级数回顾与第一道检查

给定数列 $(a_n)$，级数指的是部分和序列

$$s_n=\sum_{i=1}^{n}a_i .$$

级数收敛就是 $(s_n)$ 收敛。由 Cauchy 收敛定理，$(s_n)$ 收敛当且仅当它是 Cauchy 列。

取 $m=n-1$，Cauchy 条件给出

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

所以看到级数，第一件事不是套判别法，而是看通项是否趋于零。 若不趋于零，立刻发散。反之不成立：调和级数 $1/n\to0$ 却发散。这道检查只能快速否定，不能肯定。

时间轴

重要级数家族：$p$-级数

调和级数不是孤例，它属于

$$\sum_{n=1}^{\infty}\frac{1}{n^{\alpha}} .$$

• $\alpha=1$：调和级数，发散。
• $\alpha>1$：收敛。上一讲只证了 $\alpha=2$ 的情形，用的是与等比级数比较、再对下标 $2^n-1$ 归纳。一般 $\alpha>1$ 通常用积分判别法证明，那留到后面的课。

这个家族提供了一个「标准尺子」：任何通项与 $1/n^\alpha$ 同阶的级数，都可以用它来比较。

绝对收敛与交错级数

若 $\sum|a_i|$ 收敛，称 $\sum a_i$ 绝对收敛。上一讲已证：绝对收敛 ⇒ 收敛。

反过来不成立，反例是交错调和级数

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

它为什么收敛

把部分和想成在数轴上来回走：

• 从 $0$ 出发，$s_1=-1$，向下走了一大步；
• $s_2=-1+\tfrac12=-\tfrac12$，向上回到靠近 $0$ 的位置，但没走回原点；
• $s_3=-\tfrac12-\tfrac13=-\tfrac56$，向下，但比上一次的低点 $-1$ 高；
• $s_4=-\tfrac56+\tfrac14=-\tfrac7{12}$，向上，又比上一次的高点 $-\tfrac12$ 低。

于是偶数项部分和单调上升，奇数项部分和单调下降，两者互相夹住，而步长 $1/n\to0$，所以它们收敛到同一个数。这就是交错级数判别法的几何内容。

判别法（陈述）。 若 $a_n$ 的符号交替，$|a_n|$ 单调递减，且 $|a_n|\to0$，则 $\sum a_n$ 收敛。

在交错调和级数上取绝对值恰好得到调和级数，它发散，所以这级数收敛但不绝对收敛——这就是「收敛 ⇏ 绝对收敛」的标准例子。

非负项级数：单调有界

若所有 $a_n\ge0$，则

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

部分和单调递增。由单调收敛定理：

$$\sum a_n\ \text{收敛}\iff (s_n)\ \text{有上界}.$$

这条把「判断收敛」变成「找一个上界」，是后面所有判别法的地基——包括比较、比值、根值三种，本质上都是把待判级数压到某个等比级数下面。

比较判别法（第一形式）

定理。 设 $0\le a_n\le b_n$ 对所有 $n$ 成立。若 $\sum b_n$ 收敛，则 $\sum a_n$ 收敛；等价地，若 $\sum a_n$ 发散，则 $\sum b_n$ 发散。

证明

用单调收敛定理。 记 $s_n^a=\sum_{i=1}^{n}a_i$，$s_n^b=\sum_{i=1}^{n}b_i$。逐项比较得

$$s_n^a\le s_n^b .$$

$\sum b_n$ 收敛 ⇒ $(s_n^b)$ 收敛 ⇒ 有界，设 $s_n^b\le C$。于是 $s_n^a\le C$，$(s_n^a)$ 单调递增且有上界，故收敛。

用 Cauchy 判据。 对 $m<n$，

$$|s_n^a-s_m^a|=\sum_{i=m+1}^{n}a_i\le\sum_{i=m+1}^{n}b_i=|s_n^b-s_m^b| .$$

（非负，所以绝对值可以去掉。）$\sum b_n$ 收敛 ⇒ $(s_n^b)$ 是 Cauchy 列 ⇒ 右端可任意小 ⇒ $(s_n^a)$ 也是 Cauchy 列 ⇒ 收敛。

