. (M1)将 $\dfrac{2^{n}+3^{n}}{6^{n}} = \left(\dfrac{2}{6}\right)^{n} + \left(\dfrac{3}{6}\right)^{n} = \left(\dfrac{1}{3}\right)^{n} + \left(\dfrac{1}{2}\right)^{n}$。(M1)
Both geometric series start at $n=1$ and have ratio inside $(-1,1)$:两个等比级数均从 $n=1$ 开始，公比均在 $(-1,1)$ 内：
$$ \sum_{n=1}^{\infty}\left(\frac{1}{3}\right)^{n} = \frac{1/3}{1-1/3} = \frac{1}{2}, \qquad \sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^{n} = \frac{1/2}{1-1/2} = 1. \quad \text{(A1 each)} $$The total sum is $\dfrac{1}{2} + 1 = \dfrac{3}{2}$. (A1)两者之和为 $\dfrac{1}{2} + 1 = \dfrac{3}{2}$。(A1)
Compute the term limit: $\displaystyle\lim_{n\to\infty}\frac{n^{2}}{n^{2}+1} = \lim_{n\to\infty}\frac{1}{1+1/n^{2}} = 1 \ne 0$. (M1)计算通项极限：$\displaystyle\lim_{n\to\infty}\frac{n^{2}}{n^{2}+1} = \lim_{n\to\infty}\frac{1}{1+1/n^{2}} = 1 \ne 0$。(M1)
Because the terms do not tend to zero, the $n$th-term divergence test immediately gives divergence. (A1) The series diverges. (R1)由于通项不趋于零，第 $n$ 项发散判别法直接得出发散结论。(A1) 该级数发散。(R1)
For $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+2)}$: (a) find the partial fraction decomposition; (b) collapse the $N$th partial sum and evaluate the series.对于 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+2)}$：(a) 求部分分式分解；(b) 化简第 $N$ 个部分和并求级数的精确值。
Write $\dfrac{1}{n(n+2)} = \dfrac{A}{n} + \dfrac{B}{n+2}$, so $1 = A(n+2) + Bn$. (M1)设 $\dfrac{1}{n(n+2)} = \dfrac{A}{n} + \dfrac{B}{n+2}$，则 $1 = A(n+2) + Bn$。(M1)
Setting $n = 0$: $1 = 2A$, so $A = \tfrac{1}{2}$. Setting $n = -2$: $1 = -2B$, so $B = -\tfrac{1}{2}$. (A1)令 $n = 0$：$1 = 2A$，故 $A = \tfrac{1}{2}$。令 $n = -2$：$1 = -2B$，故 $B = -\tfrac{1}{2}$。(A1)
Write the $N$th partial sum using the decomposition: (M1)利用分解式写出第 $N$ 个部分和：(M1)
$$ S_N = \frac{1}{2}\sum_{n=1}^{N}\left(\frac{1}{n} - \frac{1}{n+2}\right). $$List the first few terms of the expanded sum:展开前几项：
$$ \left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots+\left(\frac{1}{N}-\frac{1}{N+2}\right). $$Every interior term $\tfrac{1}{k}$ (for $3 \le k \le N$) is cancelled by the later term $-\tfrac{1}{k}$. The surviving terms are $\tfrac{1}{1}$, $\tfrac{1}{2}$, $-\tfrac{1}{N+1}$, $-\tfrac{1}{N+2}$, giving (A1)所有内部项 $\tfrac{1}{k}$（$3 \le k \le N$）均与后续项 $-\tfrac{1}{k}$ 相消。保留的项为 $\tfrac{1}{1}$、$\tfrac{1}{2}$、$-\tfrac{1}{N+1}$、$-\tfrac{1}{N+2}$，得 (A1)
$$ S_N = \frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{N+1} - \frac{1}{N+2}\right). $$As $N \to \infty$, the last two terms vanish:当 $N \to \infty$ 时，最后两项趋于零：
