Unit B6: Sequences and Infinite Series第 B6 单元：数列（sequence）与无穷级数（series）

Find $\lim_{n\to\infty}\dfrac{\ln n}{n}$.

Replace $n$ with the continuous variable $x$. The limit $\dfrac{\ln x}{x}$ has the form $\tfrac{\infty}{\infty}$, so L'Hopital applies:

$$\lim_{x\to\infty}\frac{\ln x}{x}=\lim_{x\to\infty}\frac{1/x}{1}=\lim_{x\to\infty}\frac{1}{x}=0.$$

Therefore $\lim_{n\to\infty}\dfrac{\ln n}{n}=0$. The logarithm grows more slowly than any positive power of $n$.

求 $\lim_{n\to\infty}\dfrac{\ln n}{n}$。

将 $n$ 换为连续变量 $x$。极限 $\dfrac{\ln x}{x}$ 具有 $\tfrac{\infty}{\infty}$ 型，可用洛必达法则（L'Hopital's Rule）：

$$\lim_{x\to\infty}\frac{\ln x}{x}=\lim_{x\to\infty}\frac{1/x}{1}=\lim_{x\to\infty}\frac{1}{x}=0.$$

故 $\lim_{n\to\infty}\dfrac{\ln n}{n}=0$。对数函数比 $n$ 的任何正次幂增长都慢。

Show that $\lim_{n\to\infty}\dfrac{\cos n}{n}=0$.

Since $-1\le\cos n\le 1$ for every $n$, dividing by the positive quantity $n$ gives

$$-\frac{1}{n}\le\frac{\cos n}{n}\le\frac{1}{n}.$$

Both outer sequences converge to $0$, so by the squeeze theorem the middle sequence converges to $0$ as well.

证明 $\lim_{n\to\infty}\dfrac{\cos n}{n}=0$。

对一切 $n$，有 $-1\le\cos n\le 1$，除以正数 $n$ 得

$$-\frac{1}{n}\le\frac{\cos n}{n}\le\frac{1}{n}.$$

两端数列均趋于 $0$，故由夹逼定理（squeeze theorem），中间数列也趋于 $0$。

Find $\lim_{n\to\infty} n^{1/n}$.

A power with both the base and the exponent moving is best handled by taking logarithms. Set $a_n=n^{1/n}$, so $\ln a_n=\dfrac{\ln n}{n}$. By Worked Example 1.1 that quotient tends to $0$. Since $\ln$ is continuous, the limit passes through it:

$$\lim_{n\to\infty}\ln a_n=0\ \Longrightarrow\ \lim_{n\to\infty}a_n=e^{0}=1.$$

So $\lim_{n\to\infty} n^{1/n}=1$. This standard limit reappears in Section 7, where it makes the root test of a $p$-series return exactly $L=1$.

求 $\lim_{n\to\infty} n^{1/n}$。

底数和指数同时变化的幂式，最好取对数处理。令 $a_n=n^{1/n}$，则 $\ln a_n=\dfrac{\ln n}{n}$。由例题 1.1，该商趋于 $0$。由于 $\ln$ 函数连续，极限可穿过它：

$$\lim_{n\to\infty}\ln a_n=0\ \Longrightarrow\ \lim_{n\to\infty}a_n=e^{0}=1.$$

故 $\lim_{n\to\infty} n^{1/n}=1$。此标准极限在第 7 节重现，对 p 级数使用根式判别法时恰好得到 $L=1$。

Let $a_1=\sqrt 2$ and $a_{n+1}=\sqrt{2+a_n}$. Show the sequence converges and find its limit.

Bounded above by $2$. By induction: $a_1=\sqrt2<2$, and if $a_n<2$ then $a_{n+1}=\sqrt{2+a_n}<\sqrt{2+2}=2$. Increasing. Note $a_{n+1}^2-a_n^2=2+a_n-a_n^2=-(a_n-2)(a_n+1)>0$ whenever $0a_n$. The sequence is increasing and bounded above, hence convergent by the Monotone Convergence Theorem. Call the limit $L$. Taking limits in the recursion (valid because both sides converge) gives $L=\sqrt{2+L}$, so $L^2=2+L$, that is $L^2-L-2=(L-2)(L+1)=0$. Since terms are positive, $L=2$.

设 $a_1=\sqrt 2$，$a_{n+1}=\sqrt{2+a_n}$。证明数列收敛并求其极限。

有上界 $2$。用数学归纳法：$a_1=\sqrt2<2$；若 $a_n<2$，则 $a_{n+1}=\sqrt{2+a_n}<\sqrt{2+2}=2$。单调递增。注意到 $a_{n+1}^2-a_n^2=2+a_n-a_n^2=-(a_n-2)(a_n+1)>0$（当 $0a_n$。数列单调递增且有上界，由单调收敛定理（Monotone Convergence Theorem）收敛。设极限为 $L$。对递推式两端取极限（两端均收敛故成立）得 $L=\sqrt{2+L}$，即 $L^2=2+L$，即 $L^2-L-2=(L-2)(L+1)=0$。因各项为正，故 $L=2$。

Find $\lim_{n\to\infty}\dfrac{n!}{n^{\,n}}$.

Write the quotient as a product of $n$ factors and bound it. Each factor $\dfrac{k}{n}\le1$, and the first factor is $\dfrac1n$:

$$0<\frac{n!}{n^{\,n}}=\frac{1}{n}\cdot\frac{2}{n}\cdot\frac{3}{n}\cdots\frac{n}{n}\le\frac{1}{n}\cdot1\cdot1\cdots1=\frac1n.$$

The right side tends to $0$, so by the squeeze theorem $\dfrac{n!}{n^{\,n}}\to0$. The terms of $\sum n!/n^n$ therefore pass the necessary $a_n\to0$ check, and the ratio test of Worked Example 7.3 confirms the full series converges.

