上一讲留下的缺口

上一讲证明了 Cauchy 收敛定理：每个 Cauchy 列都收敛。但那个证明借用了另一个定理——

Bolzano–Weierstrass 定理。 每个有界数列 $(a_n)$ 都有收敛的子列 $(a_{n_k})$。

当时的论证路线是：Cauchy 列 ⇒ 有界（这一步当时证过）⇒ 由 Bolzano–Weierstrass 取到收敛子列 ⇒ 再证明这个子列的极限就是原数列的极限。

Bolzano–Weierstrass 本身却没证。它不能用 Cauchy 定理来证，否则就是循环论证——顺序必须是：先有 Bolzano–Weierstrass（只用到实数的完备性），再有 Cauchy 定理。本讲补上这一步。

时间轴

Bolzano–Weierstrass：用二分法逼出一个子列

思路

有界列 $(a_n)$ 被困在某个区间里，比如 $[0,1]$（换任何有界区间都一样，论证中没有任何地方依赖具体端点）。既然无穷多个点挤在一个有限区间里，总有一半（或两半）含有无穷多项。不断取「含有无穷多项的那一半」，区间越缩越小，子列就被夹在中间，最终被迫收敛。

具体地，要同时构造三样东西：

• 一个子列 $(a_{n_k})$；
• 左端点列 $(b_k)$，单调递增；
• 右端点列 $(c_k)$，单调递减；

并且满足

$$b_k\le a_{n_k}\le c_k .$$

只要 $b_k$ 与 $c_k$ 收敛到同一个数，夹在中间的 $a_{n_k}$ 也只能收敛到那个数（夹逼）。

构造

第一步。 令

$$b_1=0,\qquad c_1=1,\qquad a_{n_1}=a_1 .$$

因为 $a_1\in[0,1]$，显然 $b_1\le a_{n_1}\le c_1$。

第二步。 把 $[0,1]$ 平分成 $[0,\tfrac12]$ 与 $[\tfrac12,1]$。整个数列落在两个半区间里，所以至少有一半含有无穷多项。取含无穷多项的那一半（若两半都含无穷多项，任取其一），记

$$b_2=\text{该半区间的左端点},\qquad c_2=\text{该半区间的右端点}.$$

然后从第一个下标之后开始，取第一个落在该半区间中的项，作为 $a_{n_2}$。注意必须取新的下标 $n_2>n_1$，否则拿不到子列。

第 $k$ 步。 重复同样的操作：把当前区间 $[b_{k-1},c_{k-1}]$ 平分，挑出含无穷多项的一半作为新的 $[b_k,c_k]$；再在 $a_{n_{k-1}}$ 之后找出第一个落在新区间里的项作为 $a_{n_k}$。

每一步都能进行下去，依赖的正是「区间内始终有无限多项」这个归纳不变量：平分后两半的并集是原区间，原区间含无穷多项，所以至少一半含无穷多项。

区间长度的递归

每次平分，长度减半：

$$c_{k+1}-b_{k+1}=\frac12\,(c_k-b_k).$$

而初始长度是 $c_1-b_1=1$，于是递推得到

$$c_k-b_k=\left(\frac12\right)^{k-1}=2^{1-k}.$$

由于左端点只向右移、右端点只向左移，

$$b_1\le b_2\le b_3\le\cdots\le c_3\le c_2\le c_1 .$$

收敛性

$(b_k)$ 单调递增且有上界 $c_1$，$(c_k)$ 单调递减且有下界 $b_1$。由单调收敛定理，两者都收敛，记

$$b_k\to b,\qquad c_k\to c .$$

又因为 $b_k\le c_k$，取极限得 $b\le c$。另一方面

$$c-b=\lim_{k\to\infty}(c_k-b_k)=\lim_{k\to\infty}2^{1-k}=0,$$

所以 $b=c$。

最后验证子列。对 $l\ge k$，由构造 $a_{n_l}\in[b_l,c_l]\subseteq[b_k,c_k]$，因此

$$|a_{n_l}-b|\le c_k-b_k=2^{1-k}.$$

给定 $\varepsilon>0$，取 $k$ 使 $2^{1-k}<\varepsilon$，则对所有 $l\ge k$ 都有 $|a_{n_l}-b|<\varepsilon$。所以 $a_{n_k}\to b$。$\square$

这里用到的关键性质其实只有一条：单调有界必收敛。整条链是「完备性 ⇒ 单调收敛定理 ⇒ Bolzano–Weierstrass ⇒ Cauchy 收敛定理」，每一步都不能倒过来用。