$$ \sum_{n=1}^{\infty}\frac{1}{n(n+2)} = \frac{1}{2}\left(1+\frac{1}{2}\right) = \frac{1}{2}\cdot\frac{3}{2} = \frac{3}{4}. \quad \text{(A1)} $$Convergence with justification: (a) $\displaystyle\sum_{n=1}^{\infty}\tfrac{1}{n^{3/2}}$; (b) $\displaystyle\sum_{n=2}^{\infty}\tfrac{1}{n\ln n}$ by the integral test; (c) convergence values of $p$ for $\displaystyle\sum\tfrac{1}{n^{p}}$.附完整论证判断收敛性：(a) $\displaystyle\sum_{n=1}^{\infty}\tfrac{1}{n^{3/2}}$；(b) 用积分判别法处理 $\displaystyle\sum_{n=2}^{\infty}\tfrac{1}{n\ln n}$；(c) 确定 $\displaystyle\sum\tfrac{1}{n^{p}}$ 收敛的 $p$ 值范围。
This is a $p$-series with $p = \tfrac{3}{2} > 1$. (A1) By the $p$-series theorem, it converges. (R1)这是 $p = \tfrac{3}{2} > 1$ 的 $p$-级数。(A1) 由 $p$-级数定理，该级数收敛。(R1)
Verify hypotheses.验证前提条件。 For $x \ge 2$, $f(x) = \dfrac{1}{x\ln x}$ is positive ($\ln x > 0$ for $x > 1$), continuous, and decreasing (both $x$ and $\ln x$ are increasing, so their product is increasing, and its reciprocal is decreasing). (M1)当 $x \ge 2$ 时，$f(x) = \dfrac{1}{x\ln x}$ 为正值（$x > 1$ 时 $\ln x > 0$），连续，且单调递减（$x$ 与 $\ln x$ 均递增，故其乘积递增，倒数递减）。(M1)
Evaluate the improper integral.计算广义积分。 Substitute $u = \ln x$, $du = dx/x$: (M1)令 $u = \ln x$，$du = dx/x$：(M1)
$$ \int_{2}^{\infty}\frac{dx}{x\ln x} = \int_{\ln 2}^{\infty}\frac{du}{u} = \Big[\ln u\Big]_{\ln 2}^{\infty} = \infty. \quad \text{(A1)} $$Since the integral diverges, the integral test gives: $\displaystyle\sum_{n=2}^{\infty}\dfrac{1}{n\ln n}$ diverges. (A1)由于广义积分发散，积分判别法给出：$\displaystyle\sum_{n=2}^{\infty}\dfrac{1}{n\ln n}$ 发散。(A1)
$\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^{p}}$ converges if and only if $p > 1$. It diverges for $p \le 1$, including the boundary case $p = 1$ (the harmonic series), which diverges as established by the integral test since $\int_1^\infty x^{-1}\,dx = \infty$. (A1)$\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^{p}}$ 当且仅当 $p > 1$ 时收敛。当 $p \le 1$ 时发散，包括边界情形 $p = 1$（调和级数），由积分判别法可知 $\int_1^\infty x^{-1}\,dx = \infty$ 故发散。(A1)
With $S_N = \displaystyle\sum_{n=0}^{N} r^{n}$: (a) prove $S_N = \dfrac{1-r^{N+1}}{1-r}$; (b) prove $\displaystyle\sum_{n=0}^{\infty} r^{n} = \dfrac{1}{1-r}$ when $|r| < 1$, and explain why the formula fails for $|r| \ge 1$.设 $S_N = \displaystyle\sum_{n=0}^{N} r^{n}$：(a) 证明 $S_N = \dfrac{1-r^{N+1}}{1-r}$；(b) 证明当 $|r| < 1$ 时 $\displaystyle\sum_{n=0}^{\infty} r^{n} = \dfrac{1}{1-r}$，并解释当 $|r| \ge 1$ 时公式失效的原因。