求 $\lim_{n\to\infty}\dfrac{n!}{n^{\,n}}$。

将商写成 $n$ 个因子的乘积并估界。每个因子 $\dfrac{k}{n}\le1$，第一个因子为 $\dfrac1n$：

$$0<\frac{n!}{n^{\,n}}=\frac{1}{n}\cdot\frac{2}{n}\cdot\frac{3}{n}\cdots\frac{n}{n}\le\frac{1}{n}\cdot1\cdot1\cdots1=\frac1n.$$

右端趋于 $0$，故由夹逼定理 $\dfrac{n!}{n^{\,n}}\to0$。这说明 $\sum n!/n^n$ 的通项满足必要条件 $a_n\to0$，例题 7.3 的比值判别法将进一步证明该级数收敛。

Determine whether $\displaystyle\sum_{n=1}^{\infty}\frac{n}{2n+1}$ converges.

Examine the general term: $\dfrac{n}{2n+1}\to\dfrac12$ as $n\to\infty$, which is not zero.

$$\lim_{n\to\infty}\frac{n}{2n+1}=\frac12\ne 0.$$

By the $n$th-term test the series diverges. The partial sums increase by roughly $\tfrac12$ each step and therefore grow without bound.

判断 $\displaystyle\sum_{n=1}^{\infty}\frac{n}{2n+1}$ 是否收敛。

检验通项：$\dfrac{n}{2n+1}\to\dfrac12$（$n\to\infty$），不趋于零。

$$\lim_{n\to\infty}\frac{n}{2n+1}=\frac12\ne 0.$$

由第 $n$ 项判别法（$n$th-term test），级数发散。部分和每步增加约 $\tfrac12$，故无界增长。

Evaluate $\displaystyle\sum_{n=1}^{\infty}\frac{3^{\,n}+2^{\,n}}{6^{\,n}}$.

The summand separates into two geometric pieces. Because each piece converges on its own, the algebra of series lets us add the sums term by term:

$$\sum_{n=1}^{\infty}\frac{3^{\,n}+2^{\,n}}{6^{\,n}}=\sum_{n=1}^{\infty}\left(\frac12\right)^{n}+\sum_{n=1}^{\infty}\left(\frac13\right)^{n}=\frac{1/2}{1-1/2}+\frac{1/3}{1-1/3}=1+\frac12=\frac32.$$

The step is legitimate only because both component series converge. Splitting a series whose pieces diverge is the most common way to manufacture a wrong answer here.

求 $\displaystyle\sum_{n=1}^{\infty}\frac{3^{\,n}+2^{\,n}}{6^{\,n}}$。

被求和项可分拆为两个等比部分。由于每部分各自收敛，级数的代数法则允许逐项相加：

$$\sum_{n=1}^{\infty}\frac{3^{\,n}+2^{\,n}}{6^{\,n}}=\sum_{n=1}^{\infty}\left(\frac12\right)^{n}+\sum_{n=1}^{\infty}\left(\frac13\right)^{n}=\frac{1/2}{1-1/2}+\frac{1/3}{1-1/3}=1+\frac12=\frac32.$$

此步骤成立的前提是两个分量级数各自收敛。对各部分发散的级数强行拆分，是产生错误答案最常见的原因。

The partial sums of a series are $s_n=\dfrac{2n}{n+3}$. Decide whether $\sum a_n$ converges, and find $a_1$ and $a_3$.

By definition the series converges exactly when $\{s_n\}$ converges: $s_n=\dfrac{2n}{n+3}\to 2$, so $\sum a_n=2$. Individual terms come from differencing the partial sums, $a_n=s_n-s_{n-1}$ (with $a_1=s_1$):

$$a_1=s_1=\frac{2}{4}=\frac12,\qquad a_3=s_3-s_2=\frac{6}{6}-\frac{4}{5}=1-\frac45=\frac15.$$

This is the working definition in reverse: the terms are recovered from the partial sums by first differences.

已知某级数的部分和 $s_n=\dfrac{2n}{n+3}$。判断 $\sum a_n$ 是否收敛，并求 $a_1$ 和 $a_3$。

按定义，级数收敛当且仅当 $\{s_n\}$ 收敛：$s_n=\dfrac{2n}{n+3}\to 2$，故 $\sum a_n=2$。各项由相邻部分和之差恢复，$a_n=s_n-s_{n-1}$（$a_1=s_1$）：

$$a_1=s_1=\frac{2}{4}=\frac12,\qquad a_3=s_3-s_2=\frac{6}{6}-\frac{4}{5}=1-\frac45=\frac15.$$

这是定义的逆向应用：通项由部分和的一阶差分恢复。

What can you say about $\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{2^{\,n}}+\frac1n\right)$?

The algebra-of-series laws run in one direction only: they combine two convergent series. Here $\sum 1/2^n$ converges but $\sum 1/n$ diverges, so the laws do not apply, and in fact the sum diverges. To see it cleanly, suppose the combined series converged; subtracting the convergent geometric series $\sum 1/2^n$ would then force $\sum 1/n$ to converge, a contradiction. A convergent series plus a divergent series is always divergent.