级数：部分和序列

给定数列 $(a_n)$，可以造出另一个数列

$$s_n=a_1+a_2+\cdots+a_n=\sum_{i=1}^{n}a_i .$$

定义。 形如 $(s_n)$ 的数列称为级数（series）。若 $(s_n)$ 收敛，就说级数收敛，其极限记作

$$\sum_{i=1}^{\infty}a_i .$$

这里有一个记号上的小陷阱：符号 $\sum_{i=1}^{\infty}a_i$ 同时被用来表示「级数这个对象」和「收敛时的和」。写的时候要清楚自己指的是哪一个。级数是否收敛与从第几项开始无关（改变的只是和的值，不是收敛性）。

等比级数：唯一能算出和的重要级数

取定 $c$，考虑

$$s_n=\sum_{i=0}^{n}c^i=1+c+c^2+\cdots+c^n .$$

求和公式

把 $s_n$ 乘以 $c$：

$$c\,s_n=c+c^2+\cdots+c^{n+1}.$$

用 $s_n$ 减去它：

$$(1-c)s_n=(1+c+\cdots+c^n)-(c+\cdots+c^{n+1})=1-c^{n+1},$$

中间项两两抵消。于是当 $c\ne1$ 时

$$s_n=\frac{1-c^{n+1}}{1-c}.$$

收敛判定

定理。 等比级数 $\sum_{i=0}^{\infty}c^i$ 在 $|c|<1$ 时收敛，且和为 $\dfrac{1}{1-c}$；在 $|c|\ge1$ 时发散。

证明（$|c|<1$）。 由求和公式，

$$s_n=\frac{1-c^{n+1}}{1-c}.$$

因为 $|c^{n+1}|=|c|^{n+1}\to0$（$|c|<1$ 时几何项趋于零），由极限的代数运算，

$$s_n\to\frac{1-0}{1-c}=\frac{1}{1-c}.$$

这里能除以 $1-c$，正因为 $|c|<1$ 保证了 $c\ne1$。

证明（$|c|\ge1$）。 看相邻两项之差：

$$s_{n+1}-s_n=c^{n+1}.$$

若 $|c|\ge1$，则 $|s_{n+1}-s_n|=|c|^{n+1}\ge1$ 对一切 $n$ 成立。收敛数列必是 Cauchy 列，而 Cauchy 列要求足够靠后的任意两项都任意接近——这里连相邻两项都始终相距至少 $1$，所以 $(s_n)$ 不可能是 Cauchy 列，故发散。$\square$

这也解释了为什么等比级数是「最重要的级数」：它是少数几个能算出精确和的级数，而大多数级数只能判断收敛与否。更重要的是，它的收敛性可以直接拿来当标尺，去比较别的级数。

调和级数：发散

$$\sum_{n=1}^{\infty}\frac1n=1+\frac12+\frac13+\cdots$$

它是最有名的发散级数。这里给一个只用分组的初等证明。

定理。 调和级数发散。

证明。 记 $s_n=\sum_{i=1}^{n}1/i$。要证 $(s_n)$ 无界。考虑下标为 $2^n-1$ 的子列，证明

$$s_{2^n-1}\ge\frac{n}{2}\qquad(n\ge1).$$

基础情形 $n=1$： $s_{2^1-1}=s_1=1\ge\tfrac12$。

归纳步： 设 $s_{2^n-1}\ge n/2$。考察

$$s_{2^{n+1}-1}=s_{2^n-1}+\sum_{i=2^{n}}^{2^{n+1}-1}\frac1i .$$

后面这一段共有 $2^{n+1}-1-2^n+1=2^n$ 项，其中每一项都至少是最后一项 $\dfrac{1}{2^{n+1}-1}$，而它大于 $\dfrac{1}{2^{n+1}}$。于是

$$\sum_{i=2^{n}}^{2^{n+1}-1}\frac1i \ge 2^n\cdot\frac{1}{2^{n+1}}=\frac12 .$$

代回：

$$s_{2^{n+1}-1}\ge \frac{n}{2}+\frac12=\frac{n+1}{2}.$$

归纳完成。

于是子列 $(s_{2^n-1})$ 满足 $s_{2^n-1}\ge n/2\to\infty$，无界。若原级数收敛，则每个子列都应收敛（因而有界），矛盾。所以调和级数发散。$\square$

这个证明的要点是分组：把第 $2^n$ 项到第 $2^{n+1}-1$ 项捆成一组，每组有 $2^n$ 个数，每个数都不小于 $2^{-(n+1)}$，所以每组之和至少是 $1/2$。无穷多个 $1/2$ 加起来当然发散。

