一个判据的诞生

上一讲说明：收敛必有界，但有界不一定收敛（$(-1)^n$）。本讲补上一个真正能用的充分条件——只要数列单调，有界就足以保证收敛。

这个结论之所以重要，是因为它把「找极限」变成「验证两条性质」：单调、有界。很多数列的极限写不出闭式，但可以证明存在。

时间轴

用四则运算算极限

求

$$a_n=\frac{n^2+1}{n^2+n+1}$$

的极限。分子分母同除以 $n^2$：

$$a_n=\frac{1+\dfrac1{n^2}}{1+\dfrac1n+\dfrac1{n^2}} .$$

已知 $1/n\to0$、$1/n^2\to0$，常数列 $1\to1$。由加法规则，分子趋于 $1$，分母趋于 $1$；由乘法与加法规则，分母 $1+\frac1n+\frac1{n^2}$ 趋于 $1\ne0$，且每一项都为正，所以可以应用倒数规则；再用乘法规则：

$$a_n\to\frac{1}{1}=1 .$$

这里每一步都能指出用的是哪条规则。这也说明上一讲的四则运算不是装饰：它们把「看起来复杂的表达式」拆成已知极限的组合。

单调数列

定义。 数列 $(a_n)$ 单调递增，如果

$$a_n\le a_{n+1}\quad\text{对一切 }n .$$

若不等式严格，称为严格递增。类似地，单调递减指 $a_n\ge a_{n+1}$ 对一切 $n$。

注意「单调」本身不要求有界。$a_n=n$ 单调递增但无界，不收敛；$a_n=-n$ 单调递减但无界。

单调收敛定理

定理。 设 $(a_n)$ 单调递增且有上界，则 $(a_n)$ 收敛，且

$$\lim_{n\to\infty}a_n=\sup_n a_n .$$

对称地，若 $(a_n)$ 单调递减且有下界，则收敛到 $\inf_n a_n$。

单调递增数列天然有下界（$a_1$ 即可），所以「有界」在这条定理里只需检查上界；单调递减情形同理只需检查下界。

证明（递增情形）

设 $a_n$ 单调递增且有上界。由完备性，$a=\sup\{a_n:n\in\mathbb N\}$ 存在。要证 $a_n\to a$。

因为 $a$ 是上界，$a_n\le a$ 对一切 $n$ 成立，所以

$$|a_n-a|=a-a_n .$$

于是只需证：给定 $\varepsilon>0$，当 $n$ 足够大时 $a-a_n<\varepsilon$。

关键一步：$a-\varepsilon$ 不是上界。若它是上界，就会有一个比最小上界 $a$ 更小的上界，矛盾。所以存在 $N$，使

$$a_N>a-\varepsilon .$$

（严格大于：若所有 $a_n\le a-\varepsilon$，则 $a-\varepsilon$ 就是上界。）

再由单调性，当 $n\ge N$ 时 $a_n\ge a_N>a-\varepsilon$。结合 $a_n\le a$，

$$0\le a-a_n<a-(a-\varepsilon)=\varepsilon \quad (n\ge N).$$

所以 $a_n\to a$。$\square$

证明的核心只有一句：上确界是「刚好压住」的那个数，所以任何比它小的数（$a-\varepsilon$）都压不住，必然被数列中的某一项超过。之后单调性把这一次超过变成「从此一直超过」。

递减情形把 $a_n$ 换成 $-a_n$ 即可，$-a_n$ 递增且有上界 $-\inf a_n$。

任意有界数列都能造出单调数列

单调收敛定理看起来只适用于特殊数列。但给定任意有界数列 $(a_n)$，可以造出两个单调数列。

定义

$$b_n=\sup\{a_k:k\ge n\}, \qquad c_n=\inf\{a_k:k\ge n\}.$$

（有界保证这两个上、下确界存在。）

$b_n$ 递减。 $b_{n+1}$ 是对更小的集合 $\{a_k:k\ge n+1\}$ 取上确界，而 $b_n$ 的集合多包含一个元素 $a_n$，集合更大，上确界不会更小：

