§1 Series Review

Calculus studies two closely related ideas:

• differentiation, which describes rates of change and local approximations;
• integration, which describes accumulation through limits of sums.

For a one-variable function $f$, the derivative $f'(a)$, read “$f$ prime of $a$,” gives the slope at $x=a$. The symbol $\approx$ is read “is approximately equal to.” Near $a$, the tangent line gives the approximation

$$f(x)\approx f(a)+f'(a)(x-a).$$

If the second derivative $f''(a)$, read “$f$ double prime of $a$,” exists, a quadratic approximation is

$$f(x)\approx f(a)+f'(a)(x-a)+\frac12 f''(a)(x-a)^2.$$

In multivariable calculus, the same ideas extend to functions of several variables. For a surface $z=f(x,y)$, the partial derivatives $f_x(a,b)$ and $f_y(a,b)$, read “$f$ sub $x$” and “$f$ sub $y$,” measure change in the $x$- and $y$-directions. They determine the tangent plane, which gives a local linear approximation near $(a,b)$.

Integration also extends to several variables. A double integral adds many small contributions over a two-dimensional region. The symbol $\ge$ is read “greater than or equal to.” When $f(x,y)$ is nonnegative, written $f(x,y)\ge 0$, a double integral can represent the volume under the surface $z=f(x,y)$.

The rest of this note reviews the one-variable series tools needed later in the course.

Infinite series and partial sums

A sequence is an ordered list of numbers

$$a_0,a_1,a_2,\ldots,$$

where $a_n$ is the term with index $n$. The symbol $\ldots$ means the pattern continues.

The Greek capital letter $\Sigma$, read “sigma,” means to add terms. The symbol $\infty$, read “infinity,” indicates that the addition continues without a final term. An infinite series is written as

$$\sum_{n=0}^{\infty}a_n =a_0+a_1+a_2+\cdots.$$

We define its value through finite sums. For a nonnegative integer $N$, the $N$th partial sum is

$$S_N=\sum_{n=0}^{N}a_n.$$

The notation $\lim$, read “the limit,” describes the value an expression approaches. The arrow $\to$ is read “approaches.” If the partial sums approach a finite real number $S$, we write

$$\lim_{N\to\infty}S_N=S,$$

then the series converges to $S$. If the partial sums do not approach a finite number, the series diverges.

Changing, adding, or removing finitely many initial terms can change the value of a convergent series, but it cannot change whether the series converges.

Geometric series

A geometric series has a constant ratio $r$ between consecutive terms:

$$\sum_{n=0}^{\infty}ar^n =a+ar+ar^2+\cdots,$$

where $a$ is the first term and $r$ is the common ratio. The vertical bars in $|r|$ mean the absolute value of $r$.

The symbol $\ne$ is read “is not equal to.” When $r\ne1$, the $N$th partial sum is

$$S_N=a\frac{1-r^{N+1}}{1-r}$$

The symbols $<$ and $>$ are read “less than” and “greater than.” If $|r|<1$, then $r^{N+1}\to0$, so

$$\sum_{n=0}^{\infty}ar^n=\frac{a}{1-r}.$$

If $|r|\ge1$ and $a\ne0$, the geometric series diverges.

The divergence test

Every convergent series must have terms that approach zero. The symbol $\Longrightarrow$ is read “implies”:

$$\sum_{n=0}^{\infty}a_n\text{ converges} \quad\Longrightarrow\quad \lim_{n\to\infty}a_n=0.$$

This gives the divergence test, also called the $n$th-term test:

$$\lim_{n\to\infty}a_n\ne0 \quad\text{or the limit does not exist} \quad\Longrightarrow\quad \sum_{n=0}^{\infty}a_n\text{ diverges}.$$

The reverse is false. If $a_n\to0$, the series may still diverge. For example, the harmonic series

$$\sum_{n=1}^{\infty}\frac1n$$

diverges even though $1/n\to0$.

The integral test and p-series

Suppose $f$ is continuous, positive, and decreasing for $x\ge1$, and let $a_n=f(n)$. The symbol $\int$, read “integral,” represents continuous accumulation, and $dx$ says the integration is with respect to $x$. The symbol $\Longleftrightarrow$ is read “if and only if.” The integral test says

$$\sum_{n=1}^{\infty}a_n\text{ converges} \quad\Longleftrightarrow\quad \int_1^{\infty}f(x)\,dx\text{ converges}.$$

The series and improper integral therefore either both converge or both diverge.