Write out $S_N$ and $rS_N$ aligned by powers of $r$: (M1)将 $S_N$ 与 $rS_N$ 按 $r$ 的幂次对齐写出：(M1)
$$ S_N = 1 + r + r^{2} + \cdots + r^{N}, $$ $$ rS_N = \phantom{1 + {}}r + r^{2} + \cdots + r^{N} + r^{N+1}. $$Subtract: (M1)相减：(M1)
$$ S_N - rS_N = 1 - r^{N+1}, \quad\text{so}\quad (1-r)S_N = 1-r^{N+1}. \quad \text{(A1)} $$Since $r \ne 1$ (which holds for $|r| < 1$ in part (b), and more generally whenever $r \ne 1$), divide both sides by $1-r$:由于 $r \ne 1$（(b) 中 $|r| < 1$ 保证，更一般地在 $r \ne 1$ 时均成立），两边除以 $1-r$：
$$ S_N = \frac{1 - r^{N+1}}{1-r}. \quad \text{(A1)} $$When $|r| < 1$: we have $|r^{N+1}| = |r|^{N+1} \to 0$ as $N \to \infty$, because $|r| < 1$ implies the geometric sequence $|r|^{N+1}$ is decreasing to zero. (M1)当 $|r| < 1$ 时：$|r^{N+1}| = |r|^{N+1} \to 0$（$N \to \infty$），因为 $|r| < 1$ 意味着等比数列 $|r|^{N+1}$ 单调递减趋于零。(M1)
Therefore (A1)从而 (A1)
$$ \sum_{n=0}^{\infty} r^{n} = \lim_{N\to\infty} S_N = \lim_{N\to\infty}\frac{1-r^{N+1}}{1-r} = \frac{1-0}{1-r} = \frac{1}{1-r}. $$When $|r| = 1$: either $r = 1$ (and the formula for $S_N$ was derived by dividing by $1-r$, which is zero, so the derivation breaks down), or $|r| = 1$ with $r \ne 1$ (so $|r^{N+1}| = 1$ for all $N$, and $r^{N+1}$ oscillates without tending to zero). (R1)当 $|r| = 1$ 时：若 $r = 1$，则 $S_N$ 的推导中需除以 $1-r = 0$，推导失效；若 $|r| = 1$ 且 $r \ne 1$，则 $|r^{N+1}| = 1$ 对所有 $N$ 成立，$r^{N+1}$ 振荡而不趋于零。(R1)
When $|r| > 1$: $|r^{N+1}| = |r|^{N+1} \to \infty$, so $S_N$ diverges. In both cases $|r| \ge 1$, the limit $\displaystyle\lim_{N\to\infty}S_N$ does not exist, and the sum formula $\dfrac{1}{1-r}$ has no meaning as the value of the series. (A1)当 $|r| > 1$ 时：$|r^{N+1}| = |r|^{N+1} \to \infty$，故 $S_N$ 发散。两种情形下（$|r| \ge 1$），$\displaystyle\lim_{N\to\infty}S_N$ 均不存在，公式 $\dfrac{1}{1-r}$ 作为级数和无任何意义。(A1)
(a) Prove: if $\sum a_n$ converges, then $a_n \to 0$, and state the contrapositive. (b) Prove $\sum 1/n$ diverges using the integral test.(a) 证明：若 $\sum a_n$ 收敛，则 $a_n \to 0$，并陈述其逆否命题。(b) 用积分判别法证明 $\sum 1/n$ 发散。
Suppose $\displaystyle\sum_{n=1}^{\infty} a_n$ converges to $S$. Let $S_N = \displaystyle\sum_{n=1}^{N} a_n$. Then (M1)设 $\displaystyle\sum_{n=1}^{\infty} a_n$ 收敛于 $S$，令 $S_N = \displaystyle\sum_{n=1}^{N} a_n$。则 (M1)
$$ a_N = S_N - S_{N-1}. $$Since the series converges, $S_N \to S$ as $N \to \infty$, and $S_{N-1} \to S$ as well (replacing $N$ by $N-1$ in the same limit). (M1) Therefore (A1)由于级数收敛，$S_N \to S$（$N \to \infty$），同理 $S_{N-1} \to S$（将 $N$ 替换为 $N-1$）。(M1) 因此 (A1)