$$b_n\ge b_{n+1}.$$

$c_n$ 递增。 同理，集合变小，下确界不会更小：

$$c_n\le c_{n+1}.$$

$b_n\ge c_n$。 两者取自同一个集合，上确界当然不小于下确界。

于是 $b_n$ 单调递减有下界（被 $c_n$ 压着），$c_n$ 单调递增有上界（被 $b_n$ 压着）。由单调收敛定理，两者都收敛：

$$b_n\downarrow b, \qquad c_n\uparrow c, \qquad c\le b .$$

最后一步用到「极限保持非严格不等式」：若 $x_n\le y_n$ 对一切 $n$ 成立且两者都收敛，则 $\lim x_n\le\lim y_n$。

这组 $b,c$ 其实就是后面要定义的上极限与下极限（$\limsup$ 与 $\liminf$）：$b=\limsup a_n$，$c=\liminf a_n$。课程在这里先把它造出来，第 6 讲证明柯西准则时会用到。

用单调收敛定理重新理解 $\sqrt 2$

第 2 讲用上确界的构造证明了 $\sqrt 2$ 存在。这里换一条路：直接按「小数点后的位数」造一个递增序列，用单调收敛定理得到极限，再验证极限平方为 $2$。

构造

对每个 $n\ge1$，令 $b_n$ 是满足

$$b_n^2\le 2\cdot10^{2n-2}$$

的最大自然数。这样的最大者存在：满足条件的自然数集合非空（$1$ 在里面）且有上界，取最大元即可。

再令

$$a_n=\frac{b_n}{10^{n-1}} .$$

当 $n=1$ 时条件为 $b_1^2\le2$，最大的自然数是 $1$，所以 $a_1=1$。$n=2$ 时 $b_2^2\le200$，$b_2=14$，$a_2=1.4$。依此类推，$a_n$ 就是 $\sqrt 2$ 保留 $n$ 位有效数字的截断。

递增性

要证 $a_n\le a_{n+1}$，等价于证明

$$10b_n\le b_{n+1}.$$

为什么？$a_n\le a_{n+1}$ 即 $\dfrac{b_n}{10^{n-1}}\le\dfrac{b_{n+1}}{10^n}$，两边乘以 $10^n$ 得 $10b_n\le b_{n+1}$。

现在验证 $10b_n$ 满足定义 $b_{n+1}$ 的那条不等式。因为 $b_n^2\le2\cdot10^{2n-2}$，两边乘 $100$：

$$(10b_n)^2=100b_n^2\le200\cdot10^{2n-2}=2\cdot10^{2n}=2\cdot10^{2(n+1)-2}.$$

所以 $10b_n$ 是满足 $n+1$ 步条件的自然数。而 $b_{n+1}$ 是这类自然数中最大的，故 $10b_n\le b_{n+1}$。

有界性

取 $2\cdot10^{n-1}$，它的平方是 $4\cdot10^{2n-2}>2\cdot10^{2n-2}$，所以它不满足条件，比 $b_n$ 大。因此

$$b_n<2\cdot10^{n-1} \implies a_n=\frac{b_n}{10^{n-1}}<2 .$$

又 $a_1=1>0$，所以 $(a_n)$ 递增且有上界 $2$。由单调收敛定理，$a_n\to a$，且 $a=\sup a_n\le2$。

极限平方为 2

先证 $a^2\le2$。由 $a_n^2\le2$ 与乘积规则，$a_n^2\to a^2$。极限保持非严格不等式（若 $x_n\to x$ 且 $x_n\le2$，则 $x\le2$；否则取 $\varepsilon=(x-2)/2>0$，充分靠后时 $x_n>x-\varepsilon>2$，矛盾），所以 $a^2\le2$。

再证 $a^2\ge2$。由 $b_n$ 的最大性，$b_n+1$ 不满足条件：

$$(b_n+1)^2>2\cdot10^{2n-2}.$$

两边除以 $10^{2n-2}=10^{n-1}\cdot10^{n-1}$：

$$\left(\frac{b_n+1}{10^{n-1}}\right)^2>2 .$$

而

$$\frac{b_n+1}{10^{n-1}}=\frac{b_n}{10^{n-1}}+\frac1{10^{n-1}}=a_n+\frac1{10^{n-1}} .$$

