复习课怎么用

这是一节完整的期中复习课，目标不是重讲定义，而是把前半学期的概念连成一条可复述的链条，并给出考场上真正会用到的判定工具与反例库。

课程自己点明了主线：数学是一层层搭起来的。单个步骤往往很平凡，真正的难点在于知道「现在该调用哪一层」。比如「收敛数列有界」本身几乎是定义平移，但它后来成为乘积极限、柯西列有界等一系列结论的支点。

复习课给出的考试形式：五道主要问题加六道附加题，每题标了分值，可以先做最容易的。下面的整理按「概念 → 工具 → 陷阱」组织，最后是两道例题。

时间轴

第一条链：实数 → 数列

上确界与完备性

$A\subseteq\mathbb R$ 有上界，指存在 $M$ 使一切 $a\in A$ 满足 $a\le M$。$M$ 是最小上界（记 $\sup A$）当且仅当 $M$ 是上界，且任何上界 $\bar M$ 都满足 $M\le\bar M$。

实数的关键性质是最小上界性质：非空且有上界的集合必有上确界；有下界的集合必有下确界。整门课后来所有的收敛性结论，追根溯源都要回到这一条。

收敛的量词

$(x_n)$ 收敛到 $x$，指

$$\forall \varepsilon>0\ \exists N\in\mathbb N\ \text{使得}\ n\ge N\Rightarrow |x_n-x|<\varepsilon .$$

两个方向容易记混的地方：$\varepsilon$ 是别人给的，$N$ 是你选的；$N$ 可以依赖 $\varepsilon$，但一旦选定就对所有 $n\ge N$ 同时生效。

极限的运算

若 $a_n\to a$、$b_n\to b$，则

$$a_n+cb_n\to a+cb, \qquad a_nb_n\to ab, \qquad \frac{1}{a_n}\to\frac1a\ (\text{当 }a\ne0,\ a_n\ne0).$$

倒数那条需要「分母远离零」：由 $a_n\to a\ne0$ 可得充分靠后时 $|a_n|>|a|/2$，于是

$$\left|\frac1{a_n}-\frac1a\right|=\frac{|a_n-a|}{|a_n||a|}<\frac{2|a_n-a|}{|a|^2}\to0 .$$

第二条链：有界、单调、柯西

收敛必有界

取 $\varepsilon=1$，尾部满足 $|a_n-a|<1$，故 $|a_n|<|a|+1$；前有限项取最大即可。注意逆命题不成立：$(-1)^n$ 有界却不收敛。

子列

子列是 $x_{n_k}$，其中指标 $n_1<n_2<\cdots$ 严格递增——它保证「顺序不变、可以跳过、不能重复」。

定理。 $(x_n)$ 收敛当且仅当它的所有子列都收敛（且极限相同）。

「当」的方向是平凡的：原数列本身就是自己的子列（取指标映射为恒等）。「仅当」方向要用 $n_k\ge k$。

这个平凡定理的用处在于反证：找到两个收敛到不同极限的子列，就证明原数列发散。$(-1)^n$ 的偶数项恒为 $1$、奇数项恒为 $-1$，立刻判定发散。

单调收敛定理

• 若 $(x_n)$ 递增（$x_{n+1}\ge x_n$），则它收敛当且仅当有上界，且 $\lim x_n=\sup\{x_n\}$。
• 若 $(x_n)$ 递减，则它收敛当且仅当有下界，且 $\lim x_n=\inf\{x_n\}$。

这是「只知道有界、不知道极限是多少」时最常用的工具。它把存在性直接转成了上确界的存在性，而后者由完备性保证。

柯西列

$(x_n)$ 是柯西列，指

$$\forall\varepsilon>0\ \exists N\ \text{使得}\ m,n\ge N\Rightarrow |x_m-x_n|<\varepsilon .$$

柯西性说的是「尾部自己挤在一起」，不提及任何候选极限。它的价值在于：当你不关心极限等于多少时，验证柯西性通常比验证收敛容易。

在 $\mathbb R$ 中，柯西列必收敛（柯西收敛定理），所以两种说法等价；但在一般度量空间里只有「收敛 ⟹ 柯西」，反过来不成立。

Bolzano–Weierstrass 定理

$\mathbb R$ 中有界数列必有收敛子列。

它是柯西收敛定理的证明关键：先用 BW 取出收敛子列，再证明柯西列的整体极限就是该子列的极限。复习课特意强调：柯西收敛定理与 BW 定理的陈述在一般度量空间里都写得出来，但都不一定成立——它们依赖实数的完备性。

