What does it mean for a series to diverge?

Its sequence of partial sums has no finite limit. The sum never settles to a single fixed value, however many terms are added on.

Do the rules for combining series work on divergent series?

No. They hold only when both series converge. Applying them to divergent series can produce a meaningless result.

AP®MathsCollege BoardCalculus BCFlashcardsUnit 10: Infinite Sequences and SeriesIntroduction to Infinite Series

Introduction to Infinite Series (College Board AP® Calculus BC): Flashcards

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FrontConvergent & Divergent Infinite Series
Define a sequence and a series.

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Cards in this collection (21)

Define a sequence and a series.
A sequence is an ordered list of numbers, written $\{a_{n}\}$ .
A series is the sum of all its terms, $\sum_{n = 1}^{\infty} a_{n} = a_{1} + a_{2} + a_{3} + \dots$
Define the $n$ th partial sum of a series.
The sum of the first $n$ terms, $s_{n} = a_{1} + a_{2} + \dots + a_{n}$ .
Every series has a whole sequence of partial sums $s_{1}, s_{2}, s_{3}, \dots$
A series converges when the limit of its sequence of partial sums exists and is finite. In that case
$\sum_{n = 1}^{\infty} a_{n} = \lim_{n \to \infty}______$
The completed definition is $\sum_{n = 1}^{\infty} a_{n} = \lim_{n \to \infty} s_{n}$ .
The sum of an infinite series is defined as the limit of its partial sums, so a series converges exactly when that limit exists.
What does it mean for a series to diverge?
Its sequence of partial sums has no finite limit.
The sum never settles to a single fixed value, however many terms are added on.
True or False?
A series whose partial sums stay bounded must converge.
False.
The partial sums $1, 0, 1, 0, \dots$ never exceed $1$ , yet they never settle to a single value.
Convergence needs the limit to exist, not merely for the sums to stay bounded.
Two convergent series may be combined term by term, with constants $c$ and $d$ :
$\sum_{n = 1}^{\infty} (c a_{n} + d b_{n}) = c \sum_{n = 1}^{\infty} a_{n} +______$
The completed rule is $\sum_{n = 1}^{\infty} (c a_{n} + d b_{n}) = c \sum_{n = 1}^{\infty} a_{n} + d \sum_{n = 1}^{\infty} b_{n}$ .
A constant comes outside the summation, and a sum of two series splits into two separate sums.
Do the rules for combining series work on divergent series?
No.
They hold only when both series converge.
Applying them to divergent series can produce a meaningless result.
A sequence has $a_{n} = \frac{1}{2^{n}}$ and partial sums $s_{n} = \frac{2^{n} - 1}{2^{n}}$ . Find $\sum_{n = 1}^{\infty} a_{n}$ .
Rewrite $s_{n}$ as $1 - \frac{1}{2^{n}}$ .
As $n \to \infty$ the second term goes to zero, so the limit is $1$ and the series sums to $1$ .
Define a geometric series.
A series with a constant ratio between successive terms, $\sum_{n = 0}^{\infty} a r^{n} = a + a r + a r^{2} + \dots$
Here $a$ is the first term and $r$ is the common ratio.
A geometric series with first term $a$ and common ratio $r$ converges when $|r| < 1$ . Its sum is then
$\sum_{n = 0}^{\infty} a r^{n} = \frac{a}{______}$
The completed formula is $\sum_{n = 0}^{\infty} a r^{n} = \frac{a}{1 - r}$ .
It comes from letting $k \to \infty$ in the partial sum $\frac{a (1 - r^{k + 1})}{1 - r}$ , where the power of $r$ goes to zero.
When does a geometric series converge, and when does it diverge?
It converges when $|r| < 1$ and diverges when $|r| \geq 1$ .
The ratio has to shrink the terms for the sum to settle down.
True or False?
A geometric series with $r = 1$ converges, since every term is the same size.
False.
With $r = 1$ the series is $a + a + a + \dots$ and its partial sums are $a, 2 a, 3 a, \dots$ , which grow without bound.
Terms staying the same size is not enough.
A geometric series is written from $n = 1$ rather than $n = 0$ . Does that stop you using the formula?
No.
Only the first term and the common ratio are needed.
For $\sum_{n = 1}^{\infty} \frac{3}{5^{n}}$ they are $a = \frac{3}{5}$ and $r = \frac{1}{5}$ .
Find $\sum_{n = 1}^{\infty} \frac{1}{2^{n}}$ .
The first term and the common ratio are both $\frac{1}{2}$ , so the series converges.
The sum is $\frac{1 / 2}{1 - 1 / 2} = 1$ .
Write $0.621621621 \dots$ as a geometric series and hence as a fraction in lowest terms.
It is $\sum_{n = 0}^{\infty} \frac{621}{1000} {(\frac{1}{1000})}^{n}$ , with $a = \frac{621}{1000}$ and $r = \frac{1}{1000}$ .
The sum is $\frac{621}{999} = \frac{23}{37}$ .
Define the harmonic series, and say whether it converges.
It is $\sum_{n = 1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots$ , and it diverges.
The integral test is the usual way to show that.
A $p$ -series has the form below, where $p$ is a constant power that need not be a whole number.
$\sum_{n = 1}^{\infty} \frac{1}{______}$
The completed form is $\sum_{n = 1}^{\infty} \frac{1}{n^{p}}$ .
Taking $p = 1$ gives the harmonic series, so the harmonic series is one particular $p$ -series.
For which values of $p$ does a $p$ -series converge?
Only when $p > 1$ .
It diverges for every $p \leq 1$ , and the boundary case $p = 1$ is the harmonic series, which diverges.
What is the alternating harmonic series, and what does it sum to?
It is $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n + 1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$ , and it converges to $\ln 2$ .
True or False?
Making the harmonic series alternate is enough to turn a divergent series into a convergent one.
True.
$\sum \frac{1}{n}$ diverges, but $\sum \frac{{(- 1)}^{n + 1}}{n}$ converges to $\ln 2$ .
The alternating signs let successive terms cancel against each other.
What are the harmonic series and $p$ -series used for?
They are the standard comparison benchmarks.
The comparison and limit comparison tests decide an unknown series by measuring it against one whose behaviour is already known.

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