一个待补的漏洞

上一讲的根值判别法要求 $\sqrt[n]{|a_n|}$ 的极限存在。如果它不存在——比如在几个值之间来回摆动——判别法就直接失效。这一讲用上极限（limsup）替代极限，把判别法修好，并立刻用在幂级数上。

同一讲的另一半转向函数：先用幂级数定义指数函数，再引入连续性的 $\varepsilon$-$\delta$ 定义。两半最终会合流——连续性是把「有理数上的指数」推广到「实数上的指数」所需要的唯一性质。

limsup 与 liminf

设 $(a_n)$ 有上界。定义

$$b_n=\sup\{a_k : k\ge n\}=\sup\{a_n,a_{n+1},a_{n+2},\dots\}.$$

由于 $\{a_k:k\ge n+1\}\subseteq\{a_k:k\ge n\}$，而集合变小上确界不会变大，

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

所以 $(b_n)$ 单调递减。若它有下界，由单调收敛定理收敛，定义

$$\limsup_{n\to\infty}a_n=\inf_{n}b_n .$$

若 $(a_n)$ 无上界，则约定 $\limsup a_n=+\infty$。

对称地，若 $(a_n)$ 有下界，令

$$c_n=\inf\{a_k:k\ge n\},$$

则 $(c_n)$ 单调递增，定义

$$\liminf_{n\to\infty}a_n=\sup_{n}c_n .$$

一个重要事实：limsup 总是存在（在扩充实数系中），不管原数列有没有极限。当极限存在时三者相等：

$$\liminf a_n=\lim a_n=\limsup a_n .$$

直观上，$\limsup$ 是「最终的上界中最小者」：数列可以反复超过某个值，但只要最终会落到该值之下，这个值就不是 limsup。

时间轴

根值判别法 2.0

定理。 对任意级数 $\sum d_n$，令

$$d=\limsup_{n\to\infty}\sqrt[n]{|d_n|}.$$

则：

1. $d<1$ 时级数绝对收敛；
2. $d>1$ 时级数发散；
3. $d=1$ 时判别法失效。

与上一讲的区别只在前提：不再要求 $\sqrt[n]{|d_n|}$ 收敛，只要求它的上极限存在——而它总是存在的。

证明（$d<1$）。 取 $\bar d$ 使 $d<\bar d<1$。由 limsup 的定义，存在 $N$，当 $n\ge N$ 时

$$\sup_{k\ge n}\sqrt[k]{|d_k|}\le\bar d,$$

于是对每个 $k\ge N$ 都有 $\sqrt[k]{|d_k|}\le\bar d$，即

$$|d_k|\le\bar d^{\,k}.$$

右边是 $\bar d<1$ 的等比级数，由比较判别法，$\sum|d_k|$ 收敛。

这里的关键是：limsup 的定义本身给出了「从某项起，所有项都被 $\bar d$ 控制」，不需要极限存在。这正是它比极限好用的地方。

证明（$d>1$）。 取 $\bar d$ 使 $1<\bar d<d$。类似地，无限多项满足 $\sqrt[n]{|d_n|}>\bar d$，即 $|d_n|>\bar d^{\,n}>1$。于是通项不趋于零（甚至有无限多项的绝对值大于 $1$），级数发散。

幂级数的收敛半径

对幂级数 $\sum a_n x^n$，固定 $x$，令 $d_n=a_n x^n$。则

$$\sqrt[n]{|d_n|}=\sqrt[n]{|a_n|}\,|x|,$$

于是

$$\limsup_{n\to\infty}\sqrt[n]{|d_n|}=|x|\cdot\limsup_{n\to\infty}\sqrt[n]{|a_n|}=\frac{|x|}{R},$$

其中定义收敛半径

$$R=\frac{1}{\limsup_{n\to\infty}\sqrt[n]{|a_n|}} .$$

按约定：若 $\limsup|a_n|^{1/n}=0$，则 $R=+\infty$；若 $\limsup|a_n|^{1/n}=+\infty$，则 $R=0$。

把根值判别法 2.0 用于 $\sum a_nx^n$：

• $|x|/R<1$，即 $|x|<R$：绝对收敛；
• $|x|/R>1$，即 $|x|>R$：发散；
• $|x|=R$：判别法失效，两个端点必须单独检查。

结论（收敛区间定理）。 幂级数在 $(-R,R)$ 内收敛，在 $(-R,R)$ 外发散，两个端点 $x=\pm R$ 需要逐个判断。

这条定理排除了一个先验上完全可能、实际上不会发生的情形：级数在 $x=5$ 收敛、在 $x=10$ 发散、又在 $x=20$ 收敛。收敛点集必定是「以 $0$ 为中心、左右对称的一段」，不可能是一堆离散的点或交错的情形。

例子

• 等比级数 $\sum x^n$：$a_n=1$，$\limsup|a_n|^{1/n}=1$，故 $R=1$。端点 $x=\pm1$ 处都不收敛。
• 指数级数 $\sum x^n/n!$：上一讲用比值判别法得到对一切 $x$ 收敛，这也与 $R=\infty$ 一致（$n!$ 增长极快，$\sqrt[n]{1/n!}\to0$）。