第三条链：连续函数

定义与序列刻画

$f$ 在 $x_0$ 连续，指

$$\forall\varepsilon>0\ \exists\delta>0\ \text{使得}\ |x-x_0|<\delta\Rightarrow|f(x)-f(x_0)|<\varepsilon .$$

引理。 若 $f$ 在 $x_0$ 连续且 $x_n\to x_0$，则 $f(x_n)\to f(x_0)$。

这条引理把函数问题翻译成数列问题，是证明最值定理与介值定理的桥梁。

连续函数的运算

$f$、$g$ 连续、$c$ 为常数，则 $f+cg$、$fg$ 连续；若 $g$ 处处不为零，则 $1/g$、$f/g$ 也连续。

配合两个起点——常数函数连续、$f(x)=x$ 连续——就能归纳出：

• 所有多项式连续；
• 所有有理函数（多项式之商）在分母不为零处连续。

这正是「从简单函数造出复杂函数」的标准流水线。

两个定理

最值定理。 $f$ 在闭区间 $[a,b]$ 上连续，则 $f$ 取得最大值与最小值（从而有界）。

介值定理。 $f$ 在 $[a,b]$ 上连续，$y$ 介于 $f(a)$ 与 $f(b)$ 之间，则存在 $x\in[a,b]$ 使 $f(x)=y$。

两个定理的证明都用了同一套零件：BW 定理（取出收敛子列）＋「连续 ⟹ 保持数列极限」＋实数的上确界性质。缺一不可。

第四条链：级数

从部分和看起

给定数列 $(a_n)$，部分和 $S_n=\sum_{k=0}^{n}a_k$ 又构成一个数列。级数收敛就是部分和数列收敛，所以前面关于数列的一切工具都能直接用。

第一个必查项：若 $\sum a_n$ 收敛，则 $a_n\to0$。因为 $a_n=S_n-S_{n-1}$，两个收敛数列相减仍收敛到 $0$。

反过来不成立：调和级数 $\sum 1/n$ 的通项趋于 $0$，但级数发散。所以「通项趋于零」只是筛掉明显发散的工具，不是判据。

必须记住的三个级数

几何级数：$\sum_{n=0}^{\infty}c^n$ 在 $|c|<1$ 时收敛，且

$$\sum_{n=0}^{\infty}c^n=\frac{1}{1-c};$$

$|c|\ge1$ 时发散。

这里有两个容易丢分的细节：

1. 求和要看起点。$\sum_{n=1}^{\infty}c^n=\dfrac{c}{1-c}$，不是 $\dfrac{1}{1-c}$。
2. 收敛性与起点无关，数值与起点有关。判断敛散时可以随意挪动起点，计算和时必须扣掉被移走的那几项。

调和级数 $\sum_{n=1}^{\infty}\dfrac1n$：发散。

交错调和级数 $\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n}$：收敛（但不是绝对收敛）。

另一个常用而收敛的级数是 $\sum 1/n^2$，它是比较判别法的「标尺」。

绝对收敛

若 $\sum|a_n|$ 收敛，则称 $\sum a_n$ 绝对收敛。绝对收敛蕴含收敛，但反之不成立（交错调和级数）。

绝对收敛还带来两个额外好处：可以任意重排求和顺序而不改变和；两个绝对收敛级数的乘积级数收敛。课程在估计余项时正是靠这个性质。

三个判别法

比较判别法（A 版）。 若 $0\le a_n\le b_n$ 且 $\sum b_n$ 收敛，则 $\sum a_n$ 收敛；若 $\sum a_n$ 发散，则 $\sum b_n$ 发散。

比较判别法（B 版）。 若 $a_n,b_n>0$ 且 $\lim a_n/b_n=L$，$0<L<\infty$，则两级数同敛散。

比值判别法。 设 $a_n\ne0$，令 $c=\lim\left|\dfrac{a_{n+1}}{a_n}\right|$（若极限存在）。$c<1$ 收敛；$c>1$ 发散；$c=1$ 无法判定。