$$ \lim_{N\to\infty} a_N = \lim_{N\to\infty}(S_N - S_{N-1}) = S - S = 0. $$The logical contrapositive of "if $\sum a_n$ converges then $a_n \to 0$" is: (R1) if $a_n \not\to 0$, then $\sum a_n$ diverges. This is the $n$th-term divergence test. (A1)"若 $\sum a_n$ 收敛则 $a_n \to 0$"的逆否命题为：(R1) 若 $a_n \not\to 0$，则 $\sum a_n$ 发散。 这就是第 $n$ 项发散判别法。(A1)
Let $f(x) = \dfrac{1}{x}$ on $[1,\infty)$. Verify the three hypotheses of the integral test: (M1)设 $f(x) = \dfrac{1}{x}$ 在 $[1,\infty)$ 上。验证积分判别法的三个前提条件：(M1)
All three hold, so the integral test applies. Evaluate the improper integral: (A1)三个条件均成立，故积分判别法适用。计算广义积分：(A1)
$$ \int_{1}^{\infty}\frac{1}{x}\,dx = \lim_{t\to\infty}\int_{1}^{t}\frac{1}{x}\,dx = \lim_{t\to\infty}[\ln x]_{1}^{t} = \lim_{t\to\infty}(\ln t - 0) = \infty. \quad \text{(A1)} $$Since the integral diverges, the integral test concludes that $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n}$ diverges. (A1)由于广义积分发散，积分判别法得出 $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n}$ 发散。(A1)
(a) Direct comparison for $\displaystyle\sum_{n=1}^{\infty}\tfrac{1}{n^{2}+n+1}$; (b) limit comparison for $\displaystyle\sum_{n=1}^{\infty}\tfrac{3n^{2}-n+2}{n^{4}+5n}$.(a) 对 $\displaystyle\sum_{n=1}^{\infty}\tfrac{1}{n^{2}+n+1}$ 用直接比较判别法；(b) 对 $\displaystyle\sum_{n=1}^{\infty}\tfrac{3n^{2}-n+2}{n^{4}+5n}$ 用极限比较判别法。
For all $n \ge 1$, observe that $n^{2}+n+1 > n^{2}$, so (M1)对所有 $n \ge 1$，注意 $n^{2}+n+1 > n^{2}$，故 (M1)
$$ 0 < \frac{1}{n^{2}+n+1} < \frac{1}{n^{2}}. $$The comparison series $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^{2}}$ is a $p$-series with $p = 2 > 1$, so it converges. (A1)比较级数 $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^{2}}$ 是 $p = 2 > 1$ 的 $p$-级数，故收敛。(A1)
By the direct comparison test, since $0 \le a_n \le b_n$ for all $n$ and $\sum b_n$ converges, $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^{2}+n+1}$ converges. (R1·A1)由直接比较判别法，因对所有 $n$ 均有 $0 \le a_n \le b_n$ 且 $\sum b_n$ 收敛，故 $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^{2}+n+1}$ 收敛。(R1·A1)
The dominant behaviour of $a_n = \dfrac{3n^{2}-n+2}{n^{4}+5n}$ for large $n$ is $\dfrac{3n^{2}}{n^{4}} = \dfrac{3}{n^{2}}$. Choose the comparison series $\displaystyle\sum b_n = \sum \dfrac{1}{n^{2}}$. (M1)$a_n = \dfrac{3n^{2}-n+2}{n^{4}+5n}$ 在 $n$ 较大时的主项行为为 $\dfrac{3n^{2}}{n^{4}} = \dfrac{3}{n^{2}}$。选取比较级数 $\displaystyle\sum b_n = \sum \dfrac{1}{n^{2}}$。(M1)
Compute the limit: (M1)计算极限：(M1)