A $p$-series has the form

$$\sum_{n=1}^{\infty}\frac1{n^p},$$

where $p$ is a real number. The symbol $\le$ is read “less than or equal to.” Using the integral test,

$$\sum_{n=1}^{\infty}\frac1{n^p} \begin{cases} \text{converges}, & p>1,\\ \text{diverges}, & p\le1. \end{cases}$$

The harmonic series is the case $p=1$.

Absolute and conditional convergence

A series

$$\sum_{n=1}^{\infty}a_n$$

converges absolutely if the series of absolute values

$$\sum_{n=1}^{\infty}|a_n|$$

converges. Absolute convergence always implies convergence of the original series.

A series converges conditionally if

$$\sum_{n=1}^{\infty}a_n$$

converges but

$$\sum_{n=1}^{\infty}|a_n|$$

diverges.

The alternating series test

An alternating series switches between positive and negative terms. The factor $(-1)^{n-1}$ produces the signs $+,-,+,-,\ldots$, so an alternating series can be written as

$$\sum_{n=1}^{\infty}(-1)^{n-1}b_n,$$

where $b_n>0$.

The alternating series test says the series converges if both conditions hold:

1. The magnitudes eventually decrease: $b_{n+1}\le b_n$.
2. The magnitudes approach zero: $\lim_{n\to\infty}b_n=0$.

For example,

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

converges by the alternating series test. It does not converge absolutely because

$$\sum_{n=1}^{\infty}\left|\frac{(-1)^{n-1}}n\right| =\sum_{n=1}^{\infty}\frac1n$$

is the divergent harmonic series. Therefore, the alternating harmonic series converges conditionally.

The ratio test

For a series $\sum a_n$, suppose the following limit exists:

$$L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|.$$

The ratio test gives three cases:

The ratio test is especially useful when the terms contain factorials or powers.

Example: a factorial series

Fix a real number $x$. The notation $n!$, read “$n$ factorial,” means the product $n(n-1)\cdots2\cdot1$, with $0!=1$. Now consider

$$\sum_{n=0}^{\infty}\frac{x^n}{n!},$$

If $x=0$, the series equals $1$ and converges. For $x\ne0$, let

$$a_n=\frac{x^n}{n!}.$$

Then

$$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{x^{n+1}}{(n+1)!}\frac{n!}{x^n}\right| = \frac{|x|}{n+1} \longrightarrow0.$$

Since $0<1$, the series converges absolutely for every real $x$.

Example: checking endpoints separately

Consider

$$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}.$$

At $x=0$, every term is zero, so the series converges. For $x\ne0$, let

$$a_n=(-1)^{n-1}\frac{x^n}{n},$$

we get

$$\left|\frac{a_{n+1}}{a_n}\right| =|x|\frac{n}{n+1} \longrightarrow |x|.$$

Therefore, the series converges absolutely when $|x|<1$ and diverges when $|x|>1$. The ratio test gives no conclusion when $|x|=1$, so we check both endpoints:

• At $x=1$, the series is the alternating harmonic series, so it converges conditionally.
• At $x=-1$, the series becomes $-\sum_{n=1}^{\infty}1/n$, so it diverges.

The series therefore converges for

$$-1<x\le1.$$

The root test

The symbol $\sqrt[n]{\phantom{x}}$ means the $n$th root. For a series $\sum a_n$, suppose the following limit exists:

$$L=\lim_{n\to\infty}\sqrt[n]{|a_n|}.$$

The root test has the same three outcomes as the ratio test:

It is most useful when an entire expression is raised to the $n$th power.

Example

Fix a real number $x$ and consider

$$\sum_{n=1}^{\infty}x^n\left(\frac{n}{n+1}\right)^n.$$

Starting at $n=1$ avoids the undefined expression $0^0$ and does not affect the convergence question. Let

$$a_n=x^n\left(\frac{n}{n+1}\right)^n.$$

Then

$$\sqrt[n]{|a_n|} =|x|\frac{n}{n+1} \longrightarrow |x|.$$

The series converges absolutely when $|x|<1$ and diverges when $|x|>1$. Euler's number $e$ is the constant defined by $e=\lim_{n\to\infty}(1+1/n)^n$. At $x=1$,

$$\left(\frac{n}{n+1}\right)^n = \left(1+\frac1n\right)^{-n} \longrightarrow e^{-1}\ne0.$$

At $x=-1$, the terms alternate between values whose magnitudes approach $e^{-1}$, so their limit does not exist. In both endpoint cases, the terms fail to approach zero. The divergence test therefore shows that both endpoint series diverge.

The series converges exactly when

$$-1<x<1.$$