令 $s_n=a_n+10^{-(n-1)}$。由加法规则，$s_n\to a+0=a$；由乘积规则，$s_n^2\to a^2$。每个 $s_n^2>2$，极限保持非严格不等式，故 $a^2\ge2$。

两部分合起来，$a^2=2$。所以 $a=\sqrt 2$，而且它是递增序列 $a_n$ 的极限——这正是「$\sqrt 2$ 的十进制展开」的严格版本。

这里要注意：逐项的严格不等式 $s_n^2>2$ 只能推出极限的 $a^2\ge2$。极限可能取等号，事实上也确实取等号。

柯西数列与压缩映射（只作预告）

到目前为止，求极限都要先知道极限是什么。若极限写不出来，还能不能判断它存在？

定义。 数列 $(a_n)$ 是柯西数列，如果

$$\forall\varepsilon>0\ \exists N\ \forall n,m\ge N:\quad |a_n-a_m|<\varepsilon .$$

区别在于：收敛说「远处的项都聚在某个 $a$ 附近」，柯西说「远处的项彼此靠拢」，完全不提 $a$。

定理（第 6 讲证明）。 在 $\mathbb R$ 中，数列收敛当且仅当它是柯西数列。

这条定理使「证明极限存在」不再需要猜出极限值，后面证明微分方程解的存在性时会反复用到。

课程最后预告压缩映射定理：若 $T:\mathbb R\to\mathbb R$ 满足

$$|T(x)-T(y)|\le c|x-y|$$

对某个常数 $0<c<1$ 与一切 $x,y$ 成立，则 $T$ 有唯一不动点 $x=T(x)$。$c<1$ 是关键：$c=1$ 时结论不成立（例如平移映射 $T(x)=x+1$ 没有不动点）。这条定理在一般度量空间中同样成立，是证明常微分方程解存在唯一性的标准工具，第 24 讲的 Picard–Lindelöf 定理会用到它。

本节结论

• 单调 + 有界 ⇒ 收敛；递增时极限是上确界，递减时是下确界。
• 证明的关键：$a-\varepsilon$ 不是上界，所以某一项超过它，单调性让此后一直超过。
• 任意有界数列可造出 $b_n=\sup_{k\ge n}a_k$（递减）与 $c_n=\inf_{k\ge n}a_k$（递增），两者收敛且 $c\le b$——这就是 $\liminf$ 与 $\limsup$。
• $\sqrt 2$ 可由十进制截断 $a_n=b_n/10^{n-1}$ 定义为单调递增有界序列的极限，并验证极限平方为 $2$。
• 逐项严格不等式只能给出极限的非严格不等式。
• 柯西数列不使用极限值，$\mathbb R$ 中与收敛等价；压缩映射定理是它的重要应用。

理解检验

1. 单调收敛定理的证明中，「$a-\varepsilon$ 不是上界」这一步为什么不可省？它保证了什么？

参考答案：要证 $a-a_n<\varepsilon$，需要找到某一项 $a_N>a-\varepsilon$。若 $a-\varepsilon$ 是上界，则所有项都 $\le a-\varepsilon$，就找不到这样的 $N$。它之所以不是上界，是因为 $a$ 是最小上界：任何更小的数都不能压住整个集合。这一步把「最小性」转化为「存在某一项超过 $a-\varepsilon$」，再由单调性推广为「从 $N$ 之后一直超过」。

2. 设 $(a_n)$ 有界，$b_n=\sup_{k\ge n}a_k$，$c_n=\inf_{k\ge n}a_k$。证明 $b_n$ 递减、$c_n$ 递增，并说明为什么 $b_n\ge c_n$。

参考答案：$b_n$ 是对集合 $\{a_k:k\ge n\}$ 取上确界，而 $\{a_k:k\ge n+1\}\subseteq\{a_k:k\ge n\}$，集合更大时上确界不会更小，所以 $b_n\ge b_{n+1}$。同理 $c_n\le c_{n+1}$。对同一个集合，上确界不小于下确界，故 $b_n\ge c_n$ 对每个 $n$ 成立。