指数函数：用幂级数定义

指数函数有两种常见定义：一是满足微分方程 $y'=y$ 且 $y(0)=1$ 的函数；二是幂级数。这一讲采用后者。

定义。

$$E(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$

先确认它确实是一个函数：上一讲已证该级数对所有 $x$ 绝对收敛，所以收敛半径为 $\infty$，定义域是整个 $\mathbb R$。

几条基本性质

$E(0)=1$。 代入 $x=0$，除 $n=0$ 外每项都是 $0$，而 $0^0$ 按定义为 $1$、$0!=1$。

$x\ge0$ 时 $E(x)>0$。 此时每项 $\dfrac{x^n}{n!}\ge0$，而第一项是 $1$，所以和至少是 $1$。

$x<0$ 时 $E(x)>0$（暂借下一讲的结果）。 下一讲将证明加法公式

$$E(x+y)=E(x)E(y).$$

若先承认这一点，取 $x$ 为负、记 $-x>0$，则

$$E(x)E(-x)=E(0)=1 .$$

右边为正，$E(-x)$ 为正，所以 $E(x)=\dfrac{1}{E(-x)}>0$。于是对所有实数 $x$，$E(x)>0$。

这节课在这里只是「先借用」，加法公式本身要等到下一讲才证。这种写法本身值得注意：先列出需要什么，再明确标注尚未证明的部分。

定义 $e$ 与整数幂

定义。 $e=E(1)$。

现在检查 $E$ 与通常的「$e$ 的幂」是否一致。先看整数。由加法公式，反复使用：

$$E(1)E(1)=E(2),\quad E(1)E(1)E(1)=E(3),\quad\dots$$

所以对正整数 $m$，

$$e^m=\underbrace{e\cdot e\cdots e}_{m\text{ 个}}=E(m).$$

对 $m=0$，$e^0=1=E(0)$。对负整数 $m$，由

$$E(m)E(-m)=E(0)=1$$

得

$$E(m)=\frac{1}{E(-m)}=\frac{1}{e^{-m}}=e^{m}.$$

于是对一切整数 $m$，$e^m=E(m)$。

有理指数

设 $q=\dfrac{m}{n}$，其中 $m\in\mathbb Z$、$n\in\mathbb N$。希望

$$e^{q}=E\!\left(\frac{m}{n}\right).$$

定义 $e^q$ 为满足

$$\alpha>0,\qquad \alpha^n=e^{m}$$

的那个唯一的 $\alpha$。唯一性来自正数的幂严格单调：若 $\alpha_1^n=\alpha_2^n$ 且都为正，则 $\alpha_1=\alpha_2$。

接着验证这个 $\alpha$ 确实是 $E(m/n)$。一方面 $E(m/n)>0$（前面已证）。另一方面，由加法公式重复 $n$ 次，

$$\left(E\!\left(\frac mn\right)\right)^{n} =E\!\left(n\cdot\frac mn\right) =E(m)=e^{m}.$$

所以 $E(m/n)$ 满足定义 $e^q$ 的那两条性质，由唯一性，它就是这个 $\alpha$。于是

$$e^{q}=E(q)\qquad(q\in\mathbb Q).$$

补充：上面用到了「正数的 $n$ 次根存在」，这一步由实数的完备性保证（可用最小上界构造），课程把它留到后面；此处只用到唯一性来定义，因此论证本身是自洽的。

连续性

现在要把 $E$ 从 $\mathbb Q$ 推广到 $\mathbb R$。所需的正是连续性。

定义

设 $f$ 定义在包含 $x_0$ 的集合上。若对每个 $\varepsilon>0$，存在 $\delta>0$，使得只要 $|x-x_0|<\delta$（且 $x$ 在定义域内），就有

$$|f(x)-f(x_0)|<\varepsilon,$$

则称 $f$ 在 $x_0$ 连续。若 $f$ 在定义域内每一点都连续，就称 $f$ 连续。

两个要点：

• $\delta$ 一般同时依赖于 $\varepsilon$ 与 $x_0$；
• 定义是「给任意精度，找对应的邻域」，与数列极限的量词结构完全同型。

两个基本例子

常数函数 $f(x)=c$。 给定 $\varepsilon>0$，取 $\delta=1$（或任何正数）。当 $|x-x_0|<1$ 时，

$$|f(x)-f(x_0)|=|c-c|=0<\varepsilon .$$

所以常数函数处处连续。这里 $\delta$ 既不依赖 $\varepsilon$ 也不依赖 $x_0$。

恒等函数 $f(x)=x$。 给定 $\varepsilon>0$，取 $\delta=\varepsilon$。当 $|x-x_0|<\varepsilon$ 时，

$$|f(x)-f(x_0)|=|x-x_0|<\varepsilon .$$

所以 $f(x)=x$ 处处连续。

五条代数规则

设 $f,g$ 连续，则：

1. $f+g$ 连续；
2. $cf$ 连续（$c$ 为常数）；
3. $fg$ 连续；
4. 若 $f$ 处处不为 $0$，则 $1/f$ 连续；
5. 若复合有意义，则 $g\circ f$ 连续。