两条路线都对。Cauchy 路线的好处是不需要知道极限是多少，这正是判别法存在的意义：我们通常根本不知道 $\sum a_n$ 的和。

例子：$\sum 2^{-n}/n$

因为 $\dfrac{1}{n}\le1$，所以

$$0\le\frac{2^{-n}}{n}\le 2^{-n}.$$

右边是 $c=\tfrac12$ 的等比级数，收敛。由比较判别法，$\sum 2^{-n}/n$ 收敛。

注意这个例子的思路：把难判的级数拆出一个「已知收敛的骨架」，剩下的因子只要一致有界即可。

比较判别法（第二形式，极限比较）

定理。 设 $a_n\ge0$，$b_n>0$，且

$$\frac{a_n}{b_n}\to L, \qquad 0<L<\infty .$$

则 $\sum a_n$ 收敛当且仅当 $\sum b_n$ 收敛。

这个形式的好处是不必亲自找到逐项不等式，只要比较「阶」。

例子：$\sum \dfrac{1}{n^2-1}$

取 $a_n=\dfrac{1}{n^2-1}$（$n\ge2$），$b_n=\dfrac{1}{n^2}$。则

$$\frac{a_n}{b_n}=\frac{n^2}{n^2-1}=\frac{1}{1-1/n^2}\to1 .$$

极限 $1\in(0,\infty)$，而 $\sum 1/n^2$ 收敛，所以原级数收敛。

若用第一形式，需要构造逐项不等式（比如证 $\frac{1}{n^2-1}\le\frac{2}{n^2}$ 对 $n\ge2$ 成立），可行但更绕。极限比较正好省去这一步。

比值判别法

定理。 设 $a_n\ne0$，且

$$\left|\frac{a_{n+1}}{a_n}\right|\to a .$$

则：

1. $a<1$ 时，级数绝对收敛（因而收敛）；
2. $a>1$ 时，级数发散；
3. $a=1$ 时，判别法失效，无法下结论。

前提是极限存在；极限不存在时这个形式用不了（下一讲的 limsup 会处理这种情况）。

证明（$a<1$）

取 $\alpha$ 使 $a<\alpha<1$。因为比值趋于 $a$，存在 $N$，当 $n\ge N$ 时

$$\left|\frac{a_{n+1}}{a_n}\right|\le\alpha,$$

即

$$|a_{n+1}|\le\alpha|a_n| .$$

连续使用：

$$|a_{N+m}|\le\alpha^{m}|a_N| .$$

于是级数 $\sum|a_n|$ 的尾部被等比级数 $|a_N|\sum\alpha^m$ 控制。由比较判别法（第一形式），$\sum|a_n|$ 收敛，即原级数绝对收敛。

证明（$a>1$）

取 $\alpha$ 使 $1<\alpha<a$。存在 $N$，当 $n\ge N$ 时 $|a_{n+1}|\ge\alpha|a_n|$，于是

$$|a_{N+m}|\ge\alpha^{m}|a_N|\to\infty .$$

通项甚至不趋于零，级数发散。

关于 $a=1$

调和级数 $\sum1/n$ 与 $\sum1/n^2$ 的比值都趋于 $1$，但前者发散、后者收敛。所以 $a=1$ 时信息不足，必须换判别法。

根值判别法

定理。 设

$$\sqrt[n]{|a_n|}=|a_n|^{1/n}\to c .$$

则：

1. $c<1$ 时，级数绝对收敛；
2. $c>1$ 时，级数发散；
3. $c=1$ 时，判别法失效。

证明（$c<1$）

取 $c_0$ 使 $c<c_0<1$。存在 $N$，当 $n\ge N$ 时 $|a_n|^{1/n}\le c_0$。两边取 $n$ 次方：

$$|a_n|\le c_0^{\,n}.$$

右边是 $c_0<1$ 的等比级数，由比较判别法，$\sum|a_n|$ 收敛。

证明（$c>1$）

取 $c_0$ 使 $1<c_0<c$。当 $n\ge N$ 时 $|a_n|^{1/n}\ge c_0$，于是

$$|a_n|\ge c_0^{\,n}\to\infty .$$

通项不趋于零，发散。

两个判别法的差别

比值判别法看相邻两项，要求绝对值最终按固定比例缩小；根值判别法看整体尺度，允许个别项突然变大，只要长期增长率受控。根值判别法适用范围略宽；但在实践中，遇到阶乘、幂次时比值判别法往往更好算。