3. 在 $\sqrt 2$ 的构造中，为什么 $a_n\le a_{n+1}$ 等价于 $10b_n\le b_{n+1}$？请写出验证 $10b_n$ 满足条件的计算。

参考答案：$a_n=b_n/10^{n-1}$，$a_{n+1}=b_{n+1}/10^n$，两边乘 $10^n$ 即得等价关系。验证：由 $b_n^2\le2\cdot10^{2n-2}$，乘 $100$ 得 $(10b_n)^2\le2\cdot10^{2n}=2\cdot10^{2(n+1)-2}$，所以 $10b_n$ 是满足第 $n+1$ 步条件的自然数；而 $b_{n+1}$ 是其中最大的，故 $10b_n\le b_{n+1}$。

4. 为什么由 $s_n^2>2$（对每个 $n$）只能推出 $a^2\ge2$，而不能推出 $a^2>2$？举一个说明严格性会丢失的例子。

参考答案：极限只保留非严格不等式。若 $x_n\to x$ 且 $x_n>2$，只能得 $x\ge2$；例如 $x_n=2+1/n>2$，但极限是 $2$。在 $\sqrt2$ 的构造中，$s_n=a_n+10^{-(n-1)}$ 的平方确实对每个 $n$ 都大于 $2$，但 $s_n\to a$，而 $a^2$ 可能恰好等于 $2$——事实也正是如此。要得到严格不等式，需要额外的定量估计，而不是靠取极限。

5. 柯西数列的定义与收敛的定义有什么本质区别？为什么柯西条件在 $\mathbb Q$ 中不足以保证收敛？

参考答案：收敛定义提到极限值 $a$，要求远处的项都靠近 $a$；柯西定义只说远处的项彼此靠近，不提任何候选极限。在 $\mathbb R$ 中两者等价（完备性）。在 $\mathbb Q$ 中不等价：取 $\sqrt2$ 的有理截断序列 $1,1.4,1.41,\dots$，它是柯西序列（相邻项之差趋于 $0$，且尾部任意两项之差可任意小），但它在 $\mathbb Q$ 中没有极限，因为 $\sqrt2\notin\mathbb Q$。所以「柯西 ⇒ 收敛」正是完备性的内容。

与前后课的关系

本讲给出了第一个充分条件，也造出了 $\limsup$ 与 $\liminf$ 的雏形。下一讲证明柯西准则：在 $\mathbb R$ 中柯西与收敛等价。证明会用到本讲的「任意有界数列造两个单调数列」这一构造——先证柯西数列有界，再用 $b_n,c_n$ 夹住它。

再往后，第 7 讲的 Bolzano–Weierstrass 定理（有界数列必有收敛子列）也依赖本讲的单调收敛定理：先把子列的「峰」找出来，或者用二分法造出单调子列。

A usable criterion

The previous lecture showed that convergence implies boundedness, but that boundedness alone does not suffice: $(-1)^n$ is bounded and divergent. This lecture supplies a genuine sufficient condition — if the sequence is monotone, boundedness is enough.

The result matters because it turns "find the limit" into "check two properties": monotone and bounded. Many sequences have no closed-form limit, yet one can prove that a limit exists.

Timeline

Computing a limit with the limit laws

Find the limit of

$$a_n=\frac{n^2+1}{n^2+n+1}.$$

Divide numerator and denominator by $n^2$:

$$a_n=\frac{1+\dfrac1{n^2}}{1+\dfrac1n+\dfrac1{n^2}} .$$

We know $1/n\to0$, $1/n^2\to0$, and the constant sequence $1\to1$. By the sum rule the numerator tends to $1$ and the denominator to $1$. The denominator tends to $1\ne0$ and all its terms are positive, so the reciprocal rule applies; then the product rule gives

$$a_n\to\frac{1}{1}=1 .$$

Every step names the rule it uses. This is why the algebra of limits is not decoration: it decomposes a complicated expression into combinations of known limits.

Monotone sequences

Definition. A sequence $(a_n)$ is monotone increasing if

$$a_n\le a_{n+1}\quad\text{for all }n .$$

If the inequality is strict, it is strictly increasing. Similarly, monotone decreasing means $a_n\ge a_{n+1}$ for all $n$.