判别级数收敛的第一件事：通项必须趋于零

若级数 $\sum a_i$ 收敛，则

$$a_n=s_n-s_{n-1}\to0 .$$

所以通项不趋于零，级数一定发散，不必再试任何判别法。

调和级数恰好说明这个条件的逆命题不成立：它的通项 $1/n\to0$，级数却发散。所以「通项趋于零」只是必要条件。

非负项级数：部分和单调

若所有 $a_n\ge0$，则

$$s_{n+1}=s_n+a_{n+1}\ge s_n,$$

即 $(s_n)$ 单调递增。由单调收敛定理：非负项级数收敛，当且仅当部分和序列有上界。

这把「判断收敛」变成了「找一个上界」。下面的例子和比较判别法都建立在这条观察上。

绝对收敛

定义。 若 $\sum|a_i|$ 收敛，就说 $\sum a_i$ 绝对收敛。

定理。 绝对收敛推出收敛。

证明。 记

$$s_n=\sum_{i=1}^{n}a_i,\qquad \bar s_n=\sum_{i=1}^{n}|a_i| .$$

对 $m>n$，考察两个部分和之差：

$$|s_m-s_n|=\left|\sum_{i=n+1}^{m}a_i\right| \le\sum_{i=n+1}^{m}|a_i| =\bar s_m-\bar s_n .$$

已知 $(\bar s_n)$ 收敛，故它是 Cauchy 列：给定 $\varepsilon>0$，存在 $N$，当 $m>n\ge N$ 时 $\bar s_m-\bar s_n<\varepsilon$。由上式，此时也有 $|s_m-s_n|<\varepsilon$，所以 $(s_n)$ 是 Cauchy 列，从而收敛。$\square$

注意这里用到了三角不等式 $|\sum a_i|\le\sum|a_i|$，以及「收敛 ⇒ Cauchy」这条上一讲的定理——这条路线现在已经是完整的了。

反例：交错调和级数

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n} =-1+\frac12-\frac13+\frac14-\cdots$$

它收敛（课程在此只作陈述，证明留到后面的交错级数判别法），但加上绝对值就得到调和级数，它是发散的。所以

$$\text{绝对收敛}\Longrightarrow\text{收敛}, \qquad \text{收敛}\nRightarrow\text{绝对收敛}.$$

这种「收敛但不绝对收敛」的级数称为条件收敛。

例子：$\sum 1/n^2$ 收敛

$$\sum_{n=1}^{\infty}\frac{1}{n^2}=1+\frac14+\frac19+\cdots$$

这一项在课上是用来展示「不用积分判别法也能算」的。因为通项非负，只需证明部分和有上界。要证

$$s_{2^{n}-1}\le\sum_{i=0}^{n-1}\left(\frac12\right)^{i}\qquad(n\ge1).$$

基础情形 $n=1$： $s_1=1$，右边是 $(1/2)^0=1$，成立。

归纳步： 设 $s_{2^n-1}\le\sum_{i=0}^{n-1}2^{-i}$。则

$$s_{2^{n+1}-1} =s_{2^n-1}+\sum_{i=2^{n}}^{2^{n+1}-1}\frac{1}{i^2}.$$

第二段有 $2^n$ 项，最大的一项是第一项 $\dfrac{1}{2^{2n}}$，所以

$$\sum_{i=2^{n}}^{2^{n+1}-1}\frac{1}{i^2} \le 2^n\cdot\frac{1}{2^{2n}}=\frac{1}{2^{n}} .$$

代入归纳假设：

$$s_{2^{n+1}-1}\le\sum_{i=0}^{n-1}\frac{1}{2^{i}}+\frac{1}{2^{n}} =\sum_{i=0}^{n}\frac{1}{2^{i}} .$$

归纳完成。

于是所有形如 $s_{2^n-1}$ 的部分和被等比级数 $\sum 2^{-i}=2$ 控制，而这个子列本身单调递增，所以整个 $(s_n)$ 有上界 $2$，级数收敛。

这个证明与调和级数的证明结构完全对称：调和级数用最小的那一项给出下界，$1/n^2$ 用最大的那一项给出上界。同一个二分分组，一个证发散，一个证收敛。

常见的判别法

课上只列名字并讲了比较判别法，其余留到下一讲：

• 比较判别法（comparison test）；
• 比值判别法（ratio test）；
• 根值判别法（root test）；
• 积分判别法（integral test）；
• 交错级数判别法（alternating series test）。