这些规则与数列极限的运算法则完全平行，证明思路也相同：把误差拆成两份，各分配 $\varepsilon/2$。

乘积的证明（示范第三条）。 目标估计

$$f(x)g(x)-f(x_0)g(x_0).$$

在中间插入一项，制造两个可控的差：

$$f(x)g(x)-f(x_0)g(x_0) =g(x)\bigl(f(x)-f(x_0)\bigr)+f(x_0)\bigl(g(x)-g(x_0)\bigr).$$

于是

$$|f(x)g(x)-f(x_0)g(x_0)| \le |g(x)|\,|f(x)-f(x_0)|+|f(x_0)|\,|g(x)-g(x_0)| .$$

现在逐块控制：

• $g$ 在 $x_0$ 连续，取 $\varepsilon=1$ 得某个 $\delta_1$，使 $|x-x_0|<\delta_1$ 时 $|g(x)-g(x_0)|<1$，从而

$$|g(x)|<|g(x_0)|+1=:M .$$

• $f$ 在 $x_0$ 连续：存在 $\delta_2$，使 $|x-x_0|<\delta_2$ 时 $|f(x)-f(x_0)|<\dfrac{\varepsilon}{2M}$。
• $g$ 在 $x_0$ 连续：存在 $\delta_3$，使 $|x-x_0|<\delta_3$ 时 $|g(x)-g(x_0)|<\dfrac{\varepsilon}{2(|f(x_0)|+1)}$。

取 $\delta=\min\{\delta_1,\delta_2,\delta_3\}$。当 $|x-x_0|<\delta$ 时，

$$|f(x)g(x)-f(x_0)g(x_0)| \le M\cdot\frac{\varepsilon}{2M}+|f(x_0)|\cdot\frac{\varepsilon}{2(|f(x_0)|+1)} <\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon .$$

所以 $fg$ 在 $x_0$ 连续。$\square$

两个技术细节值得记住：先证明其中一个因子在 $x_0$ 附近有界，再分配误差；分母加 $1$ 是为了避免 $f(x_0)=0$ 时除以零。

多项式与有理函数

由两个基本例子出发，反复使用规则 1–3：

• $x$ 连续 ⇒ $x^2=x\cdot x$ 连续 ⇒ $x^n$ 连续；
• 常数 $c$ 连续 ⇒ $cx^n$ 连续；
• 有限个连续函数相加仍连续。

定理。 一切多项式都连续。

再加上规则 4：

定理。 一切有理函数（两个多项式之商）在分母不为零的点连续。

注意分母的零点只有有限多个，函数在那些点上根本没有定义，因此谈不上连续与否。有时可以通过补充定义把它延拓成连续函数，但那已经是另一个函数。

$\delta$ 对 $x_0$ 的依赖：$1/x$ 的例子

考虑 $f(x)=1/x$，定义域为 $x\ne0$（比如只看向上的那一支）。任取 $x_0>0$，要证它在 $x_0$ 连续。

直接算：

$$\left|\frac1x-\frac1{x_0}\right|=\frac{|x-x_0|}{|x|\,|x_0|}.$$

若先限制 $|x-x_0|<\dfrac{x_0}{2}$，则 $x>\dfrac{x_0}{2}$，于是

$$\left|\frac1x-\frac1{x_0}\right|<\frac{2|x-x_0|}{x_0^2}.$$

要它小于 $\varepsilon$，取 $|x-x_0|<\dfrac{\varepsilon x_0^2}{2}$ 即可。因此

$$\delta=\min\left\{\frac{x_0}{2},\ \frac{\varepsilon x_0^2}{2}\right\}$$

满足要求。

关键观察：$\delta$ 里出现了 $x_0^2$。当 $x_0$ 靠近 $0$ 时，$\delta$ 必须取得非常小——因为函数在 $0$ 附近变化极其剧烈。这正是「$\delta$ 一般依赖于 $x_0$」的具体含义。

如果某个连续函数的 $\delta$ 能取成只依赖 $\varepsilon$、与 $x_0$ 无关，这个性质有一个专门的名字：一致连续。它是后面课程的主题；课程在这里埋下伏笔，并指出 $1/x$ 的定义域不包含 $0$ 并非偶然——若定义在包含端点的闭区间上，连续函数会自动一致连续。

本节结论

• limsup 用「尾部上确界的下确界」定义，总存在；liminf 对称定义。
• 根值判别法 2.0 用 limsup 替代极限，适用于任何级数。
• 幂级数收敛半径 $R=1/\limsup|a_n|^{1/n}$；$|x|<R$ 绝对收敛，$|x|>R$ 发散，端点另判。
• 收敛区间必为以 $0$ 为中心的区间，不会出现离散的收敛点。
• $E(x)=\sum x^n/n!$ 在 $\mathbb R$ 上有定义且恒正（正性对负 $x$ 借用下一讲的加法公式）。
• $e=E(1)$；对整数与有理数 $q$，$e^q=E(q)$。
• 连续性用 $\varepsilon$-$\delta$ 定义；$\delta$ 依赖 $\varepsilon$ 与 $x_0$。
• 常数函数与 $x$ 连续；和、数乘、积、倒数、复合保持连续；多项式处处连续，有理函数在分母非零处连续。
• $1/x$ 的 $\delta$ 含 $x_0^2$，说明靠近 $0$ 时要求的邻域急剧变小。