关于 $\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{2^{\,n}}+\frac1n\right)$，你能得出什么结论？

级数代数法则只能合并两个收敛级数。此处 $\sum 1/2^n$ 收敛，但 $\sum 1/n$ 发散，法则不适用，该级数实际发散。简洁的证明：若合并后收敛，则从中减去收敛的等比级数 $\sum 1/2^n$，将迫使 $\sum 1/n$ 收敛，矛盾。一个收敛级数加一个发散级数，结果必然发散。

Suppose $\sum a_n$ converges to $S$, so $s_n\to S$ and also $s_{n-1}\to S$. The $n$th term is the difference of consecutive partial sums:

$$a_n=s_n-s_{n-1}.$$

Taking limits on both sides, $\lim a_n=\lim s_n-\lim s_{n-1}=S-S=0$. Thus convergence forces the terms to zero. The contrapositive is exactly the $n$th-term test for divergence.

设 $\sum a_n$ 收敛到 $S$，则 $s_n\to S$，$s_{n-1}\to S$。第 $n$ 项是相邻部分和之差：

$$a_n=s_n-s_{n-1}.$$

对两端取极限，$\lim a_n=\lim s_n-\lim s_{n-1}=S-S=0$。因此收敛性迫使通项趋于零。其逆否命题正是第 $n$ 项发散判别法（$n$th-term test）。

Evaluate $\displaystyle\sum_{n=1}^{\infty}\frac{2}{3^{\,n}}$.

Write out the first term: $n=1$ gives $\tfrac{2}{3}$, and each successive term multiplies by $r=\tfrac13$. So $a=\tfrac23$ and $|r|<1$.

$$\sum_{n=1}^{\infty}\frac{2}{3^{\,n}}=\frac{a}{1-r}=\frac{2/3}{1-1/3}=\frac{2/3}{2/3}=1.$$

The series converges to $1$.

求 $\displaystyle\sum_{n=1}^{\infty}\frac{2}{3^{\,n}}$。

写出首项：$n=1$ 时为 $\tfrac{2}{3}$，每后一项乘以 $r=\tfrac13$。故 $a=\tfrac23$，$|r|<1$。

$$\sum_{n=1}^{\infty}\frac{2}{3^{\,n}}=\frac{a}{1-r}=\frac{2/3}{1-1/3}=\frac{2/3}{2/3}=1.$$

级数收敛到 $1$。

Evaluate $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$.

By partial fractions, $\dfrac{1}{n(n+1)}=\dfrac1n-\dfrac{1}{n+1}$. The $n$th partial sum is

$$s_n=\left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)+\cdots+\left(\tfrac1n-\tfrac{1}{n+1}\right)=1-\frac{1}{n+1}.$$

As $n\to\infty$, $s_n\to 1$, so the series converges to $1$.

求 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+1)}$。

用部分分式分解，$\dfrac{1}{n(n+1)}=\dfrac1n-\dfrac{1}{n+1}$。第 $n$ 部分和为

$$s_n=\left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)+\cdots+\left(\tfrac1n-\tfrac{1}{n+1}\right)=1-\frac{1}{n+1}.$$

当 $n\to\infty$ 时，$s_n\to 1$，故级数收敛到 $1$。

Write the repeating decimal $0.\overline{27}=0.272727\ldots$ as a fraction.

Group the digits in pairs: the decimal is $\dfrac{27}{100}+\dfrac{27}{100^2}+\dfrac{27}{100^3}+\cdots$, a geometric series with first term $a=\tfrac{27}{100}$ and ratio $r=\tfrac{1}{100}$. Since $|r|<1$,

$$0.\overline{27}=\frac{a}{1-r}=\frac{27/100}{1-1/100}=\frac{27/100}{99/100}=\frac{27}{99}=\frac{3}{11}.$$

Every repeating decimal is a geometric series, which is why every such decimal is rational.

将循环小数 $0.\overline{27}=0.272727\ldots$ 化为分数。

将各位数字两两分组：该小数为 $\dfrac{27}{100}+\dfrac{27}{100^2}+\dfrac{27}{100^3}+\cdots$，是首项 $a=\tfrac{27}{100}$、公比 $r=\tfrac{1}{100}$ 的等比级数。由于 $|r|<1$，

$$0.\overline{27}=\frac{a}{1-r}=\frac{27/100}{1-1/100}=\frac{27/100}{99/100}=\frac{27}{99}=\frac{3}{11}.$$

每个循环小数都是一个等比级数，这正是所有循环小数都是有理数的原因。

Evaluate $\displaystyle\sum_{n=3}^{\infty}\frac{5}{2^{\,n}}$.

The closed form $\dfrac{a}{1-r}$ assumes the sum begins at the first written term, so identify that term directly rather than forcing $n=0$. The $n=3$ term is $a=\dfrac{5}{2^{3}}=\dfrac58$, and the ratio is $r=\tfrac12$:

$$\sum_{n=3}^{\infty}\frac{5}{2^{\,n}}=\frac{5/8}{1-1/2}=\frac{5/8}{1/2}=\frac54.$$

Alternatively, sum from $n=0$ to get $\dfrac{5}{1-1/2}=10$, then subtract the missing $n=0,1,2$ terms $5+\tfrac52+\tfrac54=\tfrac{35}{4}$, leaving $10-\tfrac{35}{4}=\tfrac54$. Both routes agree.

求 $\displaystyle\sum_{n=3}^{\infty}\frac{5}{2^{\,n}}$。

封闭公式 $\dfrac{a}{1-r}$ 以写出的第一项为首项，因此直接读取该项而非强行令 $n=0$。$n=3$ 时首项 $a=\dfrac{5}{2^{3}}=\dfrac58$，公比 $r=\tfrac12$：

$$\sum_{n=3}^{\infty}\frac{5}{2^{\,n}}=\frac{5/8}{1-1/2}=\frac{5/8}{1/2}=\frac54.$$

也可从 $n=0$ 求和得 $\dfrac{5}{1-1/2}=10$，再减去缺少的 $n=0,1,2$ 三项 $5+\tfrac52+\tfrac54=\tfrac{35}{4}$，余 $10-\tfrac{35}{4}=\tfrac54$。两种方法结果相同。

Evaluate $\displaystyle\sum_{n=1}^{\infty}\frac{2}{(2n-1)(2n+1)}$.