根值判别法。 令 $c=\limsup\sqrt[n]{|a_n|}$。$c<1$ 收敛；$c>1$ 发散；$c=1$ 无法判定。

三个判别法里，比值和根值都是几何级数的「后代」：想法是把通项和某个 $c^n$ 比较。根值用上极限而非极限，是为了让它在更一般的场合也有定义——$\limsup$ 总是存在，而 $\lim$ 可能不存在。

幂级数

形如 $\sum a_nx^n$ 的级数，对每个固定的 $x$ 都是一个普通级数。定义

$$R=\frac{1}{\limsup\sqrt[n]{|a_n|}}\quad(\text{约定 }1/0=\infty,\ 1/\infty=0).$$

则

• $|x|<R$ 时绝对收敛；
• $|x|>R$ 时发散；
• $|x|=R$ 的两个端点无法统一判断，必须单独讨论。

这就是为什么幂级数题目的标准答案是「收敛半径 $R=\cdots$，端点需另行判断」。

例 1： $\sum n^5x^n$。用比值法：

$$\left|\frac{a_{n+1}}{a_n}\right|=\left(\frac{n+1}{n}\right)^5|x|\to|x| ,$$

所以 $|x|<1$ 收敛、$|x|>1$ 发散，$R=1$。

例 2： $\sum \dfrac{n^5x^n}{n!}$。比值法：

$$\left|\frac{a_{n+1}}{a_n}\right| =\left(\frac{n+1}{n}\right)^5\frac{|x|}{n+1}\to 0\cdot|x|=0 ,$$

对一切 $x$ 都收敛，故 $R=\infty$。

考场上不要只写「$R=\infty$」，要把比值算出来再下结论——过程分比答案值钱。

典型题型与陷阱

题型一：用定义证明收敛

固定套路：先猜极限，再写「给定 $\varepsilon>0$，取 $N=\lceil\cdots\rceil$」，最后验证 $n\ge N$ 时的不等式。常备工具是阿基米德性质与三角不等式。

陷阱：把 $N$ 写成依赖 $n$ 的表达式。$N$ 只能依赖 $\varepsilon$。

题型二：用单调收敛定理

需要两件事：单调性（常用归纳法证明）与有界性。结论只给存在性，不给数值——除非另外求出上确界。

陷阱：只验证了有界，没验证单调；或者用「极限存在」反推有界，循环论证。

题型三：证明柯西性

估计 $|x_m-x_n|$，通常假设 $m>n$ 并写成 $|x_m-x_n|\le\sum_{k=n}^{m-1}|x_{k+1}-x_k|$，再用几何级数或积分估计尾部。

陷阱：把柯西性写成「相邻两项越来越近」。相邻差趋于零不足以保证柯西（例如 $x_n=\sqrt n$）。

题型四：幂级数收敛半径

先判断用比值还是根值。含阶乘、含 $n$ 的幂时比值通常更快；含 $n$ 次方整体结构时根值更直接。

陷阱：忘记取绝对值；忘记端点；把 $R$ 与收敛区间混为一谈。

题型五：举反例

复习课反复出现的反例清单值得背下来：

一道递推数列例题

设 $x_1=a$，$x_{n+1}=\dfrac{x_n}{n}$。求极限。

先写前几项：

$$x_1=a,\quad x_2=\frac{a}{1}=a,\quad x_3=\frac{a}{2},\quad x_4=\frac{a}{6},\quad x_5=\frac{a}{24}.$$

规律是 $x_n=\dfrac{a}{(n-1)!}$（$n\ge1$）。归纳验证：$x_{n+1}=\dfrac{x_n}{n}=\dfrac{a}{n!}$，与公式一致。于是

$$|x_n|=\frac{|a|}{(n-1)!}\le\frac{|a|}{n-1}\to0 ,$$

所以 $\lim x_n=0$，与初值 $a$ 无关。

只写出前几项不算证明；必须把通项公式归纳出来，再给出收敛论证。

一道比较判别例题

设 $0\le a_n<1/n^2$。判断 $\sum a_n$ 是否收敛。

由 $0\le a_n<1/n^2$ 且 $\sum 1/n^2$ 收敛，比较判别法直接给出 $\sum a_n$ 收敛。

注意条件里必须有 $a_n\ge0$；若允许 $a_n$ 为负，条件 $a_n<1/n^2$ 对发散级数也成立（例如 $a_n=-1$），命题为假。

本节结论

• 一切收敛性最终追溯到实数的最小上界性质。
• 收敛 ⟹ 柯西 ⟹ 有界；反向都不成立，除 $\mathbb R$ 中柯西 ⟹ 收敛。
• 单调 + 有界 ⟹ 收敛，极限是上确界或下确界。
• 连续把数列极限搬运到函数值；最值定理与介值定理靠 BW + 连续性 + 完备性。
• 级数先查通项是否趋零，再用几何级数作标尺，最后用比较、比值、根值判别。
• 幂级数先算 $R$，端点单独讨论。
• 反例与正定理同等重要：它们划定了每条定理的适用边界。

