Series and the Integral Test

Writing a number as an infinite decimal already commits us to adding infinitely many terms. The statement $π = 3.14159 \dots$ means

π = 3 + \frac{1}{10} + \frac{4}{1 0 ^{2}} + \frac{1}{1 0 ^{3}} + \frac{5}{1 0 ^{4}} + \dots,

and the $\dots$ says the sum continues forever, with each added term bringing us closer to $π$ . A sum of infinitely many terms must be defined before it can be computed.

Given a sequence ${a_{n}}$ , the expression

n = 1 \sum \infty a_{n} = a_{1} + a_{2} + a_{3} + \dots

is an infinite series. Adding infinitely many numbers is not an operation we can perform, so we define the sum through a sequence we can control.

The sum of a series is the limit of its partial sums, nothing more. So $\sum a_{n} = s$ means that by adding enough terms we can make the running total as close to $s$ as we like. Every series carries two sequences: the terms ${a_{n}}$ and the partial sums ${s_{n}}$ .

The partial sums

s_{n}

\sum 1/ 2^{n}

(dots) climb toward the sum

s = 1

(dashed line) while the terms

a_{n}

(below) shrink to zero.

The geometric series

The one series whose partial sum has a clean closed form is the geometric series, in which each term is a fixed multiple of the one before:

n = 1 \sum \infty a r^{n - 1} = a + a r + a r^{2} + a r^{3} + \dots, a \neq = 0.

The number $r$ is the common ratio. To find $s_{n}$ , write the sum and $r$ times the sum and subtract:

s_{n} = a + a r + \dots + a r^{n - 1}, r s_{n} = a r + a r^{2} + \dots + a r^{n} .

Subtracting, $s_{n} - r s_{n} = a - a r^{n}$ , so for $r \neq = 1$ ,

s_{n} = \frac{a ( 1 - r ^{n} )}{1 - r} .

Whether this converges is decided entirely by ${r^{n}}$ . If $∣ r ∣ < 1$ , then $r^{n} \to 0$ and $s_{n} \to a / (1 - r)$ . If $∣ r ∣ \geq 1$ , then ${r^{n}}$ diverges and so does the series (for $r = 1$ , $s_{n} = na \to \pm \infty$ directly).

A geometric series with

r = 1/2

: each term tiles half of what remains, so the pieces exhaust a single unit square and

\sum 1/ 2^{n} = 1

Every repeating decimal is a geometric series, hence a rational number.

A function from a series. Setting $a = 1$ , $r = x$ gives a series that represents a function on $(- 1, 1)$ :

n = 0 \sum \infty x^{n} = 1 + x + x^{2} + x^{3} + \dots = \frac{1}{1 - x}, ∣ x ∣ < 1.

This single identity generates a large fraction of the power series developed later.

Telescoping series

The other series with a computable partial sum is one whose terms cancel in pairs. For $\sum 1/ [n (n + 1)]$ , partial fractions give $\frac{1}{i ( i + 1 )} = \frac{1}{i} - \frac{1}{i + 1}$ , so

s_{n} = i = 1 \sum n (\frac{1}{i} - \frac{1}{i + 1}) = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n + 1}) = 1 - \frac{1}{n + 1} .

Every interior term cancels its neighbor, collapsing the sum to its two ends. Then $s_{n} \to 1$ , so $\sum_{n = 1}^{\infty} 1/ [n (n + 1)] = 1$ . The technique needs the general term written as a difference $b_{i} - b_{i + 1}$ ; the tool for producing that is partial fractions.

The harmonic series and the Test for Divergence

The harmonic series $\sum 1/ n$ shows that terms shrinking to zero is not enough. Group its terms in blocks of length $1, 2, 4, 8, \dots$ and bound each block below:

s_{2^{k}} = 1 + \frac{1}{2} + (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \dots + \frac{1}{8}) + \dots > 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \dots = 1 + \frac{k}{2} .

Each parenthesized block exceeds $\frac{1}{2}$ because its smallest term, repeated, already sums to $\frac{1}{2}$ . So $s_{2^{k}} \to \infty$ and the harmonic series diverges — even though $1/ n \to 0$ .

There is, however, a necessary condition. If $\sum a_{n}$ converges with sum $s$ , then $a_{n} = s_{n} - s_{n - 1} \to s - s = 0$ . Contrapositive:

The converse is not valid. If $a_{n} \to 0$ , the test is inconclusive: the series may converge or diverge (the harmonic series has terms tending to $0$ yet diverges). Only $a_{n} \neq \to 0$ is conclusive. For $\sum n^{2} / (5 n^{2} + 4)$ , the terms tend to $1/5 \neq = 0$ , so the series diverges at a glance.