幂级数

定义。 给定系数列 $(a_n)$ 与实数 $x$，形如

$$\sum_{n=0}^{\infty}a_n x^n$$

的级数称为幂级数。可以把它想成「无限次多项式」。

几个基本例子：

• 等比级数 $\sum x^n$：$a_n=1$，$|x|<1$ 时收敛，和为 $\dfrac{1}{1-x}$。
• 指数函数：$\displaystyle e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$。
• 余弦：$\displaystyle \cos x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}$。
• 正弦：$\displaystyle \sin x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}$。

并非所有函数都能写成幂级数。能写成的称为解析函数。课程提到一个反例形状的函数（在某点之前恒为零、之后才开始「有内容」），它在任何邻域内都不是解析的。补充：典型例子是 $f(x)=e^{-1/x^2}$（$x\ne0$），$f(0)=0$，它在 $0$ 处所有阶导数都是 $0$，因而若存在幂级数展开只能恒等于零，与函数本身矛盾。

中心问题：哪些 $x$ 让它收敛

固定 $x$，幂级数就变成一个普通级数，于是可以套判别法。以指数级数为例，取

$$b_n=\frac{x^n}{n!},\qquad x\ne0 .$$

计算比值：

$$\left|\frac{b_{n+1}}{b_n}\right| =\left|\frac{x^{n+1}}{(n+1)!}\cdot\frac{n!}{x^n}\right| =\frac{|x|}{n+1}\to 0 .$$

对每个固定的 $x$，极限都是 $0<1$，所以由比值判别法，级数对所有 $x$ 绝对收敛。

注意 x 在这里是固定的参数，n 才是趋于无穷的变量。先固定 $x$ 判收敛，再问「和怎样随 $x$ 变化」，是两个不同的问题——后者属于下一阶段的内容。

本节结论

• 通项不趋于零 ⇒ 级数发散；这是第一道、也是最快的检查。
• $p$-级数：$\alpha=1$ 发散，$\alpha>1$ 收敛。
• 绝对收敛 ⇒ 收敛；交错调和级数说明逆命题不成立。
• 非负项级数收敛 ⇔ 部分和有上界。
• 比较判别法第一形式：$0\le a_n\le b_n$，收敛性从大项传向小项；第二形式：$a_n/b_n\to L\in(0,\infty)$ 时两者同敛散。
• 比值判别法：$|a_{n+1}/a_n|\to a$；$a<1$ 绝对收敛，$a>1$ 发散，$a=1$ 失效。
• 根值判别法：$|a_n|^{1/n}\to c$；结论形式相同。
• 比值与根值判别法本质上都是把级数压到等比级数下，差别在于用相邻项比还是用整体 $n$ 次根。
• 幂级数：固定 $x$ 后用判别法；指数级数对所有 $x$ 收敛。

理解检验

1. 为什么「通项趋于零」只能用来否定收敛，不能用来肯定？

参考答案：若级数收敛，则 $a_n=s_n-s_{n-1}\to0$，这是必要条件。但它不是充分条件：调和级数 $\sum1/n$ 的通项趋于 $0$，级数却发散（上一讲用 $s_{2^n-1}\ge n/2$ 证明）。所以看到 $a_n\to0$ 不能下结论，必须继续用判别法；看到 $a_n\not\to0$ 则可以立刻判定发散。

2. 用比较判别法证明 $\sum 2^{-n}/n$ 收敛，并说明为什么不能反过来用它证明发散。

参考答案：因为 $1/n\le1$，故 $0\le2^{-n}/n\le2^{-n}$。$\sum2^{-n}$ 是 $c=1/2$ 的等比级数，收敛，由第一形式比较判别法得 $\sum2^{-n}/n$ 收敛。反方向不成立：即使 $a_n\le b_n$ 且 $\sum a_n$ 发散，也得不出 $\sum b_n$ 发散，因为 $b_n$ 可能比 $a_n$ 大得多却仍然收敛。要证发散，需要反过来的不等式（$0\le b_n\le a_n$ 且 $\sum b_n$ 发散）。