理解检验

1. 为什么「所有子列收敛」可以推出原数列收敛，而「存在一个收敛子列」不行？

参考答案：「所有子列收敛」包含了原数列本身——取指标映射 $g(k)=k$ 就是原数列，所以原数列当然收敛。「存在一个收敛子列」只说明数列中有无穷多项聚集在某点附近，其余项可以任意乱跑。例如 $a_n=(-1)^n$ 的偶数项子列恒为 $1$，是收敛子列，但原数列发散。

2. 证明：收敛数列必有界；并举出有界但不收敛的例子。

参考答案：设 $a_n\to a$。取 $\varepsilon=1$，存在 $N$ 使 $n\ge N$ 时 $|a_n-a|<1$，三角不等式给出 $|a_n|\le|a|+1$。令

$$M=\max\{|a_1|,\dots,|a_{N-1}|,\ |a|+1\},$$

则对一切 $n$ 有 $|a_n|\le M$。反例：$a_n=(-1)^n$，有界但两个子列分别收敛到 $1$ 与 $-1$，故发散。

3. 单调收敛定理为什么不能省掉「有界」？给出一个递增但发散的数列。

参考答案：定理的结论是「收敛 $\iff$ 有界」，其中收敛 ⟹ 有界来自一般性质，而有界 ⟹ 收敛依赖单调性：递增数列的所有项都不超过上确界，且能任意逼近它。去掉有界，递增数列可能是 $a_n=n$，它发散。反过来说，去掉单调性，仅有界也不够，如 $(-1)^n$。两条缺一不可。

4. 判断 $\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n}$ 的敛散性，并说明它是否绝对收敛。

参考答案：由交错级数判别法（$1/n$ 单调递减趋于 $0$）它收敛。但它不绝对收敛：$\sum 1/n$ 是调和级数，发散。所以它是「收敛但不绝对收敛」的标准例子，说明绝对收敛严格强于收敛。

5. 求 $\sum_{n=0}^{\infty} n^5 x^n$ 的收敛半径，并说明端点情形。

参考答案：用比值判别法，

$$\left|\frac{a_{n+1}}{a_n}\right|=\left(\frac{n+1}{n}\right)^5|x|\to|x|,$$

故 $|x|<1$ 收敛、$|x|>1$ 发散，$R=1$。端点 $x=\pm1$ 时通项为 $\pm n^5$，不趋于 $0$，级数发散。所以收敛区间是 $(-1,1)$。

6. 有人说：「因为 $a_n\to0$，所以 $\sum a_n$ 收敛。」错在哪里？

参考答案：这是把必要条件当成了充分条件。$\sum a_n$ 收敛确实蕴含 $a_n\to0$（因为 $a_n=S_n-S_{n-1}$），但反之不成立。调和级数 $a_n=1/n\to0$，部分和却无界：把项按 $2^k$ 分组，每组的和至少是 $1/2$，无穷多组相加发散。

7. 在一般度量空间中，「柯西列必收敛」为什么失效？用 $X=(0,1]$ 说明。

参考答案：柯西性只保证尾部彼此靠近，不保证那个「应该收敛到的点」属于空间。在 $X=(0,1]$ 中取 $x_n=1/n$：对任意 $\varepsilon>0$，取 $N>1/\varepsilon$，则 $m,n\ge N$ 时 $|1/m-1/n|<1/N<\varepsilon$，所以它是柯西列。但它在 $X$ 内的唯一候选极限是 $0$，而 $0\notin X$，故在 $X$ 中不收敛。$\mathbb R$ 中柯西收敛定理成立的实质是 $\mathbb R$ 完备、没有这类「缺失的点」。

与前后课的关系

本节把前半学期的内容压缩成一张网。下一讲进入度量空间：把「收敛」「柯西」「有界」从 $\mathbb R$ 搬到任意距离空间，并开始讨论开集、闭集与紧性。复习课里反复强调的「哪些结论依赖完备性」正是下一讲要精确化的分界线。

How to use a review lecture

This is a full midterm review. It does not re-derive definitions; it strings the first half of the semester into one chain and supplies the decision tools and the counterexample library you actually use in an exam.