理解检验

1. 证明 $b_n=\sup_{k\ge n}a_k$ 单调递减，并说明为什么 $\limsup$ 总存在。

参考答案：因为 $\{a_k:k\ge n+1\}\subseteq\{a_k:k\ge n\}$，在更大集合上取上确界不会更小，所以 $b_{n+1}\le b_n$。$(b_n)$ 单调递减；若它有下界，由单调收敛定理收敛，其极限即 $\limsup$；若无下界，则约定 $\limsup=-\infty$（相应地，若原数列无上界则为 $+\infty$）。无论哪种情况，在扩充实数系中 limsup 都有定义，不需要原数列收敛。

2. 若 $\limsup a_n=2$，能否断定「存在 $N$，当 $n\ge N$ 时 $a_n<2$」？

参考答案：不能。limsup 是尾部上确界的极限，只保证：对任意 $\eta>0$，存在 $N$，当 $n\ge N$ 时 $a_n<2+\eta$；也就是「最终会低于 $2+\eta$」，但可以无限次地接近甚至超过 $2$。例如 $a_n=2+1/n$ 的 limsup 是 $2$，却每项都大于 $2$。能保证的是对任何 $\eta>0$ 最终成立的不等式。

3. 为什么幂级数的收敛点集必定是一个以 $0$ 为中心的区间，而不可能像「$x=5$ 收敛、$x=10$ 发散、$x=20$ 又收敛」？

参考答案：根值判别法给出判据 $|x|\cdot\limsup|a_n|^{1/n}<1$ 或 $>1$，即 $|x|<R$ 或 $|x|>R$，其中 $R$ 是一个只与系数有关、与 $x$ 无关的数。因此收敛与否只取决于 $|x|$ 与同一个阈值 $R$ 的大小关系。若 $x=20$ 收敛，则 $|x|=20<R$，于是所有 $|y|<20$ 的 $y$ 也满足 $|y|<R$，必收敛——与 $x=10$ 发散矛盾。所以不可能出现这种交错格局。

4. 用幂级数定义 $E(x)$，说明为什么 $E(x)>0$ 对负的 $x$ 需要额外一步，而那一步依赖什么。

参考答案：$x\ge0$ 时每项非负且首项为 $1$，和显然为正。$x<0$ 时各项符号交替，不能逐项判断。需要借助加法公式 $E(x)E(-x)=E(0)=1$：由于 $-x>0$ 使 $E(-x)>0$，而乘积为 $1>0$，故 $E(x)=1/E(-x)>0$。这一步依赖加法公式，而加法公式要到下一讲才证明——课上明确标注了这一点，属于「先借用、后补证」。

5. 验证 $e^{m/n}=E(m/n)$ 时，用到了 $E$ 的哪些性质？

参考答案：用到三条：(a) $E(x)>0$，保证可以谈它的正 $n$ 次根；(b) 加法公式的迭代形式 $E(x)^n=E(nx)$，由此 $(E(m/n))^n=E(m)=e^m$；(c) 正数 $n$ 次根的唯一性，即满足 $\alpha>0$、$\alpha^n=e^m$ 的 $\alpha$ 至多一个。三条合起来说明 $E(m/n)$ 正是按定义指定的那个 $e^{m/n}$。

6. 证明常数函数连续，并说明这里的 $\delta$ 与一般情形有何不同。

参考答案：设 $f(x)=c$。给定 $\varepsilon>0$，任取 $\delta>0$（例如 $\delta=1$）。当 $|x-x_0|<\delta$ 时，$|f(x)-f(x_0)|=|c-c|=0<\varepsilon$，故处处连续。不同之处在于：这里的 $\delta$ 既不依赖 $\varepsilon$ 也不依赖 $x_0$；一般情形下 $\delta$ 通常同时依赖两者，如 $1/x$ 的例子中 $\delta$ 含 $x_0^2$。

7. 证明乘积 $fg$ 连续时，为什么要先说明 $g$ 在 $x_0$ 附近有界？

参考答案：误差估计

$$|fg-f(x_0)g(x_0)|\le|g(x)|\,|f(x)-f(x_0)|+|f(x_0)|\,|g(x)-g(x_0)|$$

中第一项含因子 $|g(x)|$，它随 $x$ 变化，不能当作常数。由 $g$ 在 $x_0$ 连续，取 $\varepsilon=1$ 可知在某个邻域内 $|g(x)|<|g(x_0)|+1=:M$。有了这个上界，才能把第一项压到 $M\cdot\frac{\varepsilon}{2M}=\frac{\varepsilon}{2}$。若不做这一步，就无法为 $|f(x)-f(x_0)|$ 指定一个固定的误差预算。