Algebra of convergent series

Convergent series inherit the Limit Laws term by term. If $\sum a_{n}$ and $\sum b_{n}$ converge and $c$ is constant, then $\sum c a_{n}$ , $\sum (a_{n} + b_{n})$ , and $\sum (a_{n} - b_{n})$ converge, with

\sum c a_{n} = c \sum a_{n}, \sum (a_{n} \pm b_{n}) = \sum a_{n} \pm \sum b_{n} .

A finite number of terms never affects convergence, only the sum: changing, dropping, or adding finitely many terms leaves a divergent series divergent and a convergent one convergent. This lets us start a series at whatever index is convenient when only convergence is in question.

The Integral Test

Most series have no closed-form partial sum. The first general test compares a series of positive terms to an improper integral, using the fact that both are limits of accumulating pieces. Consider $\sum 1/ n^{2}$ alongside the curve $y = 1/ x^{2}$ . Erect a rectangle of width $1$ and height $1/ n^{2}$ on each interval $[n, n + 1]$ ; the rectangles beyond the first lie under the curve, so their total area is less than the area under $y = 1/ x^{2}$ for $x \geq 1$ :

n = 2 \sum \infty \frac{1}{n ^{2}} < 1 \int \infty \frac{1}{x ^{2}} d x = 1.

Hence $s_{n} < 2$ for all $n$ ; the partial sums increase and are bounded, so by the Monotonic Sequence Theorem the series converges. The mirror-image picture, with rectangles rising above $y = 1/ x$ , shows $\sum 1/ n$ exceeds a divergent integral and so diverges.

Left: rectangles of height

1/ n^{2}

sit below

y = 1/ x^{2}

, so the series is under a finite integral and converges. Right: rectangles of height

1/ n

rise above

y = 1/ x

, so the series exceeds an infinite integral and diverges.

The same reasoning proves the general test.

Two cautions. The test needs $f$ eventually decreasing, not decreasing from the start, since finitely many terms are irrelevant. And it decides whether the series converges, not its value: in general $\sum a_{n} \neq = \int_{1}^{\infty} f$ . For instance $\sum 1/ n^{2} = π^{2} /6$ , while $\int_{1}^{\infty} x^{- 2} d x = 1$ .

The p-series

Applying the Integral Test to $f (x) = 1/ x^{p}$ settles a whole family at once. From the improper-integral fact that $\int_{1}^{\infty} x^{- p} d x$ converges exactly when $p > 1$ :

For $p \leq 0$ the terms do not tend to $0$ , so the Test for Divergence already settles it; for $0 < p \leq 1$ the Integral Test applies and the integral diverges; for $p > 1$ the integral converges. The threshold sits exactly at $p = 1$ , where the harmonic series ( $p = 1$ ) diverges but $\sum 1/ n^{1.001}$ converges.

The

p

-series threshold: convergent for every

p > 1

, divergent for every

p \leq 1

, with the harmonic series sitting on the dividing line at

p = 1

Estimating a sum by its remainder

When the Integral Test proves convergence, it also bounds the error in stopping at $n$ terms. Write the remainder $R_{n} = s - s_{n} = a_{n + 1} + a_{n + 2} + \dots$ . Comparing this tail of rectangles to the area under $f$ on $[n, \infty)$ from both sides gives a two-sided bound.

The remainder

R_{n} = a_{n + 1} + a_{n + 2} + \dots

is a tail of rectangles; comparing it to the area under

f

[n, \infty)

from below and above brackets it between two improper integrals.

Summary of tests

Series form	Test	Verdict
$\sum a r^{n - 1}$	geometric	converges iff $∣ r ∣ < 1$ , sum $a / (1 - r)$
$\sum (b_{i} - b_{i + 1})$	telescoping	converges iff $b_{n}$ converges
$\sum 1/ n^{p}$	$p$ -series	converges iff $p > 1$
any, with $a_{n} \neq \to 0$	Test for Divergence	diverges
$a_{n} = f (n)$ , $f$ positive decreasing	Integral Test	matches $\int_{1}^{\infty} f$

Each entry requires either a closed-form partial sum or a tractable antiderivative. Series offering neither are decided by comparison, ratio, and root tests, which read convergence off the general term alone.