The lecturer states the theme outright: mathematics is built layer on layer. Individual steps are usually trivial; the difficulty is knowing which layer to call on. "A convergent sequence is bounded" is almost a restatement of the definition, yet it later becomes the supporting point for products of limits, boundedness of Cauchy sequences, and more.

Exam format, as announced: five main problems plus six bonus problems, each with its point value, so start with whatever looks easiest. The notes below are organised as concept → tool → trap, ending with two worked examples.

Timeline

First chain: the reals → sequences

Suprema and completeness

A set $A\subseteq\mathbb R$ is bounded above if some $M$ satisfies $a\le M$ for all $a\in A$. The number $M$ is the least upper bound (written $\sup A$) exactly when $M$ bounds $A$ above and every upper bound $\bar M$ satisfies $M\le\bar M$.

The key property of the reals is the least upper bound property: every nonempty set bounded above has a supremum; every set bounded below has an infimum. Every convergence result in the course ultimately traces back to this.

The quantifiers of convergence

$(x_n)$ converges to $x$ means

$$\forall \varepsilon>0\ \exists N\in\mathbb N\ \text{such that}\ n\ge N\Rightarrow |x_n-x|<\varepsilon .$$

The two directions people mix up: $\varepsilon$ is chosen by someone else, $N$ is chosen by you; $N$ may depend on $\varepsilon$, but once fixed it must work for every $n\ge N$ simultaneously.

Algebra of limits

If $a_n\to a$ and $b_n\to b$, then

$$a_n+cb_n\to a+cb, \qquad a_nb_n\to ab, \qquad \frac{1}{a_n}\to\frac1a\ (\text{for }a\ne0,\ a_n\ne0).$$

The reciprocal needs the denominator to stay away from zero: from $a_n\to a\ne0$ we get $|a_n|>|a|/2$ eventually, hence

$$\left|\frac1{a_n}-\frac1a\right|=\frac{|a_n-a|}{|a_n||a|}<\frac{2|a_n-a|}{|a|^2}\to0 .$$

Second chain: bounded, monotone, Cauchy

Convergent implies bounded

Take $\varepsilon=1$; the tail satisfies $|a_n-a|<1$, so $|a_n|<|a|+1$. Bound the finitely many earlier terms separately and take the maximum. The converse fails: $(-1)^n$ is bounded but divergent.

Subsequences

A subsequence is $x_{n_k}$ with strictly increasing indices $n_1<n_2<\cdots$ — order preserved, terms may be skipped, no index repeated.

Theorem. $(x_n)$ converges if and only if every subsequence converges (to the same limit).

The "if" direction is trivial: the original sequence is its own subsequence (take the index map to be the identity). The "only if" direction uses $n_k\ge k$.

The trivial theorem earns its keep by contradiction: exhibit two subsequences with different limits and the original sequence diverges. For $(-1)^n$ the even terms are constantly $1$ and the odd terms constantly $-1$, so it diverges.

Monotone convergence

• If $(x_n)$ is increasing ($x_{n+1}\ge x_n$), it converges if and only if it is bounded above, and then $\lim x_n=\sup\{x_n\}$.
• If $(x_n)$ is decreasing, it converges if and only if it is bounded below, and then $\lim x_n=\inf\{x_n\}$.

This is the standard tool when you know a sequence is bounded but do not know its limit. It converts existence into the existence of a supremum, which completeness guarantees.

Cauchy sequences

$(x_n)$ is Cauchy if

$$\forall\varepsilon>0\ \exists N\ \text{such that}\ m,n\ge N\Rightarrow |x_m-x_n|<\varepsilon .$$

Cauchy-ness says the tail bunches together and mentions no candidate limit. Its value: when you do not care what the limit is, checking Cauchy is usually easier than checking convergence.

In $\mathbb R$ every Cauchy sequence converges (the Cauchy convergence theorem), so the two notions agree; in a general metric space only "convergent $\Rightarrow$ Cauchy" survives.