Monotonicity alone does not require boundedness. $a_n=n$ is increasing but unbounded, hence divergent; $a_n=-n$ is decreasing and unbounded.

The monotone convergence theorem

Theorem. If $(a_n)$ is monotone increasing and bounded above, then it converges and

$$\lim_{n\to\infty}a_n=\sup_n a_n .$$

Symmetrically, a monotone decreasing sequence bounded below converges to $\inf_n a_n$.

An increasing sequence is automatically bounded below by $a_1$, so only the upper bound needs checking; dually for decreasing sequences.

Proof (increasing case)

Let $a_n$ be increasing and bounded above. By completeness, $a=\sup\{a_n:n\in\mathbb N\}$ exists. We prove $a_n\to a$.

Since $a$ is an upper bound, $a_n\le a$ for all $n$, so

$$|a_n-a|=a-a_n .$$

It remains to show: given $\varepsilon>0$, eventually $a-a_n<\varepsilon$.

The key step: $a-\varepsilon$ is not an upper bound. If it were, there would be an upper bound smaller than the least one, a contradiction. Hence there is $N$ with

$$a_N>a-\varepsilon .$$

(The inequality is strict: if all $a_n\le a-\varepsilon$, then $a-\varepsilon$ would be an upper bound.)

By monotonicity, for $n\ge N$ we have $a_n\ge a_N>a-\varepsilon$. Together with $a_n\le a$,

$$0\le a-a_n<a-(a-\varepsilon)=\varepsilon \quad (n\ge N).$$

So $a_n\to a$. $\square$

The heart of the proof is one sentence: the supremum is exactly the number that just barely dominates the set, so anything smaller ($a-\varepsilon$) fails to dominate it and must be exceeded by some term. Monotonicity then turns that single crossing into an eventual one.

For the decreasing case apply the result to $-a_n$, which is increasing with supremum $-\inf a_n$.

Every bounded sequence yields monotone sequences

The theorem seems to apply only to special sequences. But any bounded sequence $(a_n)$ produces two monotone ones:

$$b_n=\sup\{a_k:k\ge n\}, \qquad c_n=\inf\{a_k:k\ge n\}.$$

(Boundedness guarantees these suprema and infima exist.)

$b_n$ decreases. $b_{n+1}$ is the supremum over the smaller set $\{a_k:k\ge n+1\}$, while $b_n$ ranges over that set plus $a_n$. A larger set has a supremum no smaller:

$$b_n\ge b_{n+1}.$$

$c_n$ increases. Dually, shrinking the set cannot raise the infimum:

$$c_n\le c_{n+1}.$$

$b_n\ge c_n$. Both are taken over the same set, and a supremum is at least the infimum.

So $b_n$ decreases and is bounded below by $c_n$, while $c_n$ increases and is bounded above by $b_n$. The monotone convergence theorem gives

$$b_n\downarrow b, \qquad c_n\uparrow c, \qquad c\le b .$$

The last step uses preservation of non-strict inequalities in the limit: if $x_n\le y_n$ for all $n$ and both converge, then $\lim x_n\le\lim y_n$.

These are precisely the limit superior and limit inferior: $b=\limsup a_n$, $c=\liminf a_n$. The lecture builds them here; Lecture 6 uses them to prove the Cauchy criterion.

Rethinking $\sqrt2$ with monotone convergence

Lecture 2 constructed $\sqrt2$ as a supremum. Here is another route: build an increasing sequence of decimal truncations, get a limit from monotone convergence, and verify that its square is $2$.

The construction

For each $n\ge1$, let $b_n$ be the largest natural number with

$$b_n^2\le 2\cdot10^{2n-2}.$$

Such a largest element exists: the admissible naturals are nonempty (they contain $1$) and bounded.

Set

$$a_n=\frac{b_n}{10^{n-1}} .$$

For $n=1$ the condition is $b_1^2\le2$, so $b_1=1$ and $a_1=1$. For $n=2$, $b_2^2\le200$, so $b_2=14$ and $a_2=1.4$. In general $a_n$ is $\sqrt2$ truncated to $n$ significant digits.