Split by partial fractions: $\dfrac{2}{(2n-1)(2n+1)}=\dfrac{1}{2n-1}-\dfrac{1}{2n+1}$, which is $b_n-b_{n+1}$ with $b_n=\dfrac{1}{2n-1}$. The partial sum collapses:

$$s_n=\left(1-\tfrac13\right)+\left(\tfrac13-\tfrac15\right)+\cdots+\left(\tfrac{1}{2n-1}-\tfrac{1}{2n+1}\right)=1-\frac{1}{2n+1}.$$

As $n\to\infty$ the survivor $\dfrac{1}{2n+1}\to0$, so the series converges to $1$. The art is recognizing the partial-fraction split as a difference of consecutive $b_n$.

求 $\displaystyle\sum_{n=1}^{\infty}\frac{2}{(2n-1)(2n+1)}$。

用部分分式拆分：$\dfrac{2}{(2n-1)(2n+1)}=\dfrac{1}{2n-1}-\dfrac{1}{2n+1}$，即 $b_n-b_{n+1}$，其中 $b_n=\dfrac{1}{2n-1}$。部分和化简为：

$$s_n=\left(1-\tfrac13\right)+\left(\tfrac13-\tfrac15\right)+\cdots+\left(\tfrac{1}{2n-1}-\tfrac{1}{2n+1}\right)=1-\frac{1}{2n+1}.$$

当 $n\to\infty$ 时，残留项 $\dfrac{1}{2n+1}\to0$，故级数收敛到 $1$。关键在于将部分分式拆分识别为相邻 $b_n$ 之差。

Let $s_n=a+ar+ar^2+\cdots+ar^{n-1}$ with $r\ne 1$. Multiply by $r$ and subtract:

$$s_n-rs_n=\bigl(a+ar+\cdots+ar^{n-1}\bigr)-\bigl(ar+ar^2+\cdots+ar^{n}\bigr)=a-ar^{n}.$$

The interior terms cancel in pairs, leaving $s_n(1-r)=a(1-r^{n})$, so $s_n=a\dfrac{1-r^{n}}{1-r}$. When $|r|<1$, $r^{n}\to 0$ and $s_n\to\dfrac{a}{1-r}$. When $|r|\ge 1$, $r^{n}$ does not go to zero, so the partial sums fail to settle and the series diverges. This single cancellation is the engine behind both geometric and telescoping series.

设 $s_n=a+ar+ar^2+\cdots+ar^{n-1}$，$r\ne 1$。将两端乘以 $r$ 后相减：

$$s_n-rs_n=\bigl(a+ar+\cdots+ar^{n-1}\bigr)-\bigl(ar+ar^2+\cdots+ar^{n}\bigr)=a-ar^{n}.$$

内部各项两两抵消，余 $s_n(1-r)=a(1-r^{n})$，故 $s_n=a\dfrac{1-r^{n}}{1-r}$。当 $|r|<1$ 时，$r^{n}\to 0$，$s_n\to\dfrac{a}{1-r}$。当 $|r|\ge 1$ 时，$r^{n}$ 不趋于零，部分和不能稳定，级数发散。这一次消约是等比级数与裂项级数的共同核心。

Does $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}$ converge?

Take $f(x)=\dfrac{1}{x\ln x}$, which is positive and decreasing for $x\ge 2$. Substitute $u=\ln x$, $du=\tfrac{dx}{x}$:

$$\int_2^{\infty}\frac{dx}{x\ln x}=\int_{\ln 2}^{\infty}\frac{du}{u}=\bigl[\ln u\bigr]_{\ln 2}^{\infty}=\infty.$$

The integral diverges, so by the integral test the series diverges as well.

$\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}$ 是否收敛？

取 $f(x)=\dfrac{1}{x\ln x}$，对 $x\ge 2$ 为正值递减函数。换元 $u=\ln x$，$du=\tfrac{dx}{x}$：

$$\int_2^{\infty}\frac{dx}{x\ln x}=\int_{\ln 2}^{\infty}\frac{du}{u}=\bigl[\ln u\bigr]_{\ln 2}^{\infty}=\infty.$$

积分发散，故由积分判别法（Integral Test），级数也发散。

Apply the integral test to $f(x)=x^{-p}$ with $p>0$. For $p\ne 1$,

$$\int_1^{\infty}x^{-p}\,dx=\lim_{t\to\infty}\left[\frac{x^{1-p}}{1-p}\right]_1^{t}=\lim_{t\to\infty}\frac{t^{\,1-p}-1}{1-p}.$$

If $p>1$ then $1-p<0$, so $t^{1-p}\to 0$ and the integral equals $\dfrac{1}{p-1}$, a finite value: convergence. If $00$, so $t^{1-p}\to\infty$: divergence. For $p=1$, $\int_1^{\infty}\tfrac{dx}{x}=\lim_{t\to\infty}\ln t=\infty$: divergence. This proves the $p$-series rule.

对 $f(x)=x^{-p}$（$p>0$）应用积分判别法。当 $p\ne 1$ 时，

$$\int_1^{\infty}x^{-p}\,dx=\lim_{t\to\infty}\left[\frac{x^{1-p}}{1-p}\right]_1^{t}=\lim_{t\to\infty}\frac{t^{\,1-p}-1}{1-p}.$$

若 $p>1$，则 $1-p<0$，$t^{1-p}\to 0$，积分等于 $\dfrac{1}{p-1}$（有限值）：收敛。若 $00$，$t^{1-p}\to\infty$：发散。当 $p=1$ 时，$\int_1^{\infty}\tfrac{dx}{x}=\lim_{t\to\infty}\ln t=\infty$：发散。这就证明了 p 级数规则（p-series rule）。

Does $\displaystyle\sum_{n=1}^{\infty}\frac{n}{e^{\,n}}$ converge?