$$ L = \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{3n^{2}-n+2}{n^{4}+5n}\cdot n^{2} = \lim_{n\to\infty}\frac{3n^{4}-n^{3}+2n^{2}}{n^{4}+5n}. $$Divide numerator and denominator by $n^{4}$: (A1)分子分母同除以 $n^{4}$：(A1)
$$ L = \lim_{n\to\infty}\frac{3 - 1/n + 2/n^{2}}{1+5/n^{3}} = \frac{3}{1} = 3. $$Since $0 < L = 3 < \infty$, the series $\sum a_n$ and $\sum b_n$ have the same convergence behavior. (A1) Since $\sum 1/n^{2}$ converges, $\displaystyle\sum_{n=1}^{\infty}\dfrac{3n^{2}-n+2}{n^{4}+5n}$ converges. (A1)由于 $0 < L = 3 < \infty$，级数 $\sum a_n$ 与 $\sum b_n$ 具有相同的收敛性。(A1) 因 $\sum 1/n^{2}$ 收敛，$\displaystyle\sum_{n=1}^{\infty}\dfrac{3n^{2}-n+2}{n^{4}+5n}$ 收敛。(A1)
Ratio or root test for: (a) $\displaystyle\sum_{n=1}^{\infty}\tfrac{n^{3}}{3^{n}}$; (b) $\displaystyle\sum_{n=1}^{\infty}\tfrac{4^{n}n!}{(2n)!}$; (c) $\displaystyle\sum_{n=1}^{\infty}\!\left(\tfrac{2n+1}{3n+2}\right)^{n}$.用比值判别法或根式判别法处理：(a) $\displaystyle\sum_{n=1}^{\infty}\tfrac{n^{3}}{3^{n}}$；(b) $\displaystyle\sum_{n=1}^{\infty}\tfrac{4^{n}n!}{(2n)!}$；(c) $\displaystyle\sum_{n=1}^{\infty}\!\left(\tfrac{2n+1}{3n+2}\right)^{n}$。
The series has both a power of $n$ and an exponential, making the ratio test natural. Let $a_n = n^{3}/3^{n}$. (M1)该级数同时含 $n$ 的幂次与指数，适合用比值判别法。设 $a_n = n^{3}/3^{n}$。(M1)
$$ \frac{a_{n+1}}{a_n} = \frac{(n+1)^{3}}{3^{n+1}}\cdot\frac{3^{n}}{n^{3}} = \frac{1}{3}\left(\frac{n+1}{n}\right)^{3} = \frac{1}{3}\left(1+\frac{1}{n}\right)^{3} \longrightarrow \frac{1}{3}\cdot 1 = \frac{1}{3}. $$Since $L = \tfrac{1}{3} < 1$, the ratio test gives convergence. (A1·R1)由于 $L = \tfrac{1}{3} < 1$，比值判别法给出收敛结论。(A1·R1)
Factorials respond well to the ratio test. Let $a_n = 4^{n}n!/(2n)!$. (M1)阶乘适合用比值判别法。设 $a_n = 4^{n}n!/(2n)!$。(M1)
$$ \frac{a_{n+1}}{a_n} = \frac{4^{n+1}(n+1)!}{(2n+2)!}\cdot\frac{(2n)!}{4^{n}n!} = \frac{4(n+1)}{(2n+2)(2n+1)} = \frac{4(n+1)}{2(n+1)(2n+1)} = \frac{2}{2n+1} \longrightarrow 0. $$Since $L = 0 < 1$, the series converges. (A1·R1)由于 $L = 0 < 1$，该级数收敛。(A1·R1)
The $n$th power makes the root test the natural choice. Let $a_n = \left(\dfrac{2n+1}{3n+2}\right)^{n}$. (M1)含 $n$ 次幂，根式判别法是自然选择。设 $a_n = \left(\dfrac{2n+1}{3n+2}\right)^{n}$。(M1)
$$ \sqrt[n]{a_n} = \frac{2n+1}{3n+2} \longrightarrow \frac{2}{3}. $$Since $L = \tfrac{2}{3} < 1$, the root test gives convergence. (A1·R1)由于 $L = \tfrac{2}{3} < 1$，根式判别法给出收敛结论。(A1·R1)