比较判别法的最简形式：若

$$0\le a_n\le b_n$$

对所有足够大的 $n$ 成立，且 $\sum b_n$ 收敛，则 $\sum a_n$ 收敛。

它有两个方向：拿一个复杂级数去和已知收敛的简单级数比较，可证收敛；也可以反过来说，若 $0\le b_n\le a_n$ 且 $\sum b_n$ 发散，则 $\sum a_n$ 发散。等比级数之所以重要，正是因为它是最常用的那把标尺。

本节结论

• Bolzano–Weierstrass：有界列必有收敛子列。证明靠二分构造嵌套区间，长度 $2^{1-k}\to0$，再用单调收敛定理。
• 级数 = 部分和序列；级数收敛就是 $(s_n)$ 收敛。
• 等比级数 $\sum c^i$：$|c|<1$ 时收敛，和为 $1/(1-c)$；$|c|\ge1$ 时发散。
• 调和级数发散；分组可得 $s_{2^n-1}\ge n/2$。
• 通项不趋于零 ⇒ 发散；通项趋于零不足以保证收敛。
• 非负项级数收敛 ⇔ 部分和有上界。
• 绝对收敛 ⇒ 收敛；交错调和级数说明反过来不成立。
• $\sum 1/n^2$ 收敛；与调和级数用同一套分组，只是取上界而非下界。

理解检验

1. 为什么 Bolzano–Weierstrass 不能用 Cauchy 收敛定理来证明？

参考答案：Cauchy 收敛定理的证明本身就用了 Bolzano–Weierstrass（Cauchy 列有界 → 取收敛子列 → 证明整列收敛到同一极限）。若再用 Cauchy 定理去证 Bolzano–Weierstrass，就构成循环论证。逻辑顺序必须是：完备性 ⇒ 单调收敛定理 ⇒ Bolzano–Weierstrass ⇒ Cauchy 收敛定理。本讲给出的证明只用到「单调有界必收敛」，与 Cauchy 定理无关。

2. 在二分构造中，为什么每一步都能挑到「含无穷多项的那一半」？如果两半都含无穷多项，会出问题吗？

参考答案：原区间含无穷多项，而两个半区间的并集就是原区间，所以二者中至少有一个含无穷多项（否则总项数有限，矛盾）。若两半都含无穷多项，任取其一即可，不影响后续论证，因为构造只要求新区间含有无穷多项，并不要求唯一。取定之后，下一个下标必须严格大于前一个，这也能做到，因为选中的区间里有无穷多项，必然还有当前下标之后的项。

3. 用求和公式证明：$|c|<1$ 时 $\sum_{i=0}^{\infty}c^i=1/(1-c)$，并说明 $c=1$ 时公式为何不能用。

参考答案：对 $c\ne1$，$(1-c)\sum_{i=0}^{n}c^i=1-c^{n+1}$，故 $s_n=(1-c^{n+1})/(1-c)$。当 $|c|<1$ 时 $|c|^{n+1}\to0$，于是 $s_n\to1/(1-c)$。$c=1$ 时推导中要除以 $1-c$，而 $1-c=0$，除法不合法；直接看原式，$s_n=n+1\to\infty$，级数发散，和根本不存在。

4. 证明调和级数发散时，为什么选下标 $2^n-1$ 而不是别的子列？

参考答案：分组需要每组项数固定、且每项有统一的下界。取从 $2^n$ 到 $2^{n+1}-1$ 这一组，恰好有 $2^n$ 项，而每一项都 $\ge 2^{-(n+1)}$，所以该组之和 $\ge 2^n\cdot2^{-(n+1)}=1/2$。下标 $2^n-1$ 正是这些分组端点的自然选择，使「组号 $n$ 对应 $n$ 个 $1/2$」。若随便取一个子列，未必能保证每组贡献一个与 $n$ 无关的正下界。

5. 绝对收敛为什么能推出收敛？证明中哪一步用到了完备性？

参考答案：设 $\bar s_n=\sum_{i=1}^{n}|a_i|$ 收敛。对 $m>n$，

$$|s_m-s_n|=\left|\sum_{i=n+1}^{m}a_i\right|\le\sum_{i=n+1}^{m}|a_i|=\bar s_m-\bar s_n .$$

$(\bar s_n)$ 收敛 ⇒ 它是 Cauchy 列 ⇒ 右端可任意小 ⇒ $(s_n)$ 也是 Cauchy 列 ⇒ 收敛。完备性用在最后一步：从「Cauchy 列」推出「收敛」，正是 Cauchy 收敛定理，而它依赖实数的完备性（经由 Bolzano–Weierstrass）。