Bolzano–Weierstrass

In $\mathbb R$, every bounded sequence has a convergent subsequence.

This is the key ingredient in the Cauchy convergence theorem: extract a convergent subsequence by BW, then show the whole Cauchy sequence converges to that subsequence's limit. The review stresses that both the Cauchy convergence theorem and BW can be stated in any metric space but need not hold there — they depend on completeness of the reals.

Third chain: continuous functions

Definition and sequential form

$f$ is continuous at $x_0$ if

$$\forall\varepsilon>0\ \exists\delta>0\ \text{such that}\ |x-x_0|<\delta\Rightarrow|f(x)-f(x_0)|<\varepsilon .$$

Lemma. If $f$ is continuous at $x_0$ and $x_n\to x_0$, then $f(x_n)\to f(x_0)$.

This lemma translates function questions into sequence questions and is the bridge to the extreme value and intermediate value theorems.

Algebra of continuous functions

If $f,g$ are continuous and $c$ is constant, then $f+cg$ and $fg$ are continuous; if $g$ never vanishes, so are $1/g$ and $f/g$.

With two starting points — constant functions are continuous, and $f(x)=x$ is continuous — induction gives:

• every polynomial is continuous;
• every rational function is continuous wherever its denominator does not vanish.

That is the standard assembly line for building complicated continuous functions from simple ones.

Two theorems

Extreme value theorem. If $f$ is continuous on a closed interval $[a,b]$, then $f$ attains a maximum and a minimum (and is therefore bounded).

Intermediate value theorem. If $f$ is continuous on $[a,b]$ and $y$ lies between $f(a)$ and $f(b)$, then some $x\in[a,b]$ satisfies $f(x)=y$.

Both proofs use the same parts: BW (extract a convergent subsequence), "continuity preserves sequential limits", and the supremum property of the reals. All three are needed.

Fourth chain: series

Partial sums

Given $(a_n)$, the partial sums $S_n=\sum_{k=0}^{n}a_k$ form another sequence. A series converges exactly when its partial sums converge, so every tool for sequences applies immediately.

The first thing to check: if $\sum a_n$ converges, then $a_n\to0$, because $a_n=S_n-S_{n-1}$ and the difference of two convergent sequences tends to $0$.

The converse fails: the harmonic series $\sum 1/n$ has terms tending to $0$ yet diverges. "Terms tend to zero" filters out obvious divergence; it is not a criterion.

Three series to memorise

Geometric series. $\sum_{n=0}^{\infty}c^n$ converges for $|c|<1$, and

$$\sum_{n=0}^{\infty}c^n=\frac{1}{1-c};$$

it diverges for $|c|\ge1$.

Two details that cost marks:

1. The sum depends on where you start. $\sum_{n=1}^{\infty}c^n=\dfrac{c}{1-c}$, not $\dfrac{1}{1-c}$.
2. Convergence is independent of the starting index; the value is not. Shift the index freely when testing convergence, but subtract the removed terms when computing the sum.

Harmonic series $\sum_{n=1}^{\infty}\dfrac1n$: divergent.

Alternating harmonic series $\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n}$: convergent, but not absolutely convergent.

Another useful convergent series is $\sum 1/n^2$, the yardstick for the comparison test.

Absolute convergence

$\sum a_n$ is absolutely convergent if $\sum|a_n|$ converges. Absolute convergence implies convergence, but not conversely (the alternating harmonic series).

Absolute convergence buys two extra properties: the sum is unchanged under any reordering, and the product of two absolutely convergent series converges. The course uses exactly this when estimating remainder terms.

Three tests

Comparison test (version A). If $0\le a_n\le b_n$ and $\sum b_n$ converges, then $\sum a_n$ converges; if $\sum a_n$ diverges, then $\sum b_n$ diverges.

Comparison test (version B). If $a_n,b_n>0$ and $\lim a_n/b_n=L$ with $0<L<\infty$, the two series share their convergence behaviour.

Ratio test. Suppose $a_n\ne0$ and $c=\lim\left|\dfrac{a_{n+1}}{a_n}\right|$ exists. If $c<1$ the series converges; if $c>1$ it diverges; if $c=1$ the test is inconclusive.

Root test. Let $c=\limsup\sqrt[n]{|a_n|}$. If $c<1$ the series converges; if $c>1$ it diverges; if $c=1$ the test is inconclusive.