Take $f(x)=xe^{-x}$. It is positive on $[1,\infty)$, and $f'(x)=e^{-x}(1-x)<0$ for $x>1$, so $f$ is decreasing there and the integral test applies. Integrate by parts with $u=x$, $dv=e^{-x}\,dx$:

$$\int_1^{\infty}xe^{-x}\,dx=\Bigl[-xe^{-x}\Bigr]_1^{\infty}+\int_1^{\infty}e^{-x}\,dx=\frac1e+\Bigl[-e^{-x}\Bigr]_1^{\infty}=\frac1e+\frac1e=\frac2e.$$

The improper integral is finite, so the series converges. (The exact value $2/e$ is the integral's value, not the series sum; the test reports only finiteness.)

$\displaystyle\sum_{n=1}^{\infty}\frac{n}{e^{\,n}}$ 是否收敛？

取 $f(x)=xe^{-x}$。在 $[1,\infty)$ 上为正值，且 $f'(x)=e^{-x}(1-x)<0$（$x>1$），故 $f$ 在该区间递减，积分判别法适用。用分部积分，令 $u=x$，$dv=e^{-x}\,dx$：

$$\int_1^{\infty}xe^{-x}\,dx=\Bigl[-xe^{-x}\Bigr]_1^{\infty}+\int_1^{\infty}e^{-x}\,dx=\frac1e+\Bigl[-e^{-x}\Bigr]_1^{\infty}=\frac1e+\frac1e=\frac2e.$$

反常积分有限，故级数收敛。（$2/e$ 是积分值，不是级数的和；判别法只给出有限或无限的结论。）

Does $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^2}$ converge? Compare with the divergent $\sum 1/(n\ln n)$ of Worked Example 4.1, where the extra log factor is decisive.

Let $f(x)=\dfrac{1}{x(\ln x)^2}$, positive and decreasing for $x\ge2$. Substitute $u=\ln x$, $du=dx/x$:

$$\int_2^{\infty}\frac{dx}{x(\ln x)^2}=\int_{\ln 2}^{\infty}\frac{du}{u^2}=\left[-\frac1u\right]_{\ln 2}^{\infty}=\frac{1}{\ln 2}<\infty.$$

The integral is finite, so the series converges. The single change from exponent $1$ to exponent $2$ on the logarithm flips divergence to convergence, mirroring the $p$-series boundary at $p=1$.

$\displaystyle\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^2}$ 是否收敛？与例题 4.1 中发散的 $\sum 1/(n\ln n)$ 比较，多出的对数因子起到关键作用。

令 $f(x)=\dfrac{1}{x(\ln x)^2}$，对 $x\ge2$ 为正值递减。换元 $u=\ln x$，$du=dx/x$：

$$\int_2^{\infty}\frac{dx}{x(\ln x)^2}=\int_{\ln 2}^{\infty}\frac{du}{u^2}=\left[-\frac1u\right]_{\ln 2}^{\infty}=\frac{1}{\ln 2}<\infty.$$

积分有限，故级数收敛。对数指数从 $1$ 改为 $2$ 这一变化，将发散转为收敛，恰好对应 p 级数在 $p=1$ 处的临界现象。

Does $\displaystyle\sum_{n=1}^{\infty}\frac{1}{(2n+1)^{3/2}}$ converge?

Convergence is unaffected by the linear shift inside the power, because the term still behaves like a $p$-series with $p=\tfrac32$. Confirm with $f(x)=(2x+1)^{-3/2}$, substituting $u=2x+1$, $du=2\,dx$:

$$\int_1^{\infty}\frac{dx}{(2x+1)^{3/2}}=\frac12\int_3^{\infty}u^{-3/2}\,du=\frac12\Bigl[-2u^{-1/2}\Bigr]_3^{\infty}=\frac{1}{\sqrt3}<\infty.$$

The integral is finite, so the series converges. As a faster route, limit comparison against $1/n^{3/2}$ gives the finite positive limit $\lim n^{3/2}/(2n+1)^{3/2}=(1/2)^{3/2}$, the same conclusion in one line.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{(2n+1)^{3/2}}$ 是否收敛？

幂次内的线性平移不影响收敛性，因为通项仍与 $p=\tfrac32$ 的 p 级数行为相同。用 $f(x)=(2x+1)^{-3/2}$，换元 $u=2x+1$，$du=2\,dx$ 验证：

$$\int_1^{\infty}\frac{dx}{(2x+1)^{3/2}}=\frac12\int_3^{\infty}u^{-3/2}\,du=\frac12\Bigl[-2u^{-1/2}\Bigr]_3^{\infty}=\frac{1}{\sqrt3}<\infty.$$

积分有限，故级数收敛。更快的方法：与 $1/n^{3/2}$ 做极限比较，得有限正极限 $\lim n^{3/2}/(2n+1)^{3/2}=(1/2)^{3/2}$，一步得到相同结论。

Let $f$ be positive and decreasing with $a_n=f(n)$. Draw rectangles of width $1$ and height $a_n$. Right endpoints sit under the curve and left endpoints sit above it, giving the two-sided trap

$$\int_1^{N+1}f(x)\,dx\ \le\ \sum_{n=1}^{N}a_n\ \le\ a_1+\int_1^{N}f(x)\,dx.$$

If the improper integral converges, the right inequality bounds the partial sums above by a fixed number, so the increasing partial sums converge. If the integral diverges, the left inequality pushes the partial sums to infinity. The same picture bounds the tail remainder $R_N=\sum_{n>N}a_n$ after $N$ terms:

$$\int_{N+1}^{\infty}f(x)\,dx\ \le\ R_N\ \le\ \int_{N}^{\infty}f(x)\,dx.$$

For $\sum 1/n^2$ this gives $R_N\le\int_N^{\infty}x^{-2}\,dx=1/N$, a concrete accuracy estimate the integral test hands you for free.