6. 比较判别法为什么既可以用来证明收敛，也可以用来证明发散？

参考答案：因为它是双向的。若有 $0\le a_n\le b_n$ 且 $\sum b_n$ 收敛，由非负项级数的单调性，$s_n(a)\le s_n(b)$ 有上界，故 $\sum a_n$ 收敛。反过来，若有 $0\le b_n\le a_n$ 且 $\sum b_n$ 发散，则 $s_n(a)\ge s_n(b)$ 无界，故 $\sum a_n$ 发散。两个方向只是把大小关系反过来用。

与前后课的关系

Bolzano–Weierstrass 补上了上一讲 Cauchy 定理的最后一块拼图，实数系的三条基本定理（单调收敛、Cauchy 收敛、Bolzano–Weierstrass）至此互相等价、全部证完。级数这一支才刚开头：下一讲会系统讲比值、根值、积分与交错级数判别法，并用它们研究幂级数——那是把级数用于表示函数的入口。

后面讲连续函数、极限与导数时，会反复回到本讲的两个工具：夹逼（子列被夹在 $b_k,c_k$ 之间）与分组估计（把无穷和压到几何级数上）。

The gap left by the last lecture

The previous lecture proved the Cauchy convergence theorem: every Cauchy sequence converges. But that proof borrowed another result —

Bolzano–Weierstrass theorem. Every bounded sequence $(a_n)$ has a convergent subsequence $(a_{n_k})$.

The route was: a Cauchy sequence is bounded (proved there) ⇒ Bolzano–Weierstrass supplies a convergent subsequence ⇒ show that the subsequence's limit is the limit of the whole sequence.

Bolzano–Weierstrass itself was never proved. It cannot be proved from the Cauchy theorem, since that would be circular: the correct order is Bolzano–Weierstrass first (using only completeness), Cauchy's theorem afterwards. This lecture fills the gap.

Timeline

Bolzano–Weierstrass: squeezing out a subsequence by bisection

The idea

A bounded sequence $(a_n)$ is trapped in some interval, say $[0,1]$ (any bounded interval works; nothing in the argument uses the endpoints). Infinitely many points crowd into a finite interval, so at least one half contains infinitely many terms. Keep choosing the half that contains infinitely many, and the intervals shrink until the subsequence has nowhere to go but a single limit.

Three objects are constructed simultaneously:

• a subsequence $(a_{n_k})$;
• left endpoints $(b_k)$, increasing;
• right endpoints $(c_k)$, decreasing;

with

$$b_k\le a_{n_k}\le c_k .$$

If $b_k$ and $c_k$ converge to the same number, the squeezed subsequence must converge to it too.

The construction

Step 1. Set

$$b_1=0,\qquad c_1=1,\qquad a_{n_1}=a_1 .$$

Since $a_1\in[0,1]$, certainly $b_1\le a_{n_1}\le c_1$.

Step 2. Split $[0,1]$ into $[0,\tfrac12]$ and $[\tfrac12,1]$. The whole sequence lies in the union, so at least one half contains infinitely many terms. Choose such a half (if both qualify, pick either) and set

$$b_2=\text{its left endpoint},\qquad c_2=\text{its right endpoint}.$$

Then choose as $a_{n_2}$ the first term after $a_{n_1}$ that lies in this half. The index must satisfy $n_2>n_1$; otherwise the result is not a subsequence.

Step $k$. Repeat: bisect $[b_{k-1},c_{k-1}]$, keep a half containing infinitely many terms as $[b_k,c_k]$, and take the first term after $a_{n_{k-1}}$ that lands inside it.

The construction never stalls, thanks to an inductive invariant: after bisecting, the two halves cover the previous interval, which contains infinitely many terms, so at least one half does too.

The length recursion

Each bisection halves the length:

$$c_{k+1}-b_{k+1}=\frac12\,(c_k-b_k).$$

Starting from $c_1-b_1=1$,

$$c_k-b_k=\left(\frac12\right)^{k-1}=2^{1-k}.$$

Endpoints move inward, so

$$b_1\le b_2\le b_3\le\cdots\le c_3\le c_2\le c_1 .$$

Convergence

$(b_k)$ is increasing and bounded above by $c_1$; $(c_k)$ is decreasing and bounded below by $b_1$. By the monotone convergence theorem both converge; write

$$b_k\to b,\qquad c_k\to c .$$

Since $b_k\le c_k$, passing to the limit gives $b\le c$. But

$$c-b=\lim_{k\to\infty}(c_k-b_k)=\lim_{k\to\infty}2^{1-k}=0,$$

so $b=c$.