3. 极限比较判别法为什么要求极限 $L$ 既不为 $0$ 也不为 $\infty$？举一个 $L=0$ 时结论会变的例子。

参考答案：$L\in(0,\infty)$ 保证 $a_n$ 与 $b_n$ 同阶，从而同敛散。若 $L=0$，说明 $a_n$ 比 $b_n$ 小得多，只能推出「$\sum b_n$ 收敛 ⇒ $\sum a_n$ 收敛」，不能反向。例如 $a_n=1/n^2$，$b_n=1/n$，则 $a_n/b_n=1/n\to0$，$\sum b_n$ 发散而 $\sum a_n$ 收敛——两者并不同敛散。

4. 比值判别法在 $a=1$ 时为什么失效？用两个级数说明。

参考答案：$a=1$ 时比值只说明「相邻项大致相当」，不能区分缩小得够快还是不够快。$\sum1/n$ 的比值 $\frac{n}{n+1}\to1$，发散；$\sum1/n^2$ 的比值 $\frac{n^2}{(n+1)^2}\to1$，收敛。两者极限同为 $1$，结论却相反，所以这一步没有信息。

5. 证明根值判别法 $c<1$ 的情形，并指出它在哪一步用了比较判别法。

参考答案：取 $c<c_0<1$。因 $|a_n|^{1/n}\to c$，存在 $N$，当 $n\ge N$ 时 $|a_n|^{1/n}\le c_0$，两边取 $n$ 次方得 $|a_n|\le c_0^n$。对 $n\ge N$ 逐项比较 $\sum|a_n|$ 与等比级数 $\sum c_0^n$：后者收敛（$c_0<1$），故 $\sum|a_n|$ 收敛。比较判别法用在这一步——把待判级数压到等比级数之下。有限多个前项不影响收敛性。

6. 对指数级数 $\sum x^n/n!$，为什么比值判别法给出的极限对每个 $x$ 都是 $0$？这说明收敛范围是什么？

参考答案：比值为

$$\left|\frac{x^{n+1}/(n+1)!}{x^n/n!}\right|=\frac{|x|}{n+1}\to0$$

对任意固定的 $x$ 都成立，因为分子是常数而分母趋于无穷。极限 $0<1$，故对每个 $x$ 级数绝对收敛。收敛范围是整个实数轴（事实上对复数也成立）。这不同于等比级数只在 $|x|<1$ 收敛——分母的阶乘增长得比任何指数都快。

与前后课的关系

本讲把「判断级数收敛」变成了一套可操作的流程：先看通项，再看是否非负，然后依次尝试比较、比值、根值。所有判别法的共同结构是「压到等比级数以下」，这也解释了为什么上一讲要不惜篇幅把等比级数算清楚。

下一讲补上两个缺口：当比值或 $n$ 次根的极限不存在时，用 limsup 与 liminf 代替极限，判别法仍能工作；随后转向连续函数，并把幂级数当作构造指数函数与三角函数的工具。

Series recap and the first check

Given a sequence $(a_n)$, the series is the sequence of partial sums

$$s_n=\sum_{i=1}^{n}a_i .$$

The series converges exactly when $(s_n)$ converges, which by the Cauchy convergence theorem happens exactly when $(s_n)$ is Cauchy.

Taking $m=n-1$ in the Cauchy condition gives

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

So the first thing to check is not a test but the terms themselves. If $a_n$ does not tend to zero, the series diverges immediately. The converse fails: the harmonic series has $1/n\to0$ yet diverges. This check can only rule series out, never in.

Timeline

A family of important series: the $p$-series

The harmonic series is not isolated; it sits in

$$\sum_{n=1}^{\infty}\frac{1}{n^{\alpha}} .$$

• $\alpha=1$: the harmonic series, divergent.
• $\alpha>1$: convergent. The previous lecture proved only $\alpha=2$, by comparison with a geometric series and induction at the indices $2^n-1$. The general case $\alpha>1$ is normally handled by the integral test, which comes later.

This family is a standard yardstick: any series whose terms have the same order as $1/n^\alpha$ can be compared with it.

Absolute convergence and alternating series

If $\sum|a_i|$ converges, the series $\sum a_i$ converges absolutely. The previous lecture proved that absolute convergence implies convergence.