Both the ratio and root tests are descendants of the geometric series: the idea is to compare the terms with some $c^n$. The root test uses the limsup rather than the limit so that it is defined in more situations — a limsup always exists, a limit need not.

Power series

A series $\sum a_nx^n$ is an ordinary series for each fixed $x$. Define

$$R=\frac{1}{\limsup\sqrt[n]{|a_n|}}\quad(\text{with }1/0=\infty,\ 1/\infty=0).$$

Then

• for $|x|<R$ the series converges absolutely;
• for $|x|>R$ it diverges;
• at the two endpoints $|x|=R$ nothing general can be said; they must be checked separately.

That is why the standard answer to a power series problem reads "radius $R=\cdots$; endpoints must be handled separately".

Example 1: $\sum n^5x^n$. The ratio test gives

$$\left|\frac{a_{n+1}}{a_n}\right|=\left(\frac{n+1}{n}\right)^5|x|\to|x| ,$$

so the series converges for $|x|<1$, diverges for $|x|>1$, and $R=1$.

Example 2: $\sum \dfrac{n^5x^n}{n!}$. The ratio test gives

$$\left|\frac{a_{n+1}}{a_n}\right| =\left(\frac{n+1}{n}\right)^5\frac{|x|}{n+1}\to 0\cdot|x|=0 ,$$

so it converges for every $x$ and $R=\infty$.

In an exam, do not just write "$R=\infty$": compute the ratio and draw the conclusion. Partial credit rewards the process.

Problem types and traps

Type 1: prove convergence from the definition

Template: guess the limit, then "given $\varepsilon>0$, take $N=\lceil\cdots\rceil$", and verify the inequality for $n\ge N$. The Archimedean property and the triangle inequality do most of the work.

Trap: letting $N$ depend on $n$. $N$ may depend only on $\varepsilon$.

Type 2: use monotone convergence

You need two things: monotonicity (often by induction) and boundedness. The conclusion gives existence only, not the value — unless you also compute the supremum.

Trap: proving boundedness but not monotonicity, or assuming convergence to deduce boundedness (circular).

Type 3: prove the Cauchy property

Estimate $|x_m-x_n|$, usually with $m>n$, by writing $|x_m-x_n|\le\sum_{k=n}^{m-1}|x_{k+1}-x_k|$ and bounding the tail with a geometric series or an integral.

Trap: confusing Cauchy-ness with "successive differences shrink". Shrinking successive differences does not imply Cauchy (e.g. $x_n=\sqrt n$).

Type 4: radius of convergence

Decide between ratio and root first. Factorials and powers of $n$ usually favour the ratio test; an overall $n$-th power structure favours the root test.

Trap: forgetting the absolute value; forgetting the endpoints; confusing $R$ with the interval of convergence.

Type 5: give a counterexample

The counterexample list that recurs throughout the review is worth memorising:

A recursive sequence

Let $x_1=a$ and $x_{n+1}=\dfrac{x_n}{n}$. Find the limit.

Write the first few terms:

$$x_1=a,\quad x_2=\frac{a}{1}=a,\quad x_3=\frac{a}{2},\quad x_4=\frac{a}{6},\quad x_5=\frac{a}{24}.$$

The pattern is $x_n=\dfrac{a}{(n-1)!}$ for $n\ge1$. Induction confirms it: $x_{n+1}=\dfrac{x_n}{n}=\dfrac{a}{n!}$. Hence

$$|x_n|=\frac{|a|}{(n-1)!}\le\frac{|a|}{n-1}\to0 ,$$

so $\lim x_n=0$ regardless of $a$.

Listing the first few terms is not a proof; you must derive and verify the closed form, then argue convergence.

A comparison test

Suppose $0\le a_n<1/n^2$. Does $\sum a_n$ converge?

Since $0\le a_n<1/n^2$ and $\sum 1/n^2$ converges, the comparison test gives convergence immediately.

The hypothesis $a_n\ge0$ is essential; without it, $a_n<1/n^2$ holds for divergent series too (take $a_n=-1$), and the claim is false.