与前后课的关系

这一讲把两条线接到了一起。级数一线：limsup 补全了根值判别法，幂级数因此有了收敛半径这一干净的结构。函数一线：指数函数先用幂级数定义，再借助连续性从 $\mathbb Q$ 延拓到 $\mathbb R$。

下一讲会做三件事：证明加法公式 $E(x+y)=E(x)E(y)$（本讲两处借用过它）；证明「两个连续函数若在 $\mathbb Q$ 上相等则处处相等」，从而完成指数函数的定义；并开始系统研究连续函数的性质。再往后是极值定理、介值定理与度量空间——那时第一讲提出的介值定理终于会被真正证明。

A gap to close

The root test of the previous lecture required that $\sqrt[n]{|a_n|}$ have a limit. If it does not — for instance if it oscillates between several values — the test simply fails. This lecture replaces the limit by the limit superior, repairs the test, and applies it immediately to power series.

The second half turns to functions: the exponential is defined by a power series, and continuity is introduced through an $\varepsilon$-$\delta$ definition. The two halves eventually meet — continuity is exactly what is needed to extend "the exponential on the rationals" to "the exponential on the reals".

Limsup and liminf

Let $(a_n)$ be bounded above and set

$$b_n=\sup\{a_k : k\ge n\}=\sup\{a_n,a_{n+1},a_{n+2},\dots\}.$$

Since $\{a_k:k\ge n+1\}\subseteq\{a_k:k\ge n\}$ and a supremum over a smaller set cannot be larger,

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

So $(b_n)$ decreases. If it is bounded below, monotone convergence gives a limit, and one defines

$$\limsup_{n\to\infty}a_n=\inf_{n}b_n .$$

If $(a_n)$ is unbounded above, set $\limsup a_n=+\infty$.

Symmetrically, if $(a_n)$ is bounded below, let

$$c_n=\inf\{a_k:k\ge n\},$$

which increases, and define

$$\liminf_{n\to\infty}a_n=\sup_{n}c_n .$$

A key fact: the limsup always exists (in the extended real line), whether or not the sequence converges. When the limit exists,

$$\liminf a_n=\lim a_n=\limsup a_n .$$

Intuitively, the limsup is the smallest eventual upper bound: a sequence may exceed a value repeatedly, but if it eventually stays below that value, the value is not the limsup.

Timeline

Root test 2.0

Theorem. For any series $\sum d_n$, let

$$d=\limsup_{n\to\infty}\sqrt[n]{|d_n|}.$$

Then:

1. if $d<1$ the series converges absolutely;
2. if $d>1$ the series diverges;
3. if $d=1$ the test is inconclusive.

The only change from the previous lecture is the hypothesis: the $n$-th roots need not converge — only their limsup matters, and that always exists.

Proof ($d<1$). Choose $\bar d$ with $d<\bar d<1$. By the definition of limsup there is $N$ such that for $n\ge N$,

$$\sup_{k\ge n}\sqrt[k]{|d_k|}\le\bar d,$$

hence $\sqrt[k]{|d_k|}\le\bar d$ for every $k\ge N$, i.e.

$$|d_k|\le\bar d^{\,k}.$$

The right side is a geometric series with ratio $\bar d<1$, so by comparison $\sum|d_k|$ converges.

The point is that the definition of limsup itself provides "from some index on, every term is controlled by $\bar d$", with no limit required. That is exactly what makes it more useful than the limit.

Proof ($d>1$). Choose $\bar d$ with $1<\bar d<d$. Then infinitely many $n$ satisfy $\sqrt[n]{|d_n|}>\bar d$, i.e. $|d_n|>\bar d^{\,n}>1$. So the terms do not tend to zero (infinitely many have absolute value exceeding $1$), and the series diverges.

Radius of convergence of a power series

For $\sum a_n x^n$, fix $x$ and set $d_n=a_n x^n$. Then

$$\sqrt[n]{|d_n|}=\sqrt[n]{|a_n|}\,|x|,$$

so

$$\limsup_{n\to\infty}\sqrt[n]{|d_n|}=|x|\cdot\limsup_{n\to\infty}\sqrt[n]{|a_n|}=\frac{|x|}{R},$$

where the radius of convergence is defined by

$$R=\frac{1}{\limsup_{n\to\infty}\sqrt[n]{|a_n|}} .$$

Conventions: if $\limsup|a_n|^{1/n}=0$ then $R=+\infty$; if $\limsup|a_n|^{1/n}=+\infty$ then $R=0$.

Applying root test 2.0 to $\sum a_nx^n$:

• $|x|/R<1$, i.e. $|x|<R$: absolute convergence;
• $|x|/R>1$, i.e. $|x|>R$: divergence;
• $|x|=R$: inconclusive; both endpoints must be checked separately.

Theorem (convergence interval). A power series converges on $(-R,R)$, diverges outside $(-R,R)$, and needs a separate check at $x=\pm R$.