Finally, for $l\ge k$ the construction gives $a_{n_l}\in[b_l,c_l]\subseteq[b_k,c_k]$, hence

$$|a_{n_l}-b|\le c_k-b_k=2^{1-k}.$$

Given $\varepsilon>0$, choose $k$ with $2^{1-k}<\varepsilon$; then $|a_{n_l}-b|<\varepsilon$ for all $l\ge k$. Therefore $a_{n_k}\to b$. $\square$

Only one deep fact is used: monotone and bounded implies convergent. The chain is completeness ⇒ monotone convergence ⇒ Bolzano–Weierstrass ⇒ Cauchy convergence, and no link may be used backwards.

Series: the sequence of partial sums

Given a sequence $(a_n)$, form another:

$$s_n=a_1+a_2+\cdots+a_n=\sum_{i=1}^{n}a_i .$$

Definition. A sequence of this form $(s_n)$ is called a series. If $(s_n)$ converges, the series is said to converge, and its limit is written

$$\sum_{i=1}^{\infty}a_i .$$

One notational trap: the symbol $\sum_{i=1}^{\infty}a_i$ denotes both "the series as an object" and, when it converges, "its sum". Convergence does not depend on where the indexing starts (only the value of the sum does).

The geometric series: the one important series we can sum

Fix $c$ and consider

$$s_n=\sum_{i=0}^{n}c^i=1+c+c^2+\cdots+c^n .$$

The sum formula

Multiply by $c$:

$$c\,s_n=c+c^2+\cdots+c^{n+1}.$$

Subtract:

$$(1-c)s_n=(1+c+\cdots+c^n)-(c+\cdots+c^{n+1})=1-c^{n+1},$$

because all intermediate terms cancel. Hence for $c\ne1$,

$$s_n=\frac{1-c^{n+1}}{1-c}.$$

Convergence

Theorem. The geometric series $\sum_{i=0}^{\infty}c^i$ converges for $|c|<1$ with sum $\dfrac{1}{1-c}$, and diverges for $|c|\ge1$.

Proof ($|c|<1$). By the sum formula,

$$s_n=\frac{1-c^{n+1}}{1-c}.$$

Since $|c^{n+1}|=|c|^{n+1}\to0$, the algebra of limits gives

$$s_n\to\frac{1-0}{1-c}=\frac{1}{1-c}.$$

Division by $1-c$ is legitimate precisely because $|c|<1$ forces $c\ne1$.

Proof ($|c|\ge1$). Look at consecutive differences:

$$s_{n+1}-s_n=c^{n+1}.$$

If $|c|\ge1$, then $|s_{n+1}-s_n|=|c|^{n+1}\ge1$ for every $n$. A convergent sequence is Cauchy, so any two terms sufficiently far out must be arbitrarily close. Here even consecutive terms stay at distance at least $1$, so $(s_n)$ cannot be Cauchy and diverges. $\square$

This is why the geometric series is called the most important series: it is one of the few whose sum can be computed exactly, and its convergence serves as a yardstick for comparing other series.

The harmonic series diverges

$$\sum_{n=1}^{\infty}\frac1n=1+\frac12+\frac13+\cdots$$

Theorem. The harmonic series diverges.

Proof. Let $s_n=\sum_{i=1}^{n}1/i$. We show the subsequence at indices $2^n-1$ is unbounded:

$$s_{2^n-1}\ge\frac{n}{2}\qquad(n\ge1).$$

Base case $n=1$: $s_{2^1-1}=s_1=1\ge\tfrac12$.

Inductive step: assume $s_{2^n-1}\ge n/2$. Then

$$s_{2^{n+1}-1}=s_{2^n-1}+\sum_{i=2^{n}}^{2^{n+1}-1}\frac1i .$$

The block has $2^{n+1}-1-2^n+1=2^n$ terms, and each is at least the last one, $\dfrac{1}{2^{n+1}-1}$, which exceeds $\dfrac{1}{2^{n+1}}$. Therefore

$$\sum_{i=2^{n}}^{2^{n+1}-1}\frac1i \ge 2^n\cdot\frac{1}{2^{n+1}}=\frac12 .$$

Substituting,

$$s_{2^{n+1}-1}\ge \frac{n}{2}+\frac12=\frac{n+1}{2}.$$

The induction is complete. Hence $s_{2^n-1}\ge n/2\to\infty$ is unbounded. A convergent series would have every subsequence convergent, hence bounded — contradiction. So the harmonic series diverges. $\square$

The engine is grouping: terms $2^n$ through $2^{n+1}-1$ form a block of $2^n$ terms, each at least $2^{-(n+1)}$, so each block contributes at least $1/2$. Infinitely many halves diverge.

