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Defining Convergent and Divergent Infinite Series

Samuel Baker

Samuel Baker

9 min read

Study Guide Overview

This study guide introduces sequences and series, covering terminology like nth partial sum and infinite series. It explains how to find terms and limits of sequences, classifying them as convergent or divergent. The guide also defines increasing, decreasing, monotonic, and bounded sequences, and presents a theorem connecting monotonicity and boundedness to convergence. Finally, it discusses series convergence and divergence, partial sums, telescoping series, and properties of convergent series, with practice problems on finding partial sums.

10.1 Defining Convergent and Divergent Infinite Series

Welcome to unit 10 of AP Calculus BC! This is the last one. πŸ₯³

This first, longer part of the guide concerns the behavior of numerical series and whether they have an actual sum as the number of terms approaches infinity. There are many tests and methods that we can use to answer this question, each working on specific cases. However, before we can answer this, we need to first talk about what sequences and series are.

Before we start talking about series, we need to talk about sequences and some terminology about sequences.

πŸ€” What is a Sequence?

A sequence is just a list of terms related by a common pattern to each other.

Here is how we represent sequences:

an1∞{a_n}^\infty_1

πŸ“Β Terms in a Sequence

List a1,a2,a3,ana_1, a_2, a_3, a_n, and anβ€…+β€…1a_{nβ€…+β€…1} for the following sequences. We can go through this together!

✏️ Finding Terms in A Sequence: Example 1

1n1∞{\dfrac1n}^\infty_1

This is nothing more than just plugging in n = 1, 2, 3, n, and n+1 into ana_n.

1,12,13,...,1n,1n+1,...{1, \dfrac12, \dfrac13,...,\frac{1}{n}, \frac{1}{n+1}, ...}

We use the ... between the 3rd and the nthn^{th} term and after the (n+1)th(n+1)^{th} term because there are an indefinite number of terms in these intervals.

This is a special sequence called the harmonic sequence. We will learn more about the following study guide later in this unit: 10.5 Harmonic Series and p-Series.

✏️ Finding Terms in A Sequence: Example 2

(βˆ’1)nβ‹…n!2n1∞{\frac{(-1)^n \cdot n!}{2^n}}^\infty_1

This is the same process as the last example, but here, we can use some algebra to simplify an+1a_{n+1} bit!

a1=βˆ’12a_1 = -\dfrac12

a2=12a_2 = \dfrac12

an=(βˆ’1)nβ‹…n!2na_n = \frac{(-1)^n \cdot n!}{2^n}

an+1=(βˆ’1)n+1β‹…(n+1)!2n+1=βˆ’(βˆ’1)n(n+1)n!2(2n)=βˆ’anβ‹…n+12a_{n+1} = \frac{(-1)^{n+1}\cdot(n+1)!}{2^n+1} = -\frac{(-1)^n(n+1)n!}{2(2^n)} = -a_n\cdot\frac{n+1}{2}

(βˆ’1)n(-1)^n

This is a special type of sequence called an alternating sequence. You can identify this by the (βˆ’1)n(-1)^n in the sequence formula, and we will learn more about alternating sequences and sequences in the following study guide: 10.7 Alternating Ser...

Question 1 of 11

What are the first three terms of the sequence an=1n+1a_n = \frac{1}{n+1}?

1, 1/2, 1/3

1/2, 1/3, 1/4

1, 2, 3

2, 3, 4