What this lecture establishes

• Every convergence statement ultimately rests on the least upper bound property.
• Convergent ⟹ Cauchy ⟹ bounded; neither arrow reverses, except Cauchy ⟹ convergent in $\mathbb R$.
• Monotone plus bounded implies convergence, with the limit equal to the supremum or infimum.
• Continuity transports sequential limits to function values; the extreme value and intermediate value theorems use BW, continuity, and completeness.
• For series: check the terms first, use the geometric series as a yardstick, then apply comparison, ratio or root tests.
• For power series: compute $R$ first, then treat the endpoints separately.
• Counterexamples matter as much as theorems: they mark the boundary of each theorem's validity.

Comprehension check

1. Why does "all subsequences converge" imply the original sequence converges, while "one convergent subsequence exists" does not?

Reference answer: "All subsequences converge" includes the original sequence itself — take the index map $g(k)=k$. "One convergent subsequence exists" only says infinitely many terms cluster near a point while the rest may wander. For example, the even-indexed subsequence of $a_n=(-1)^n$ is constantly $1$ and converges, yet the original sequence diverges.

2. Prove that a convergent sequence is bounded, and give a bounded but divergent example.

Reference answer: Let $a_n\to a$. With $\varepsilon=1$ there is $N$ such that $n\ge N$ implies $|a_n-a|<1$, so $|a_n|\le|a|+1$ by the triangle inequality. Let

$$M=\max\{|a_1|,\dots,|a_{N-1}|,\ |a|+1\};$$

then $|a_n|\le M$ for all $n$. Counterexample: $a_n=(-1)^n$ is bounded, but its even and odd subsequences converge to $1$ and $-1$, so it diverges.

3. Why can the monotone convergence theorem not drop "bounded"? Give an increasing divergent sequence.

Reference answer: The theorem says "convergent $\iff$ bounded". Convergence implies boundedness in general, while boundedness implies convergence only with monotonicity: every term of an increasing sequence is at most the supremum and the sequence approaches it. Without boundedness, $a_n=n$ is increasing and divergent. Conversely, without monotonicity boundedness alone fails: $(-1)^n$. Both hypotheses are needed.

4. Determine whether $\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n}$ converges, and whether it converges absolutely.

Reference answer: It converges by the alternating series test ($1/n$ decreases monotonically to $0$). It is not absolutely convergent, because $\sum 1/n$ is the harmonic series and diverges. So it is the standard example of "convergent but not absolutely convergent", showing absolute convergence is strictly stronger.

5. Find the radius of convergence of $\sum_{n=0}^{\infty} n^5 x^n$ and discuss the endpoints.

Reference answer: The ratio test gives

$$\left|\frac{a_{n+1}}{a_n}\right|=\left(\frac{n+1}{n}\right)^5|x|\to|x|,$$

so the series converges for $|x|<1$, diverges for $|x|>1$, and $R=1$. At $x=\pm1$ the terms are $\pm n^5$, which do not tend to zero, so the series diverges there. The interval of convergence is $(-1,1)$.

6. Someone argues: "Since $a_n\to0$, the series $\sum a_n$ converges." What is wrong?

Reference answer: This mistakes a necessary condition for a sufficient one. Convergence of $\sum a_n$ does imply $a_n\to0$ (because $a_n=S_n-S_{n-1}$), but the converse fails. The harmonic series has $a_n=1/n\to0$ while its partial sums are unbounded: grouping terms in blocks of size $2^k$ gives at least $1/2$ per block, and infinitely many blocks diverge.

7. Why does "Cauchy implies convergent" fail in a general metric space? Illustrate with $X=(0,1]$.

Reference answer: Cauchy-ness only says the tail bunches together; it does not say the point it is heading for belongs to the space. In $X=(0,1]$ take $x_n=1/n$. For any $\varepsilon>0$, choose $N>1/\varepsilon$; then $m,n\ge N$ gives $|1/m-1/n|<1/N<\varepsilon$, so the sequence is Cauchy. Its only candidate limit in $X$ would be $0$, which is not in $X$, so it does not converge in $X$. The Cauchy convergence theorem in $\mathbb R$ holds precisely because $\mathbb R$ is complete and has no such missing points.

Relation to neighbouring lectures

This lecture compresses the first half of the semester into one network. The next lecture moves to metric spaces: convergence, Cauchy-ness and boundedness are transported from $\mathbb R$ to arbitrary distance spaces, and open sets, closed sets and compactness enter. The review's repeated emphasis on "which results depend on completeness" is exactly the dividing line the next lecture makes precise.