First check: terms must tend to zero

If $\sum a_i$ converges, then

$$a_n=s_n-s_{n-1}\to0 .$$

So if the terms do not tend to zero, the series diverges — no further test is needed. The converse fails: the harmonic series has $1/n\to0$ yet diverges.

Non-negative series: partial sums are monotone

If all $a_n\ge0$, then

$$s_{n+1}=s_n+a_{n+1}\ge s_n,$$

so $(s_n)$ is increasing. By monotone convergence: a non-negative-term series converges if and only if its partial sums are bounded above.

This turns "decide convergence" into "find a bound", the basis for the examples below and for the comparison test.

Absolute convergence

Definition. $\sum a_i$ converges absolutely if $\sum|a_i|$ converges.

Theorem. Absolute convergence implies convergence.

Proof. Let

$$s_n=\sum_{i=1}^{n}a_i,\qquad \bar s_n=\sum_{i=1}^{n}|a_i| .$$

For $m>n$,

$$|s_m-s_n|=\left|\sum_{i=n+1}^{m}a_i\right| \le\sum_{i=n+1}^{m}|a_i| =\bar s_m-\bar s_n .$$

Since $(\bar s_n)$ converges, it is Cauchy: given $\varepsilon>0$ there is $N$ with $\bar s_m-\bar s_n<\varepsilon$ whenever $m>n\ge N$. The display then gives $|s_m-s_n|<\varepsilon$, so $(s_n)$ is Cauchy and converges. $\square$

The proof uses the triangle inequality and the implication "convergent ⇒ Cauchy", whose own proof is now complete.

A counterexample: the alternating harmonic series

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n} =-1+\frac12-\frac13+\frac14-\cdots$$

It converges (stated here; proved with the alternating series test later), but taking absolute values gives the harmonic series, which diverges. Hence

$$\text{absolute convergence}\Longrightarrow\text{convergence}, \qquad \text{convergence}\nRightarrow\text{absolute convergence}.$$

A series that converges but not absolutely is called conditionally convergent.

Example: $\sum 1/n^2$ converges

$$\sum_{n=1}^{\infty}\frac{1}{n^2}=1+\frac14+\frac19+\cdots$$

The lecture uses this to show that convergence can be settled without the integral test. Since terms are non-negative, it suffices to bound the partial sums. Claim:

$$s_{2^{n}-1}\le\sum_{i=0}^{n-1}\left(\frac12\right)^{i}\qquad(n\ge1).$$

Base case $n=1$: $s_1=1$ and the right side is $(1/2)^0=1$.

Inductive step: assume the claim for $n$. Then

$$s_{2^{n+1}-1} =s_{2^n-1}+\sum_{i=2^{n}}^{2^{n+1}-1}\frac{1}{i^2}.$$

The second block has $2^n$ terms and its largest term is the first, $\dfrac{1}{2^{2n}}$, so

$$\sum_{i=2^{n}}^{2^{n+1}-1}\frac{1}{i^2} \le 2^n\cdot\frac{1}{2^{2n}}=\frac{1}{2^{n}} .$$

Substituting the inductive hypothesis,

$$s_{2^{n+1}-1}\le\sum_{i=0}^{n-1}\frac{1}{2^{i}}+\frac{1}{2^{n}} =\sum_{i=0}^{n}\frac{1}{2^{i}} .$$

Induction complete.

Thus every partial sum of the form $s_{2^n-1}$ is controlled by the geometric series $\sum2^{-i}=2$, and that subsequence is increasing, so the whole sequence is bounded by $2$ and the series converges.

The proof is the exact mirror of the harmonic argument: the harmonic series uses the smallest term of each block to get a lower bound, $\sum1/n^2$ uses the largest to get an upper bound. One bisection scheme, two opposite conclusions.

The standard tests

The lecture names them and proves only the comparison test; the rest come next time:

• comparison test;
• ratio test;
• root test;
• integral test;
• alternating series test.

Simplest form of the comparison test: if

$$0\le a_n\le b_n$$

for all sufficiently large $n$ and $\sum b_n$ converges, then $\sum a_n$ converges.

It works in both directions: compare a complicated series with a known convergent one to prove convergence, or with a known divergent one (with the inequality reversed) to prove divergence. The geometric series matters precisely because it is the usual yardstick.

