Walkthrough

What does it mean to sum infinitely many numbers?

Imagine trying to add an infinite list of numbers: $\sum_{n=0}^{\infty} a_n = a_0 + a_1 + a_2 + \cdots$ You cannot perform infinitely many additions at once. Instead, you add finite portions one term at a time. The running total after index $N$ is called the $N$th partial sum: $S_N = \sum_{n=0}^N a_n = a_0 + a_1 + \cdots + a_N.$

As you add more terms, you generate a sequence of partial totals: $S_0, S_1, S_2, \ldots$. If these partial sums stabilize and approach a finite value $S$ as $N \to \infty$, the series converges to $S$. If the partial sums do not settle on a single finite number, the series diverges.

Because convergence depends solely on the long-term behavior of the tail, changing, adding, or removing finitely many initial terms can change the final numerical sum, but it never changes whether the series converges or diverges.

Terms approaching zero are not enough

Before doing advanced tests, check what the individual terms $a_n$ are doing. If the sequence $a_n$ does not converge to $0$—because it approaches a nonzero value or has no limit—the series diverges: $a_n\not\to0 \implies \sum_{n=0}^{\infty}a_n\text{ diverges}.$ This is the divergence test (or $n$th-term test).

A frequent point of confusion is assuming the converse is true. If $a_n \to 0$, does the series converge? Not necessarily. Consider the harmonic series: $\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots$ The individual terms $1/n$ approach $0$, yet their sum grows without bound and diverges. Terms shrinking to zero is a required baseline, not a guarantee of convergence.

Two baseline benchmarks: geometric and $p$-series

To determine whether shrinking terms decay quickly enough to converge, we compare them against two standard families:

1. Geometric series ($\sum ar^n$): Each term is multiplied by a fixed ratio $r$. When $|r| < 1$, the partial sums approach the finite sum $a/(1-r)$. When $|r| \ge 1$ (with $a \ne 0$), the terms fail to approach $0$, so the series diverges.
2. $p$-series ($\sum 1/n^p$): Here the power $p$ is fixed while the base $n$ grows. For $p>0$, the integral test connects this sum to $\int_1^{\infty}x^{-p}\,dx$, giving convergence when $p>1$ and divergence when $0<p\le1$. If $p\le0$, the terms do not approach $0$, so the divergence test applies. (The harmonic series is the boundary case $p=1$.)

Signs and structure: absolute vs. conditional convergence

When a series alternates between positive and negative values, cancellation can pull an otherwise divergent sum toward a finite limit.

• A series converges absolutely if the series of magnitudes, $\sum |a_n|$, converges. Absolute convergence guarantees that the original series $\sum a_n$ also converges.
• A series converges conditionally if $\sum a_n$ converges, but stripping the signs makes it diverge ($\sum |a_n|$ diverges).

For an alternating series $\sum (-1)^{n-1}b_n$ with $b_n\ge0$, the alternating series test guarantees convergence when two sufficient conditions hold:

1. The sizes are eventually nonincreasing: $b_{n+1}\le b_n$ from some index onward.
2. The sizes vanish: $\lim_{n\to\infty} b_n = 0$.

For example, the alternating harmonic series $\sum (-1)^{n-1}/n$ converges by this test. However, taking absolute values yields the standard harmonic series $\sum 1/n$, which diverges. Therefore, the alternating harmonic series converges conditionally.

Ratio and root tests: measuring long-term growth

When terms involve factorials or expressions raised to the $n$th power, we often measure a limiting growth rate: $L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| \quad \text{or} \quad L = \lim_{n\to\infty} \sqrt[n]{|a_n|}.$

When the corresponding displayed limit exists, the ratio or root test gives three cases:

• If $L < 1$, the terms decay faster than a convergent geometric series, ensuring absolute convergence.
• If $L > 1$ (or $L = \infty$), the terms grow or fail to vanish, meaning the series diverges.
• If $L = 1$, the test is inconclusive.

When $L=1$, the ratio or root test gives no answer. Analyze the series with another applicable tool, such as the divergence test or the alternating series test.

Calculus and series

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 $f'(a)$ gives the slope at $x=a$. Near $a$, the tangent line gives the approximation

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

If the second derivative $f''(a)$ 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 $f_y(a,b)$ 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. When $f(x,y)\ge 0$, 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 $n$.

An infinite series represents the formal sum of all terms in a sequence:

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

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

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

If the partial sums approach a finite real number $S$,

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

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.

When $r\ne1$, the $N$th partial sum is

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

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:

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

Taking the contrapositive 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. Even 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 integral test connects the series to an improper integral:

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

The series and the 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. Applying the integral test to $f(x)=1/x^p$ shows that

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

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

The graph below compares the partial sums for $p=2$ with the harmonic series. Move the slider to include more terms. The $p=2$ curve levels off, while the harmonic partial sums continue to grow.

N =

30

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 sign pattern $+,-,+,-,\ldots$, 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 states that 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, the alternating harmonic series

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

converges by the alternating series test. However, taking absolute values yields

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

which 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 classifies the behavior into three cases:

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

Example: a factorial series

Fix a real number $x$. Recall that $n!=n(n-1)\cdots2\cdot1$, with $0!=1$. Consider

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

If $x=0$, every term after the first is zero, so the series equals $1$ and converges. For $x\ne0$, let

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

Computing the ratio of consecutive terms,

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

Since the limit is $0<1$, 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 $x\ne0$, setting

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

gives

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

By the ratio test, the series converges absolutely when $|x|<1$ and diverges when $|x|>1$. Because the ratio test gives no conclusion when $|x|=1$, we test both boundary points directly:

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

The series therefore converges for

$$-1<x\le1.$$

The root test

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

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

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

It is especially convenient 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 alter whether the series converges. Setting

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

allows us to compute the $n$th root directly:

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

The root test shows that the series converges absolutely when $|x|<1$ and diverges when $|x|>1$. At $|x|=1$, the root test is inconclusive, so we evaluate the terms directly. 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 in sign with magnitudes approaching $e^{-1}$, so their limit does not exist. Because the terms fail to approach zero in both cases, the divergence test shows that both endpoint series diverge.

The series converges exactly when

$$-1<x<1.$$