Monotonicity

We claim $a_n\le a_{n+1}$, which is equivalent to

$$10b_n\le b_{n+1}.$$

Indeed $a_n\le a_{n+1}$ reads $\dfrac{b_n}{10^{n-1}}\le\dfrac{b_{n+1}}{10^n}$, and multiplying by $10^n$ gives the claim.

Now check that $10b_n$ satisfies the defining inequality for step $n+1$. From $b_n^2\le2\cdot10^{2n-2}$, multiply by $100$:

$$(10b_n)^2=100b_n^2\le200\cdot10^{2n-2}=2\cdot10^{2n}=2\cdot10^{2(n+1)-2}.$$

So $10b_n$ is an admissible natural number at step $n+1$. Since $b_{n+1}$ is the largest admissible one, $10b_n\le b_{n+1}$.

Boundedness

Take $2\cdot10^{n-1}$; its square is $4\cdot10^{2n-2}>2\cdot10^{2n-2}$, so it is not admissible and exceeds $b_n$. Hence

$$b_n<2\cdot10^{n-1} \implies a_n=\frac{b_n}{10^{n-1}}<2 .$$

Since $a_1=1>0$, the sequence is increasing and bounded above by $2$. By monotone convergence, $a_n\to a$ with $a=\sup a_n\le2$.

The limit squares to 2

First, $a^2\le2$. Since $a_n^2\le2$ and $a_n^2\to a^2$ by the product rule, and limits preserve non-strict inequalities (if $x_n\to x$ and $x_n\le2$ then $x\le2$; otherwise with $\varepsilon=(x-2)/2>0$ we would eventually have $x_n>x-\varepsilon>2$), we get $a^2\le2$.

Second, $a^2\ge2$. By maximality of $b_n$, the number $b_n+1$ is not admissible:

$$(b_n+1)^2>2\cdot10^{2n-2}.$$

Divide by $10^{2n-2}=10^{n-1}\cdot10^{n-1}$:

$$\left(\frac{b_n+1}{10^{n-1}}\right)^2>2 .$$

But

$$\frac{b_n+1}{10^{n-1}}=\frac{b_n}{10^{n-1}}+\frac1{10^{n-1}}=a_n+\frac1{10^{n-1}} .$$

Let $s_n=a_n+10^{-(n-1)}$. The sum rule gives $s_n\to a+0=a$, and the product rule gives $s_n^2\to a^2$. Each $s_n^2>2$, and limits preserve non-strict inequalities, so $a^2\ge2$.

Combining, $a^2=2$. Thus $a=\sqrt2$, obtained as the limit of the increasing truncations — the rigorous version of "the decimal expansion of $\sqrt2$".

Note carefully: termwise strict inequalities $s_n^2>2$ only yield $a^2\ge2$. The limit may take equality, and here it does.

Cauchy sequences and contraction (preview)

So far every limit required knowing the limit in advance. Can existence be decided without it?

Definition. A sequence $(a_n)$ is Cauchy if

$$\forall\varepsilon>0\ \exists N\ \forall n,m\ge N:\quad |a_n-a_m|<\varepsilon .$$

The difference: convergence says distant terms cluster around some $a$; Cauchy says distant terms cluster around each other, with no mention of $a$.

Theorem (proved in Lecture 6). In $\mathbb R$, a sequence converges if and only if it is Cauchy.

This removes the need to guess the limit in order to prove that one exists — repeatedly useful when proving existence of solutions to differential equations.

The lecture closes with the contraction mapping theorem: if $T:\mathbb R\to\mathbb R$ satisfies

$$|T(x)-T(y)|\le c|x-y|$$

for some constant $0<c<1$ and all $x,y$, then $T$ has a unique fixed point $x=T(x)$. The condition $c<1$ is essential: with $c=1$ the conclusion fails (the translation $T(x)=x+1$ has no fixed point). The theorem holds in general metric spaces and is the standard tool for existence and uniqueness of ODE solutions; Lecture 24's Picard–Lindelöf theorem uses it.