What this lecture establishes

• Bolzano–Weierstrass: every bounded sequence has a convergent subsequence, via nested bisected intervals of length $2^{1-k}\to0$ and monotone convergence.
• A series is its partial-sum sequence; convergence of one means convergence of the other.
• Geometric series: converges for $|c|<1$ with sum $1/(1-c)$; diverges for $|c|\ge1$.
• The harmonic series diverges, since $s_{2^n-1}\ge n/2$.
• Terms not tending to zero ⇒ divergence; tending to zero is not sufficient.
• Non-negative series converge iff partial sums are bounded above.
• Absolute convergence ⇒ convergence; the alternating harmonic series shows the converse fails.
• $\sum1/n^2$ converges, by the same block trick used for the harmonic series but with an upper bound.

Comprehension check

1. Why can Bolzano–Weierstrass not be proved from the Cauchy convergence theorem?

Reference answer: The Cauchy theorem's proof already uses Bolzano–Weierstrass (a Cauchy sequence is bounded → take a convergent subsequence → show the whole sequence converges to that limit). Using Cauchy's theorem to prove Bolzano–Weierstrass would be circular. The logical order is completeness ⇒ monotone convergence ⇒ Bolzano–Weierstrass ⇒ Cauchy convergence. The proof given here uses only "monotone and bounded implies convergent".

2. In the bisection, why can a half containing infinitely many terms always be found? What if both halves qualify?

Reference answer: The previous interval contains infinitely many terms, and the two halves cover it, so at least one half contains infinitely many (otherwise the total would be finite). If both do, either may be chosen; the construction only needs the new interval to contain infinitely many terms, not to be unique. The next index can always be chosen strictly larger than the previous one, because infinitely many terms remain beyond any given index.

3. Use the sum formula to prove $\sum_{i=0}^{\infty}c^i=1/(1-c)$ for $|c|<1$, and explain why the formula fails at $c=1$.

Reference answer: For $c\ne1$, $(1-c)\sum_{i=0}^{n}c^i=1-c^{n+1}$, so $s_n=(1-c^{n+1})/(1-c)$. If $|c|<1$ then $|c|^{n+1}\to0$, hence $s_n\to1/(1-c)$. At $c=1$ the derivation divides by $1-c=0$, which is illegal; directly, $s_n=n+1\to\infty$, so the series diverges and no sum exists.

4. In the harmonic proof, why choose the subsequence at indices $2^n-1$?

Reference answer: Grouping needs a fixed number of terms per block with a uniform lower bound. The block from $2^n$ to $2^{n+1}-1$ has exactly $2^n$ terms, each at least $2^{-(n+1)}$, so the block contributes at least $2^n\cdot2^{-(n+1)}=1/2$. The index $2^n-1$ is exactly the endpoint of these blocks, making block number $n$ contribute $n$ halves in total. An arbitrary subsequence need not give a block contribution bounded below by a constant independent of $n$.

5. Why does absolute convergence imply convergence, and where is completeness used?

Reference answer: Let $\bar s_n=\sum_{i=1}^{n}|a_i|$ converge. For $m>n$,

$$|s_m-s_n|=\left|\sum_{i=n+1}^{m}a_i\right|\le\sum_{i=n+1}^{m}|a_i|=\bar s_m-\bar s_n .$$

A convergent $(\bar s_n)$ is Cauchy, so the right side becomes arbitrarily small, hence $(s_n)$ is Cauchy and converges. Completeness is used in the last step: "Cauchy ⇒ convergent" is the Cauchy convergence theorem, which rests on completeness via Bolzano–Weierstrass.

6. Why can the comparison test prove both convergence and divergence?

Reference answer: It is two-sided. If $0\le a_n\le b_n$ and $\sum b_n$ converges, monotonicity of non-negative partial sums gives $s_n(a)\le s_n(b)$, which is bounded, so $\sum a_n$ converges. Conversely, if $0\le b_n\le a_n$ and $\sum b_n$ diverges, then $s_n(a)\ge s_n(b)$ is unbounded, so $\sum a_n$ diverges. The two directions just reverse the inequality.

Relation to neighbouring lectures

Bolzano–Weierstrass completes the previous lecture's Cauchy theorem; the three fundamental theorems of the real line (monotone convergence, Cauchy convergence, Bolzano–Weierstrass) are now proved and mutually equivalent. The series story has only begun: next time come the ratio, root, integral and alternating tests, and then power series — the gateway to representing functions by series.

When continuity, limits and derivatives arrive, two tools from this lecture will keep returning: squeezing (the subsequence trapped between $b_k$ and $c_k$) and block estimates (pushing an infinite sum under a geometric bound).