Absolute convergence, conditional convergence, or divergence for: (a) $\displaystyle\sum_{n=1}^{\infty}\tfrac{(-1)^{n}}{n^{2}}$; (b) $\displaystyle\sum_{n=1}^{\infty}\tfrac{(-1)^{n+1}}{\sqrt{n}}$; (c) $\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\tfrac{n}{n+1}$.判断以下级数绝对收敛、条件收敛还是发散：(a) $\displaystyle\sum_{n=1}^{\infty}\tfrac{(-1)^{n}}{n^{2}}$；(b) $\displaystyle\sum_{n=1}^{\infty}\tfrac{(-1)^{n+1}}{\sqrt{n}}$；(c) $\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\tfrac{n}{n+1}$。
Test $\displaystyle\sum_{n=1}^{\infty}\left|\frac{(-1)^{n}}{n^{2}}\right| = \sum_{n=1}^{\infty}\frac{1}{n^{2}}$, which is a $p$-series with $p = 2 > 1$, hence convergent. (M1)检验 $\displaystyle\sum_{n=1}^{\infty}\left|\frac{(-1)^{n}}{n^{2}}\right| = \sum_{n=1}^{\infty}\frac{1}{n^{2}}$，这是 $p = 2 > 1$ 的 $p$-级数，故收敛。(M1)
Since the series of absolute values converges, the original series converges absolutely. (A1·R1)由于绝对值级数收敛，原级数绝对收敛。(A1·R1)
Absolute convergence fails.绝对收敛不成立。 $\displaystyle\sum_{n=1}^{\infty}\left|\frac{(-1)^{n+1}}{\sqrt{n}}\right| = \sum_{n=1}^{\infty}\frac{1}{n^{1/2}}$, a $p$-series with $p = \tfrac{1}{2} < 1$, hence divergent. (M1)$\displaystyle\sum_{n=1}^{\infty}\left|\frac{(-1)^{n+1}}{\sqrt{n}}\right| = \sum_{n=1}^{\infty}\frac{1}{n^{1/2}}$，$p = \tfrac{1}{2} < 1$ 的 $p$-级数，故发散。(M1)
Leibniz test.莱布尼茨判别法。 Let $b_n = \dfrac{1}{\sqrt{n}}$. Then: (i) $b_n = 1/\sqrt{n} > 1/\sqrt{n+1} = b_{n+1}$, so the terms are decreasing; (ii) $b_n = 1/\sqrt{n} \to 0$. Both hypotheses of the Alternating Series (Leibniz) Test are satisfied, so the series converges. (A1)设 $b_n = \dfrac{1}{\sqrt{n}}$。则：(i) $b_n = 1/\sqrt{n} > 1/\sqrt{n+1} = b_{n+1}$，各项单调递减；(ii) $b_n = 1/\sqrt{n} \to 0$。莱布尼茨交错级数判别法的两个条件均满足，故级数收敛。(A1)
Since it converges but does not converge absolutely, it converges conditionally. (R1)由于级数收敛但非绝对收敛，故为条件收敛。(R1)
The terms are $a_n = (-1)^{n}\dfrac{n}{n+1}$. The Leibniz test requires $b_n = \dfrac{n}{n+1} \to 0$, but in fact (M1)通项为 $a_n = (-1)^{n}\dfrac{n}{n+1}$。莱布尼茨判别法要求 $b_n = \dfrac{n}{n+1} \to 0$，但实际上 (M1)
$$ \frac{n}{n+1} = \frac{1}{1+1/n} \longrightarrow 1 \ne 0. $$Since the underlying terms $n/(n+1)$ do not tend to zero, neither do $a_n$ (they oscillate near $\pm 1$). (A1) By the $n$th-term divergence test, the series diverges. (R1)由于 $n/(n+1)$ 不趋于零，$a_n$ 也不趋于零（在 $\pm 1$ 附近振荡）。(A1) 由第 $n$ 项发散判别法，该级数发散。(R1)