设 $f$ 为正值递减函数，$a_n=f(n)$。画出宽度为 $1$、高度为 $a_n$ 的矩形。右端点矩形在曲线下方，左端点矩形在曲线上方，得到双边夹逼：

$$\int_1^{N+1}f(x)\,dx\ \le\ \sum_{n=1}^{N}a_n\ \le\ a_1+\int_1^{N}f(x)\,dx.$$

若反常积分收敛，右侧不等式将部分和从上方限制为某常数，递增的部分和因而收敛。若积分发散，左侧不等式将部分和推向无穷。同一图形还给出 $N$ 项后尾余项 $R_N=\sum_{n>N}a_n$ 的估计：

$$\int_{N+1}^{\infty}f(x)\,dx\ \le\ R_N\ \le\ \int_{N}^{\infty}f(x)\,dx.$$

对 $\sum 1/n^2$，此式给出 $R_N\le\int_N^{\infty}x^{-2}\,dx=1/N$，这是积分判别法附赠的具体精度估计。

Classify $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=1-\tfrac12+\tfrac13-\tfrac14+\cdots$.

Here $b_n=\tfrac1n$ is positive, decreasing, and tends to $0$, so the alternating series test gives convergence. But the series of absolute values is the harmonic series $\sum 1/n$, which diverges.

$$\sum\left|\frac{(-1)^{n-1}}{n}\right|=\sum\frac{1}{n}=\infty.$$

Therefore the alternating harmonic series is conditionally convergent. Its sum is $\ln 2$.

对 $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=1-\tfrac12+\tfrac13-\tfrac14+\cdots$ 分类。

此处 $b_n=\tfrac1n$ 为正值、递减且趋于 $0$，交错级数判别法给出收敛。但绝对值级数是调和级数 $\sum 1/n$，发散。

$$\sum\left|\frac{(-1)^{n-1}}{n}\right|=\sum\frac{1}{n}=\infty.$$

故交错调和级数是条件收敛（conditionally convergent）级数。其和为 $\ln 2$。

How many terms of $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}$ guarantee an error below $0.01$?

The remainder after $n$ terms is at most $b_{n+1}=\dfrac{1}{(n+1)^2}$. Require

$$\frac{1}{(n+1)^2}\le 0.01\iff (n+1)^2\ge 100\iff n+1\ge 10\iff n\ge 9.$$

So summing the first $9$ terms approximates the total to within $0.01$.

$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}$ 至少需要取多少项，才能保证误差小于 $0.01$？

取前 $n$ 项后的余项至多为 $b_{n+1}=\dfrac{1}{(n+1)^2}$。要求

$$\frac{1}{(n+1)^2}\le 0.01\iff (n+1)^2\ge 100\iff n+1\ge 10\iff n\ge 9.$$

故取前 $9$ 项即可将误差控制在 $0.01$ 以内。

Classify $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}$.

Test the absolute series first, since absolute convergence is the stronger property and, once secured, settles ordinary convergence automatically. Here $\sum\left|\dfrac{(-1)^{n-1}}{n^2}\right|=\sum\dfrac{1}{n^2}$, a convergent $p$-series ($p=2>1$). So the series is absolutely convergent, hence convergent. Unlike the alternating harmonic series, this one may be rearranged at will without changing its sum.

对 $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}$ 分类。

先检验绝对级数，因为绝对收敛是更强的性质，一旦成立即自动确保普通收敛。此处 $\sum\left|\dfrac{(-1)^{n-1}}{n^2}\right|=\sum\dfrac{1}{n^2}$，是收敛的 p 级数（$p=2>1$）。故级数绝对收敛，从而收敛。与交错调和级数不同，此级数可以任意调换求和顺序而不改变其和。

Examine $\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\frac{n}{n+1}$.

It is tempting to invoke the alternating series test, but its hypotheses fail at the very first gate: $b_n=\dfrac{n}{n+1}\to 1\ne 0$. The terms of the full series therefore do not tend to $0$ (they oscillate near $\pm1$), so the $n$th-term test of Section 2 forces divergence. The alternating sign is irrelevant once the terms fail to shrink.

分析 $\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\frac{n}{n+1}$。

容易想到使用交错级数判别法，但其前提条件在第一关就失败了：$b_n=\dfrac{n}{n+1}\to 1\ne 0$。因此级数的通项不趋于 $0$（在 $\pm1$ 附近振荡），第 2 节的第 $n$ 项判别法立即给出发散。通项不趋于零时，交错符号毫无意义。

Classify $\displaystyle\sum_{n=2}^{\infty}\frac{(-1)^{n}}{\sqrt n}$.

Set $b_n=1/\sqrt n$. It is positive, decreasing, and tends to $0$, so the alternating series test gives convergence. For absolute convergence, test $\sum 1/\sqrt n$, a $p$-series with $p=\tfrac12\le1$, which diverges. So the series converges but not absolutely: it is conditionally convergent. By the remainder bound, truncating after $N$ terms leaves an error at most $1/\sqrt{N+1}$, which shrinks slowly, a hallmark of conditional convergence.