The converse fails, and the standard counterexample is the alternating harmonic series

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

Why it converges

Picture the partial sums walking back and forth on the line:

• starting at $0$, $s_1=-1$ takes a big step down;
• $s_2=-1+\tfrac12=-\tfrac12$ climbs back toward $0$, but not all the way;
• $s_3=-\tfrac12-\tfrac13=-\tfrac56$ goes down, yet stays above the previous low $-1$;
• $s_4=-\tfrac56+\tfrac14=-\tfrac7{12}$ goes up, yet stays below the previous high $-\tfrac12$.

So the even partial sums increase, the odd ones decrease, each set brackets the other, and the step size $1/n\to0$; both converge to the same number. This is the geometric content of the alternating series test.

Test (stated). If the signs of $a_n$ alternate, $|a_n|$ decreases monotonically, and $|a_n|\to0$, then $\sum a_n$ converges.

Applying absolute values to the alternating harmonic series produces the harmonic series, which diverges. Hence the series converges but not absolutely — the standard illustration that convergence does not imply absolute convergence.

Non-negative series: monotone and bounded

If all $a_n\ge0$, then

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

so the partial sums increase. By monotone convergence:

$$\sum a_n\ \text{converges}\iff (s_n)\ \text{is bounded above}.$$

This turns "decide convergence" into "find an upper bound" and underlies every test that follows — comparison, ratio and root all amount to pushing the series under some geometric series.

Comparison test, first form

Theorem. Let $0\le a_n\le b_n$ for all $n$. If $\sum b_n$ converges, then $\sum a_n$ converges; equivalently, if $\sum a_n$ diverges, then $\sum b_n$ diverges.

Proof

Via monotone convergence. Write $s_n^a=\sum_{i=1}^{n}a_i$ and $s_n^b=\sum_{i=1}^{n}b_i$. Termwise,

$$s_n^a\le s_n^b .$$

Convergence of $\sum b_n$ makes $(s_n^b)$ bounded, say $s_n^b\le C$. Then $s_n^a\le C$, and the increasing sequence $(s_n^a)$ converges.

Via Cauchy. For $m<n$,

$$|s_n^a-s_m^a|=\sum_{i=m+1}^{n}a_i\le\sum_{i=m+1}^{n}b_i=|s_n^b-s_m^b| ,$$

where the absolute values on the left may be dropped because terms are non-negative. If $\sum b_n$ converges, $(s_n^b)$ is Cauchy, so the right side is arbitrarily small, so $(s_n^a)$ is Cauchy and converges.

Both routes work. The Cauchy route has the advantage of never needing to know the sum — which is the whole point of a test, since the sum is usually unknowable.

Example: $\sum 2^{-n}/n$

Since $\dfrac1n\le1$,

$$0\le\frac{2^{-n}}{n}\le 2^{-n}.$$

The right side is a geometric series with $c=\tfrac12$, hence convergent. By comparison, $\sum2^{-n}/n$ converges.

The pattern: extract a known-convergent skeleton and check that the remaining factor is uniformly bounded.

Comparison test, second form (limit comparison)

Theorem. Let $a_n\ge0$, $b_n>0$, and

$$\frac{a_n}{b_n}\to L, \qquad 0<L<\infty .$$

Then $\sum a_n$ converges if and only if $\sum b_n$ converges.

Its advantage: no termwise inequality is needed, only a comparison of order.

Example: $\sum \dfrac{1}{n^2-1}$

Take $a_n=\dfrac{1}{n^2-1}$ (for $n\ge2$) and $b_n=\dfrac{1}{n^2}$. Then

$$\frac{a_n}{b_n}=\frac{n^2}{n^2-1}=\frac{1}{1-1/n^2}\to1 .$$

Since $1\in(0,\infty)$ and $\sum1/n^2$ converges, the given series converges.

The first form would require an explicit inequality such as $\frac{1}{n^2-1}\le\frac{2}{n^2}$ for $n\ge2$ — doable, but more roundabout.

Ratio test

Theorem. Suppose $a_n\ne0$ and

$$\left|\frac{a_{n+1}}{a_n}\right|\to a .$$

Then:

1. if $a<1$ the series converges absolutely (hence converges);
2. if $a>1$ the series diverges;
3. if $a=1$ the test is inconclusive.