What this lecture establishes

• Monotone plus bounded implies convergence; the limit is the supremum (increasing) or infimum (decreasing).
• The proof turns "the supremum is least" into "some term exceeds $a-\varepsilon$", then uses monotonicity.
• Any bounded sequence yields $b_n=\sup_{k\ge n}a_k$ (decreasing) and $c_n=\inf_{k\ge n}a_k$ (increasing), converging with $c\le b$ — the liminf and limsup.
• $\sqrt2$ is the limit of the increasing bounded truncations $a_n=b_n/10^{n-1}$, and its square is $2$.
• Termwise strict inequalities give only non-strict inequalities in the limit.
• Cauchy sequences avoid the limit value and are equivalent to convergence in $\mathbb R$; the contraction mapping theorem is a major application.

Comprehension check

1. In the proof of monotone convergence, why is the step "$a-\varepsilon$ is not an upper bound" indispensable? What does it guarantee?

Reference answer: To get $a-a_n<\varepsilon$ we need some term $a_N>a-\varepsilon$. If $a-\varepsilon$ were an upper bound, every term would satisfy $a_n\le a-\varepsilon$ and no such $N$ would exist. It fails to be an upper bound precisely because $a$ is the least upper bound: nothing smaller can dominate the set. This converts minimality into "some term exceeds $a-\varepsilon$", and monotonicity extends it to "every later term exceeds it".

2. For a bounded sequence, let $b_n=\sup_{k\ge n}a_k$ and $c_n=\inf_{k\ge n}a_k$. Prove $b_n$ decreases and $c_n$ increases, and explain why $b_n\ge c_n$.

Reference answer: $b_n$ is the supremum over $\{a_k:k\ge n\}$, while $\{a_k:k\ge n+1\}\subseteq\{a_k:k\ge n\}$; a larger set has a supremum no smaller, so $b_n\ge b_{n+1}$. Dually $c_n\le c_{n+1}$. Over the same set, the supremum is at least the infimum, so $b_n\ge c_n$ for each $n$.

3. In the construction of $\sqrt2$, why is $a_n\le a_{n+1}$ equivalent to $10b_n\le b_{n+1}$? Write out the verification that $10b_n$ is admissible.

Reference answer: With $a_n=b_n/10^{n-1}$ and $a_{n+1}=b_{n+1}/10^n$, multiplying by $10^n$ gives the equivalence. Verification: from $b_n^2\le2\cdot10^{2n-2}$, multiplying by $100$ gives $(10b_n)^2\le2\cdot10^{2n}=2\cdot10^{2(n+1)-2}$, so $10b_n$ is an admissible natural number at step $n+1$; since $b_{n+1}$ is the largest admissible one, $10b_n\le b_{n+1}$.

4. Why do the strict inequalities $s_n^2>2$ yield only $a^2\ge2$ rather than $a^2>2$? Give an example where strictness is lost.

Reference answer: Limits preserve only non-strict inequalities. If $x_n\to x$ and $x_n>2$, one can only conclude $x\ge2$; for instance $x_n=2+1/n>2$ but the limit is $2$. In the construction, $s_n=a_n+10^{-(n-1)}$ does satisfy $s_n^2>2$ for every $n$, yet $s_n\to a$ and $a^2$ may equal $2$ — and it does. A strict conclusion would require an additional quantitative estimate, not merely a passage to the limit.

5. What is the essential difference between Cauchy and convergent? Why is the Cauchy condition insufficient in $\mathbb Q$?

Reference answer: Convergence names a limit $a$ and requires distant terms to approach it; Cauchy requires only that distant terms approach each other, with no candidate limit. In $\mathbb R$ the two are equivalent (completeness). In $\mathbb Q$ they are not: the rational truncations $1,1.4,1.41,\dots$ of $\sqrt2$ form a Cauchy sequence, but they have no limit in $\mathbb Q$ because $\sqrt2\notin\mathbb Q$. So "Cauchy implies convergent" is exactly the content of completeness.

Relation to neighbouring lectures

This lecture gives the first sufficient condition and sketches limsup/liminf. The next lecture proves the Cauchy criterion — equivalence of Cauchy and convergent in $\mathbb R$ — using the construction of two monotone sequences to squeeze a Cauchy sequence.

Later, Lecture 7's Bolzano–Weierstrass theorem (every bounded sequence has a convergent subsequence) also rests on monotone convergence: either locate peaks or bisect to build a monotone subsequence.