对 $\displaystyle\sum_{n=2}^{\infty}\frac{(-1)^{n}}{\sqrt n}$ 分类。

令 $b_n=1/\sqrt n$，为正值、递减且趋于 $0$，交错级数判别法给出收敛。检验绝对收敛：$\sum 1/\sqrt n$ 是 $p=\tfrac12\le1$ 的 p 级数，发散。故级数收敛但非绝对收敛：为条件收敛。由余项估计，取前 $N$ 项后误差至多为 $1/\sqrt{N+1}$，缩小缓慢，这是条件收敛的典型特征。

Let $b_n>0$ be decreasing with $b_n\to0$, and $s_n$ the partial sums of $\sum(-1)^{n-1}b_n$. Group the even partial sums in nonnegative pairs:

$$s_{2n}=(b_1-b_2)+(b_3-b_4)+\cdots+(b_{2n-1}-b_{2n}),$$

every bracket is $\ge0$, so $\{s_{2n}\}$ is nondecreasing. Regrouping the other way, $s_{2n}=b_1-(b_2-b_3)-\cdots-b_{2n}\le b_1$, so it is bounded above. By monotone convergence $s_{2n}\to S$ for some $S$. Since $s_{2n+1}=s_{2n}+b_{2n+1}$ and $b_{2n+1}\to0$, the odd sums converge to the same $S$, so $s_n\to S$. For the error bound, $S$ lies between any two consecutive partial sums (the sums oscillate inward toward $S$), so the remainder satisfies

$$|S-s_n|\le|s_{n+1}-s_n|=b_{n+1}.$$

That is exactly the remainder estimate used in Worked Example 6.2: the truncation error is no larger than the first omitted term.

设 $b_n>0$ 递减且 $b_n\to0$，$s_n$ 为 $\sum(-1)^{n-1}b_n$ 的部分和。将偶数部分和按非负括号分组：

$$s_{2n}=(b_1-b_2)+(b_3-b_4)+\cdots+(b_{2n-1}-b_{2n}),$$

每个括号 $\ge0$，故 $\{s_{2n}\}$ 单调不减。换一种分组方式，$s_{2n}=b_1-(b_2-b_3)-\cdots-b_{2n}\le b_1$，从而有上界。由单调收敛定理，$s_{2n}\to S$。由于 $s_{2n+1}=s_{2n}+b_{2n+1}$，$b_{2n+1}\to0$，奇数部分和收敛到同一个 $S$，故 $s_n\to S$。对于误差估计，$S$ 介于任意相邻两个部分和之间（部分和向内振荡逼近 $S$），故余项满足

$$|S-s_n|\le|s_{n+1}-s_n|=b_{n+1}.$$

这正是例题 6.2 中使用的余项估计：截断误差不超过第一个省略项。

Test $\displaystyle\sum_{n=1}^{\infty}\frac{2^{\,n}}{n!}$.

Form the ratio of consecutive terms:

$$\left|\frac{a_{n+1}}{a_n}\right|=\frac{2^{\,n+1}}{(n+1)!}\cdot\frac{n!}{2^{\,n}}=\frac{2}{n+1}.$$

As $n\to\infty$ this ratio tends to $0$, so $L=0<1$ and the series converges absolutely. The factorial in the denominator overwhelms the exponential in the numerator.

判别 $\displaystyle\sum_{n=1}^{\infty}\frac{2^{\,n}}{n!}$。

构造相邻项之比：

$$\left|\frac{a_{n+1}}{a_n}\right|=\frac{2^{\,n+1}}{(n+1)!}\cdot\frac{n!}{2^{\,n}}=\frac{2}{n+1}.$$

当 $n\to\infty$ 时比值趋于 $0$，故 $L=0<1$，级数绝对收敛。分母的阶乘压倒了分子的指数。

Test $\displaystyle\sum_{n=1}^{\infty}\left(\frac{3n+1}{4n+5}\right)^{n}$.

The term is a perfect $n$th power, so the root test is natural:

$$\sqrt[n]{|a_n|}=\frac{3n+1}{4n+5}\xrightarrow[n\to\infty]{}\frac{3}{4}.$$

Since $L=\tfrac34<1$, the series converges absolutely.

判别 $\displaystyle\sum_{n=1}^{\infty}\left(\frac{3n+1}{4n+5}\right)^{n}$。

通项是完整的 $n$ 次幂，故根式判别法是自然选择：

$$\sqrt[n]{|a_n|}=\frac{3n+1}{4n+5}\xrightarrow[n\to\infty]{}\frac{3}{4}.$$

由于 $L=\tfrac34<1$，级数绝对收敛。

Test $\displaystyle\sum_{n=1}^{\infty}\frac{n!}{n^{\,n}}$.

Form the ratio of consecutive terms, simplifying the factorials and the powers separately:

$$\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^{\,n}}{n!}=\frac{(n+1)\,n^{\,n}}{(n+1)^{n+1}}=\frac{n^{\,n}}{(n+1)^{n}}=\left(\frac{n}{n+1}\right)^{n}.$$

That last expression is $\left(1+\tfrac1n\right)^{-n}\to e^{-1}$. So $L=1/e\approx0.368<1$ and the series converges absolutely. The cancellation that turns the messy ratio into the standard limit $\left(1+\tfrac1n\right)^n\to e$ is the whole trick.

判别 $\displaystyle\sum_{n=1}^{\infty}\frac{n!}{n^{\,n}}$。

构造相邻项之比，分别化简阶乘部分和幂次部分：

$$\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^{\,n}}{n!}=\frac{(n+1)\,n^{\,n}}{(n+1)^{n+1}}=\frac{n^{\,n}}{(n+1)^{n}}=\left(\frac{n}{n+1}\right)^{n}.$$

最后的表达式为 $\left(1+\tfrac1n\right)^{-n}\to e^{-1}$，故 $L=1/e\approx0.368<1$，级数绝对收敛。将繁杂的比值化简为标准极限 $\left(1+\tfrac1n\right)^n\to e$ 正是解题的全部技巧。

Test $\displaystyle\sum_{n=1}^{\infty}\frac{2^{\,n}}{n^{\,n}}$.