For $S = \displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^{3}}$: (a) verify the Leibniz hypotheses; (b) compute $S_4$ exactly; (c) find the smallest $N$ such that $|S - S_N| < 0.005$.对于 $S = \displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n^{3}}$：(a) 验证莱布尼茨条件；(b) 精确计算 $S_4$；(c) 求最小的 $N$ 使 $|S - S_N| < 0.005$。
Let $b_n = \dfrac{1}{n^{3}}$.设 $b_n = \dfrac{1}{n^{3}}$。
Decreasing:递减： $b_{n+1} = \dfrac{1}{(n+1)^{3}} < \dfrac{1}{n^{3}} = b_n$ for all $n \ge 1$, since $(n+1)^3 > n^3$. (A1)$b_{n+1} = \dfrac{1}{(n+1)^{3}} < \dfrac{1}{n^{3}} = b_n$ 对所有 $n \ge 1$ 成立，因 $(n+1)^3 > n^3$。(A1)
Limit:极限： $\displaystyle\lim_{n\to\infty}\frac{1}{n^{3}} = 0$. (A1)$\displaystyle\lim_{n\to\infty}\frac{1}{n^{3}} = 0$。(A1)
Both hypotheses of the Alternating Series Test are satisfied, so the series converges.交错级数判别法的两个条件均成立，故级数收敛。
$S_4 = \dfrac{1}{1^{3}} - \dfrac{1}{2^{3}} + \dfrac{1}{3^{3}} - \dfrac{1}{4^{3}} = 1 - \dfrac{1}{8} + \dfrac{1}{27} - \dfrac{1}{64}$. (M1)$S_4 = \dfrac{1}{1^{3}} - \dfrac{1}{2^{3}} + \dfrac{1}{3^{3}} - \dfrac{1}{4^{3}} = 1 - \dfrac{1}{8} + \dfrac{1}{27} - \dfrac{1}{64}$。(M1)
Find a common denominator. $\text{lcm}(1, 8, 27, 64)$: since $8 = 2^3$, $27 = 3^3$, $64 = 2^6$, the lcm is $2^6 \cdot 3^3 = 64 \cdot 27 = 1728$. (M1)求公分母。$\text{lcm}(1, 8, 27, 64)$：因 $8 = 2^3$，$27 = 3^3$，$64 = 2^6$，最小公倍数为 $2^6 \cdot 3^3 = 64 \cdot 27 = 1728$。(M1)
$$ S_4 = \frac{1728}{1728} - \frac{216}{1728} + \frac{64}{1728} - \frac{27}{1728} = \frac{1728 - 216 + 64 - 27}{1728} = \frac{1549}{1728}. \quad \text{(A1)} $$State the theorem.陈述定理。 If $\sum (-1)^{n+1}b_n$ satisfies the Leibniz hypotheses, then the error of the $N$th partial sum is bounded by the first omitted term: (M1)若 $\sum (-1)^{n+1}b_n$ 满足莱布尼茨条件，则第 $N$ 个部分和的误差由第一个被省略的项控制：(M1)
$$ |S - S_N| \le b_{N+1} = \frac{1}{(N+1)^{3}}. $$The error bound is valid because the partial sums alternately over- and underestimate $S$, and the gaps shrink monotonically. (A1)误差界成立是因为部分和交替地高估和低估 $S$，且差距单调递减。(A1)
Find the smallest $N$.求最小的 $N$。 We need $\dfrac{1}{(N+1)^{3}} < 0.005 = \dfrac{1}{200}$, which rearranges to $(N+1)^{3} > 200$. (M1)需要 $\dfrac{1}{(N+1)^{3}} < 0.005 = \dfrac{1}{200}$，即 $(N+1)^{3} > 200$。(M1)
Compute: $5^3 = 125 < 200$ and $6^3 = 216 > 200$. So $N+1 \ge 6$, giving $N \ge 5$. (A1)计算：$5^3 = 125 < 200$，$6^3 = 216 > 200$。故 $N+1 \ge 6$，即 $N \ge 5$。(A1)
Verify.验证。 For $N = 5$: the first omitted term is $b_6 = \dfrac{1}{6^{3}} = \dfrac{1}{216} \approx 0.00463 < 0.005$. The bound is satisfied. (A1)取 $N = 5$：第一个被省略的项为 $b_6 = \dfrac{1}{6^{3}} = \dfrac{1}{216} \approx 0.00463 < 0.005$。误差界满足。(A1)