This rules out something that is a priori conceivable but never happens: convergence at $x=5$, divergence at $x=10$, convergence again at $x=20$. The convergence set is always a symmetric interval about $0$, never a scattered or alternating pattern.

Examples

• Geometric $\sum x^n$: $a_n=1$, $\limsup|a_n|^{1/n}=1$, so $R=1$; neither endpoint converges.
• Exponential $\sum x^n/n!$: the previous lecture used the ratio test to get convergence for all $x$, consistent with $R=\infty$ (since $n!$ grows so fast that $\sqrt[n]{1/n!}\to0$).

The exponential function defined by a power series

There are two common definitions: the solution of $y'=y$ with $y(0)=1$, or a power series. This lecture takes the second.

Definition.

$$E(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$

This really is a function: the previous lecture proved absolute convergence for every $x$, so the radius is $\infty$ and the domain is all of $\mathbb R$.

Basic properties

$E(0)=1$. At $x=0$ every term with $n\ge1$ vanishes, while $0^0=1$ by convention and $0!=1$.

$E(x)>0$ for $x\ge0$. Every term $\dfrac{x^n}{n!}\ge0$ and the first is $1$, so the sum is at least $1$.

$E(x)>0$ for $x<0$ (borrowing from the next lecture). The next lecture proves

$$E(x+y)=E(x)E(y).$$

Granting it, take $x<0$ so that $-x>0$. Then

$$E(x)E(-x)=E(0)=1 .$$

The right side is positive and $E(-x)$ is positive, hence $E(x)=\dfrac{1}{E(-x)}>0$. So $E(x)>0$ for every real $x$.

The lecture explicitly flags this as a loan to be repaid: it states what is needed and marks what is not yet proved. The pattern is worth noticing.

Defining $e$ and integer powers

Definition. $e=E(1)$.

Now check that $E$ agrees with the usual powers of $e$. For positive integers, iterating the addition formula gives

$$E(1)E(1)=E(2),\quad E(1)E(1)E(1)=E(3),\quad\dots$$

so for $m\in\mathbb N$,

$$e^m=\underbrace{e\cdot e\cdots e}_{m\text{ copies}}=E(m).$$

For $m=0$, $e^0=1=E(0)$. For negative integers $m$, from

$$E(m)E(-m)=E(0)=1$$

we get

$$E(m)=\frac{1}{E(-m)}=\frac{1}{e^{-m}}=e^{m}.$$

So $e^m=E(m)$ for every integer $m$.

Rational powers

Let $q=\dfrac{m}{n}$ with $m\in\mathbb Z$, $n\in\mathbb N$. We want

$$e^{q}=E\!\left(\frac{m}{n}\right).$$

Define $e^q$ as the unique $\alpha$ with

$$\alpha>0,\qquad \alpha^n=e^{m}.$$

Uniqueness comes from strict monotonicity of powers of positive numbers: if $\alpha_1,\alpha_2>0$ and $\alpha_1^n=\alpha_2^n$, then $\alpha_1=\alpha_2$.

Then check that this $\alpha$ really is $E(m/n)$. First, $E(m/n)>0$ as shown above. Second, applying the addition formula $n$ times,

$$\left(E\!\left(\frac mn\right)\right)^{n} =E\!\left(n\cdot\frac mn\right) =E(m)=e^{m}.$$

So $E(m/n)$ satisfies both defining properties of $e^q$, and uniqueness forces it to be $\alpha$:

$$e^{q}=E(q)\qquad(q\in\mathbb Q).$$

Note: the argument uses the existence of the positive $n$-th root, which completeness guarantees (a least-upper-bound construction) but which the course defers. Only uniqueness is used in the definition itself, so the reasoning is self-consistent.

Continuity

To promote $E$ from $\mathbb Q$ to $\mathbb R$, continuity is exactly what is needed.

Definition

Let $f$ be defined on a set containing $x_0$. It is continuous at $x_0$ if for every $\varepsilon>0$ there is $\delta>0$ such that whenever $|x-x_0|<\delta$ (and $x$ lies in the domain),

$$|f(x)-f(x_0)|<\varepsilon .$$

If $f$ is continuous at every point of its domain, it is continuous.

Two points to keep in mind:

• $\delta$ generally depends on both $\varepsilon$ and $x_0$;
• the definition has the same quantified shape as the limit of a sequence: an arbitrary precision is given, and a corresponding neighbourhood must be produced.

Two basic examples

Constant $f(x)=c$. Given $\varepsilon>0$, take $\delta=1$ (or any positive number). If $|x-x_0|<1$ then

$$|f(x)-f(x_0)|=|c-c|=0<\varepsilon .$$

So constant functions are continuous everywhere. Here $\delta$ depends on neither $\varepsilon$ nor $x_0$.

Identity $f(x)=x$. Given $\varepsilon>0$, take $\delta=\varepsilon$. If $|x-x_0|<\varepsilon$ then

$$|f(x)-f(x_0)|=|x-x_0|<\varepsilon .$$

So the identity is continuous everywhere.