Because the whole term is essentially an $n$th power, the root test is the efficient choice. Using $n^{1/n}\to1$ from Worked Example 1.3 is not even needed here:

$$\sqrt[n]{|a_n|}=\sqrt[n]{\frac{2^{\,n}}{n^{\,n}}}=\frac{2}{n}\xrightarrow[n\to\infty]{}0.$$

Since $L=0<1$, the series converges absolutely. The ratio test would also work, but it would require simplifying $\left(\tfrac{n}{n+1}\right)^n$ again; the root test reads the answer off in one line.

判别 $\displaystyle\sum_{n=1}^{\infty}\frac{2^{\,n}}{n^{\,n}}$。

整个通项本质上是一个 $n$ 次幂，故根式判别法是高效选择。此处甚至不需要用到例题 1.3 的 $n^{1/n}\to1$：

$$\sqrt[n]{|a_n|}=\sqrt[n]{\frac{2^{\,n}}{n^{\,n}}}=\frac{2}{n}\xrightarrow[n\to\infty]{}0.$$

由于 $L=0<1$，级数绝对收敛。比值判别法也可行，但需再次化简 $\left(\tfrac{n}{n+1}\right)^n$；根式判别法一行即得答案。

For which $x>0$ does $\displaystyle\sum_{n=1}^{\infty}\frac{x^{\,n}}{n!}$ converge?

This is the series that defines $e^{x}$, and the ratio test settles it for every $x$. Form the ratio:

$$\left|\frac{a_{n+1}}{a_n}\right|=\frac{x^{\,n+1}}{(n+1)!}\cdot\frac{n!}{x^{\,n}}=\frac{x}{n+1}\xrightarrow[n\to\infty]{}0.$$

The limit is $L=0<1$ regardless of how large $x$ is, so the series converges absolutely for all $x$. The factorial denominator dominates any fixed-base power, which is exactly why the exponential series converges everywhere, a fact carried forward into the power-series unit.

$\displaystyle\sum_{n=1}^{\infty}\frac{x^{\,n}}{n!}$ 对哪些 $x>0$ 收敛？

这是定义 $e^{x}$ 的级数，比值判别法对每个 $x$ 均可解决。构造比值：

$$\left|\frac{a_{n+1}}{a_n}\right|=\frac{x^{\,n+1}}{(n+1)!}\cdot\frac{n!}{x^{\,n}}=\frac{x}{n+1}\xrightarrow[n\to\infty]{}0.$$

无论 $x$ 多大，极限均为 $L=0<1$，故级数对所有 $x$ 绝对收敛。分母的阶乘压倒任何固定底数的幂，这正是指数级数在全体实数上收敛的原因，这一结论将在幂级数单元中延续。

Apply the root test to $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{\,3}}$, then resolve it.

The $n$th root of the term uses the standard limit $n^{1/n}\to1$ from Worked Example 1.3:

$$\sqrt[n]{\frac{1}{n^{3}}}=\frac{1}{\bigl(n^{1/n}\bigr)^{3}}\xrightarrow[n\to\infty]{}\frac{1}{1^{3}}=1.$$

So $L=1$ and the root test is inconclusive, exactly as warned. The series is a $p$-series with $p=3>1$, so it converges by the $p$-series rule of Section 4. This is the prototype of why $L=1$ cannot be read as a verdict.

对 $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{\,3}}$ 使用根式判别法，再用其他方法解决。

通项的 $n$ 次根利用例题 1.3 的标准极限 $n^{1/n}\to1$：

$$\sqrt[n]{\frac{1}{n^{3}}}=\frac{1}{\bigl(n^{1/n}\bigr)^{3}}\xrightarrow[n\to\infty]{}\frac{1}{1^{3}}=1.$$

故 $L=1$，根式判别法无结论，正如事先警告的那样。该级数是 $p=3>1$ 的 p 级数，由第 4 节的 p 级数规则知其收敛。这是 $L=1$ 不能作为结论的典型例子。

Suppose $L=\lim|a_{n+1}/a_n|<1$. Pick any $r$ with $L $$|a_{N+k}|So the tail $\sum_{k\ge0}|a_{N+k}|$ is dominated by the convergent geometric series $|a_N|\sum_{k\ge0} r^{k}=\dfrac{|a_N|}{1-r}$. By direct comparison $\sum|a_n|$ converges, hence $\sum a_n$ converges absolutely. If instead $L>1$, then eventually $|a_{n+1}|>|a_n|$, so the terms grow and cannot tend to $0$; the $n$th-term test forces divergence. The ratio test is comparison against a geometric series whose ratio is squeezed just above $L$.

设 $L=\lim|a_{n+1}/a_n|<1$。取任意 $r$ 满足 $L $$|a_{N+k}|故尾部 $\sum_{k\ge0}|a_{N+k}|$ 被收敛等比级数 $|a_N|\sum_{k\ge0} r^{k}=\dfrac{|a_N|}{1-r}$ 控制。由直接比较判别法，$\sum|a_n|$ 收敛，从而 $\sum a_n$ 绝对收敛。若 $L>1$，则最终有 $|a_{n+1}|>|a_n|$，项递增不趋于零；第 $n$ 项判别法给出发散。比值判别法本质上是与一个公比被夹紧在 $L$ 稍上方的等比级数做直接比较。