The hypothesis requires that the limit exists; when it does not, this form cannot be used (limsup in the next lecture handles that case).

Proof ($a<1$)

Choose $\alpha$ with $a<\alpha<1$. Since the ratio tends to $a$, there is $N$ such that for $n\ge N$,

$$\left|\frac{a_{n+1}}{a_n}\right|\le\alpha,$$

that is,

$$|a_{n+1}|\le\alpha|a_n| .$$

Iterating,

$$|a_{N+m}|\le\alpha^{m}|a_N| .$$

So the tail of $\sum|a_n|$ is dominated by the geometric series $|a_N|\sum\alpha^m$. By the first comparison test, $\sum|a_n|$ converges, i.e. the series converges absolutely.

Proof ($a>1$)

Choose $\alpha$ with $1<\alpha<a$. For $n\ge N$, $|a_{n+1}|\ge\alpha|a_n|$, hence

$$|a_{N+m}|\ge\alpha^{m}|a_N|\to\infty .$$

The terms do not even tend to zero, so the series diverges.

Why $a=1$ is inconclusive

Both $\sum1/n$ and $\sum1/n^2$ have ratios tending to $1$, yet the first diverges and the second converges. At $a=1$ the ratio test carries no information.

Root test

Theorem. Suppose

$$\sqrt[n]{|a_n|}=|a_n|^{1/n}\to c .$$

Then:

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

Proof ($c<1$)

Choose $c<c_0<1$. For $n\ge N$, $|a_n|^{1/n}\le c_0$. Raising to the $n$-th power,

$$|a_n|\le c_0^{\,n}.$$

The right side is a geometric series with $c_0<1$, so by comparison $\sum|a_n|$ converges.

Proof ($c>1$)

Choose $1<c_0<c$. For $n\ge N$, $|a_n|^{1/n}\ge c_0$, so

$$|a_n|\ge c_0^{\,n}\to\infty .$$

The terms do not tend to zero; the series diverges.

How the two tests differ

The ratio test looks at consecutive terms and asks whether the absolute size shrinks by a fixed factor. The root test looks at the overall scale, tolerating occasional jumps provided the long-run growth is controlled. The root test is slightly more general; the ratio test is often easier to compute for factorials and powers.

Power series

Definition. Given coefficients $(a_n)$ and a real $x$, a series of the form

$$\sum_{n=0}^{\infty}a_n x^n$$

is a power series — an "infinite-degree polynomial".

Basic examples:

• Geometric: $\sum x^n$ with $a_n=1$; converges for $|x|<1$ with sum $\dfrac{1}{1-x}$.
• Exponential: $\displaystyle e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$.
• Cosine: $\displaystyle \cos x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}$.
• Sine: $\displaystyle \sin x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}$.

Not every function admits a power series. Those that do are called analytic. The lecture points to a function that is "dead zero" up to a point and only then becomes interesting, which is analytic nowhere. Note: the standard example is $f(x)=e^{-1/x^2}$ for $x\ne0$ and $f(0)=0$; every derivative at $0$ vanishes, so any power series expansion would have to be identically zero, contradicting the function.

The central question: for which $x$ does it converge?

Fix $x$; the power series becomes an ordinary series, and the tests apply. For the exponential series, set

$$b_n=\frac{x^n}{n!},\qquad x\ne0 .$$

The ratio is

$$\left|\frac{b_{n+1}}{b_n}\right| =\left|\frac{x^{n+1}}{(n+1)!}\cdot\frac{n!}{x^n}\right| =\frac{|x|}{n+1}\to 0 .$$

For every fixed $x$ the limit is $0<1$, so by the ratio test the series converges absolutely for all $x$.

Note carefully: $x$ is a fixed parameter and $n$ is what tends to infinity. First decide convergence for a fixed $x$; asking how the sum varies with $x$ is a separate question, taken up later.