Five algebraic rules

If $f,g$ are continuous, then:

1. $f+g$ is continuous;
2. $cf$ is continuous for constant $c$;
3. $fg$ is continuous;
4. if $f$ never vanishes, $1/f$ is continuous;
5. whenever the composition makes sense, $g\circ f$ is continuous.

These parallel the algebra of limits of sequences, and so do the proofs: split the error into two parts and allot $\varepsilon/2$ to each.

Proof of the product rule (rule 3). Estimate

$$f(x)g(x)-f(x_0)g(x_0).$$

Insert a cross term to create two controllable differences:

$$f(x)g(x)-f(x_0)g(x_0) =g(x)\bigl(f(x)-f(x_0)\bigr)+f(x_0)\bigl(g(x)-g(x_0)\bigr).$$

Hence

$$|f(x)g(x)-f(x_0)g(x_0)| \le |g(x)|\,|f(x)-f(x_0)|+|f(x_0)|\,|g(x)-g(x_0)| .$$

Control each piece:

• Continuity of $g$ at $x_0$ with $\varepsilon=1$ gives $\delta_1$ such that $|x-x_0|<\delta_1$ implies $|g(x)-g(x_0)|<1$, hence

$$|g(x)|<|g(x_0)|+1=:M .$$

• Continuity of $f$: there is $\delta_2$ with $|f(x)-f(x_0)|<\dfrac{\varepsilon}{2M}$ when $|x-x_0|<\delta_2$.
• Continuity of $g$: there is $\delta_3$ with $|g(x)-g(x_0)|<\dfrac{\varepsilon}{2(|f(x_0)|+1)}$ when $|x-x_0|<\delta_3$.

Take $\delta=\min\{\delta_1,\delta_2,\delta_3\}$. For $|x-x_0|<\delta$,

$$|f(x)g(x)-f(x_0)g(x_0)| \le M\cdot\frac{\varepsilon}{2M}+|f(x_0)|\cdot\frac{\varepsilon}{2(|f(x_0)|+1)} <\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon .$$

So $fg$ is continuous at $x_0$. $\square$

Two technical habits are worth keeping: first show one factor is bounded near $x_0$, then allocate the error; and add $1$ in a denominator to avoid dividing by zero when $f(x_0)=0$.

Polynomials and rational functions

Starting from the two basic examples and applying rules 1–3 repeatedly:

• $x$ is continuous ⇒ $x^2=x\cdot x$ is continuous ⇒ $x^n$ is continuous;
• constants are continuous ⇒ $cx^n$ is continuous;
• finite sums of continuous functions are continuous.

Theorem. Every polynomial is continuous.

Adding rule 4:

Theorem. Every rational function (a quotient of two polynomials) is continuous wherever the denominator does not vanish.

The zeros of the denominator are finitely many; the function is not defined there, so continuity is not even in question. Sometimes it can be extended continuously to such a point, but that produces a different function.

How $\delta$ depends on $x_0$: the example $1/x$

Take $f(x)=1/x$ on $x\ne0$ (say the positive branch). Fix $x_0>0$ and prove continuity there. Directly,

$$\left|\frac1x-\frac1{x_0}\right|=\frac{|x-x_0|}{|x|\,|x_0|}.$$

If first we require $|x-x_0|<\dfrac{x_0}{2}$, then $x>\dfrac{x_0}{2}$ and

$$\left|\frac1x-\frac1{x_0}\right|<\frac{2|x-x_0|}{x_0^2}.$$

Making this smaller than $\varepsilon$ needs $|x-x_0|<\dfrac{\varepsilon x_0^2}{2}$. So

$$\delta=\min\left\{\frac{x_0}{2},\ \frac{\varepsilon x_0^2}{2}\right\}$$

works.

The key observation: $\delta$ contains $x_0^2$. As $x_0$ approaches $0$, $\delta$ must be taken very small, because the function changes violently near $0$. That is exactly what "$\delta$ depends on $x_0$" means in practice.

If a continuous function admits a $\delta$ depending only on $\varepsilon$, uniformly in $x_0$, it is called uniformly continuous. That is a later topic; the lecture plants the hook and notes that excluding $0$ from the domain of $1/x$ is not an accident — a continuous function on a closed interval would automatically be uniformly continuous.

What this lecture establishes

• The limsup is the infimum of tail suprema and always exists; the liminf is defined symmetrically.
• Root test 2.0 uses the limsup instead of a limit and applies to any series.
• Radius of convergence $R=1/\limsup|a_n|^{1/n}$; absolute convergence for $|x|<R$, divergence for $|x|>R$, endpoints checked separately.
• The convergence set is always an interval centred at $0$, never a scattered set.
• $E(x)=\sum x^n/n!$ is defined on all of $\mathbb R$ and is everywhere positive (positivity at negative $x$ borrows the next lecture's addition formula).
• $e=E(1)$; for integers and rationals $q$, $e^q=E(q)$.
• Continuity is an $\varepsilon$-$\delta$ condition with $\delta$ depending on $\varepsilon$ and $x_0$.
• Constants and the identity are continuous; sums, scalar multiples, products, reciprocals and compositions preserve continuity; polynomials are continuous everywhere and rational functions wherever the denominator is nonzero.
• The $\delta$ for $1/x$ contains $x_0^2$, showing the neighbourhood must shrink drastically near $0$.

Comprehension check

1. Prove that $b_n=\sup_{k\ge n}a_k$ decreases, and explain why the limsup always exists.

Reference answer: Since $\{a_k:k\ge n+1\}\subseteq\{a_k:k\ge n\}$, a supremum over the smaller set cannot exceed one over the larger, so $b_{n+1}\le b_n$. The sequence $(b_n)$ decreases; if bounded below, monotone convergence gives a limit, which is the limsup; if unbounded below, one sets it to $-\infty$ (dually $+\infty$ when the original sequence is unbounded above). Either way the limsup is defined in the extended real line without requiring convergence of $(a_n)$.

2. If $\limsup a_n=2$, can we conclude that $a_n<2$ for all sufficiently large $n$?

Reference answer: No. The limsup controls tail suprema in the limit: for every $\eta>0$ there is $N$ such that $a_n<2+\eta$ for $n\ge N$, i.e. the sequence eventually stays below $2+\eta$, but it may approach or exceed $2$ infinitely often. For example $a_n=2+1/n$ has limsup $2$ yet every term exceeds $2$. What is guaranteed is the inequality with an arbitrary $\eta>0$, eventually.

3. Why must the convergence set of a power series be an interval centred at $0$, and not something like "converges at $x=5$, diverges at $x=10$, converges again at $x=20$"?

Reference answer: The root test gives the criterion $|x|\cdot\limsup|a_n|^{1/n}<1$ or $>1$, i.e. $|x|<R$ or $|x|>R$, where $R$ depends only on the coefficients, not on $x$. Convergence therefore depends only on how $|x|$ compares with the single threshold $R$. If $x=20$ converges then $20<R$, so every $y$ with $|y|<20$ also has $|y|<R$ and converges — contradicting divergence at $x=10$. Such an alternating pattern is impossible.

4. With $E$ defined by its power series, why does positivity at negative $x$ need an extra step, and what does that step rely on?

Reference answer: For $x\ge0$ every term is non-negative and the first is $1$, so the sum is positive. For $x<0$ the terms alternate and no termwise argument applies. One uses the addition formula $E(x)E(-x)=E(0)=1$: since $-x>0$ gives $E(-x)>0$ and the product is $1>0$, we get $E(x)=1/E(-x)>0$. This depends on the addition formula, proved only in the next lecture; the lecture flags it explicitly as a loan.

5. When verifying $e^{m/n}=E(m/n)$, which properties of $E$ are used?

Reference answer: Three: (a) $E(x)>0$, so its positive $n$-th root makes sense; (b) the iterated addition formula $E(x)^n=E(nx)$, giving $(E(m/n))^n=E(m)=e^m$; (c) uniqueness of the positive $n$-th root, i.e. at most one $\alpha>0$ satisfies $\alpha^n=e^m$. Together these identify $E(m/n)$ with the number called $e^{m/n}$.

6. Prove that constant functions are continuous, and explain how their $\delta$ differs from the general case.

Reference answer: Let $f(x)=c$. Given $\varepsilon>0$, take any $\delta>0$, say $\delta=1$. If $|x-x_0|<\delta$ then $|f(x)-f(x_0)|=|c-c|=0<\varepsilon$, so $f$ is continuous everywhere. The difference: here $\delta$ depends on neither $\varepsilon$ nor $x_0$, whereas in general $\delta$ depends on both — for $1/x$ it contains $x_0^2$.

7. Why must boundedness of $g$ near $x_0$ be established before proving that $fg$ is continuous?

Reference answer: The estimate

$$|fg-f(x_0)g(x_0)|\le|g(x)|\,|f(x)-f(x_0)|+|f(x_0)|\,|g(x)-g(x_0)|$$

contains the varying factor $|g(x)|$, which cannot be treated as a constant. Continuity of $g$ at $x_0$ with $\varepsilon=1$ gives $|g(x)|<|g(x_0)|+1=:M$ on a neighbourhood. With this bound the first term can be pushed below $M\cdot\frac{\varepsilon}{2M}=\frac{\varepsilon}{2}$. Without it, no fixed error budget can be assigned to $|f(x)-f(x_0)|$.

Relation to neighbouring lectures

Two threads meet here. On the series side, the limsup completes the root test and gives power series the clean structure of a radius of convergence. On the function side, the exponential is defined by a power series and extended from $\mathbb Q$ to $\mathbb R$ by continuity.

The next lecture does three things: it proves the addition formula $E(x+y)=E(x)E(y)$ (borrowed twice here); it proves that two continuous functions agreeing on $\mathbb Q$ agree everywhere, completing the definition of the exponential; and it begins the systematic study of continuous functions. After that come the extreme value theorem, the intermediate value theorem and metric spaces — at which point the theorem from the very first lecture is finally proved.