What this lecture establishes

• Terms not tending to zero ⇒ divergence; this is the fastest first check.
• $p$-series: $\alpha=1$ diverges, $\alpha>1$ converges.
• Absolute convergence ⇒ convergence; the alternating harmonic series shows the converse fails.
• Non-negative series converge iff partial sums are bounded above.
• Comparison test, first form: $0\le a_n\le b_n$ transfers convergence downward; second form: $a_n/b_n\to L\in(0,\infty)$ makes the two series share their fate.
• Ratio test: $|a_{n+1}/a_n|\to a$; $a<1$ absolute convergence, $a>1$ divergence, $a=1$ inconclusive.
• Root test: $|a_n|^{1/n}\to c$; same three cases.
• Both tests push the series under a geometric series; one uses consecutive ratios, the other an $n$-th root.
• Power series: fix $x$ and apply a test; the exponential series converges for every $x$.

Comprehension check

1. Why can "terms tend to zero" only rule out convergence, never establish it?

Reference answer: If a series converges then $a_n=s_n-s_{n-1}\to0$, a necessary condition. It is not sufficient: the harmonic series has $1/n\to0$ yet diverges (shown last lecture via $s_{2^n-1}\ge n/2$). So $a_n\to0$ settles nothing and further tests are needed, while $a_n\not\to0$ immediately proves divergence.

2. Use the comparison test to prove $\sum2^{-n}/n$ converges, and explain why the test cannot be used in the opposite direction.

Reference answer: Since $1/n\le1$, we have $0\le2^{-n}/n\le2^{-n}$; the right side is the geometric series with $c=1/2$, which converges, so by the first comparison test $\sum2^{-n}/n$ converges. The reverse use fails because $a_n\le b_n$ together with divergence of $\sum a_n$ says nothing about $\sum b_n$ — the larger series could still converge. Proving divergence requires the inequality to go the other way: $0\le b_n\le a_n$ with $\sum b_n$ divergent.

3. Why does limit comparison require $L$ to be neither $0$ nor $\infty$? Give an example where $L=0$ changes the conclusion.

Reference answer: $L\in(0,\infty)$ means the terms have the same order, hence share convergence. If $L=0$, $a_n$ is much smaller than $b_n$, and only "$\sum b_n$ converges ⇒ $\sum a_n$ converges" survives. Example: $a_n=1/n^2$, $b_n=1/n$ gives $a_n/b_n=1/n\to0$; $\sum b_n$ diverges while $\sum a_n$ converges, so they do not share their fate.

4. Why is the ratio test inconclusive at $a=1$? Illustrate with two series.

Reference answer: At $a=1$ the ratio only says consecutive terms are comparable in size; it cannot distinguish sufficiently fast decay from insufficient decay. For $\sum1/n$ the ratio is $\frac{n}{n+1}\to1$ and the series diverges; for $\sum1/n^2$ the ratio is $\frac{n^2}{(n+1)^2}\to1$ and the series converges. The same limit gives opposite conclusions, so no information is available.

5. Prove the root test for $c<1$, and identify where the comparison test is used.

Reference answer: Choose $c<c_0<1$. Since $|a_n|^{1/n}\to c$, there is $N$ such that $|a_n|^{1/n}\le c_0$ for $n\ge N$; raising to the $n$-th power gives $|a_n|\le c_0^n$. Comparing $\sum|a_n|$ termwise with the convergent geometric series $\sum c_0^n$ shows $\sum|a_n|$ converges. The comparison test is used precisely at that step — pushing the series under a geometric series. Finitely many initial terms do not affect convergence.

6. For the exponential series $\sum x^n/n!$, why is the ratio limit $0$ for every $x$, and what does that say about the range of convergence?

Reference answer: The ratio is

$$\left|\frac{x^{n+1}/(n+1)!}{x^n/n!}\right|=\frac{|x|}{n+1}\to0$$

for any fixed $x$, since the numerator is constant and the denominator grows without bound. The limit $0<1$ gives absolute convergence for every $x$, so the range is the entire real line (indeed the complex plane). This differs from the geometric series, confined to $|x|<1$: the factorial in the denominator outgrows every exponential.

Relation to neighbouring lectures

This lecture turns "decide whether a series converges" into a workflow: check the terms, check non-negativity, then try comparison, ratio, root. Every test has the same skeleton — dominate by a geometric series — which is why the previous lecture laboured over the geometric sum.

The next lecture closes two gaps: when the ratio or the $n$-th root has no limit, limsup and liminf replace the limit and the tests still work; then the course turns to continuous functions and uses power series to construct the exponential and trigonometric functions.