AP Calculus BC 10.1 Defining Convergent and Divergent Infinite Series Study Notes - New Syllabus
AP Calculus BC 10.1 Defining Convergent and Divergent Infinite Series Study Notes- New syllabus
AP Calculus BC 10.1 Defining Convergent and Divergent Infinite Series Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Applying limits may allow us to determine the finite sum of infinitely many terms.
Key Concepts:
- Defining Convergent and Divergent Infinite Series
Defining Convergent and Divergent Infinite Series
Defining Convergent and Divergent Infinite Series
An infinite series is the sum of infinitely many terms written as
\( S = a_1 + a_2 + a_3 + \dots \)
To determine whether such a sum has a meaningful value, we examine the sequence of partial sums:
\( S_n = a_1 + a_2 + \dots + a_n \)
We then investigate what happens as \( n \to \infty \).
Convergent Infinite Series
An infinite series is called convergent if the sequence of its partial sums approaches a finite limit as \( n \to \infty \). In other words, there exists a real number \( L \) such that:
\( \lim_{n \to \infty} S_n = L \), where \( L \) is finite.
This means that adding more terms makes the sum get closer and closer to \( L \), without exceeding it in a way that grows indefinitely.
Example
Determine whether the series \( \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dots \) converges or diverges.
▶️ Answer/Explanation
This is a geometric series with \( a = \dfrac{1}{2} \) and \( r = \dfrac{1}{2} \).
For an infinite geometric series, if \( |r| < 1 \), the sum is given by:
\( S = \dfrac{a}{1 – r} \)
Substituting the values: \( S = \dfrac{\dfrac{1}{2}}{1 – \dfrac{1}{2}} = 1 \).
Since the sum approaches \( 1 \) as more terms are added, the series is convergent.
Final Answer: Convergent, with sum \( S = 1 \).
Divergent Infinite Series
An infinite series is called divergent if the sequence of its partial sums does not approach a finite limit as \( n \to \infty \).
This can happen if:
- The partial sums grow without bound (tend to \( +\infty \) or \( -\infty \)).
- The partial sums oscillate and fail to settle at a single value.
If a series is divergent, it cannot be assigned a finite sum in the standard sense.
Example
Determine whether the series \( 1 + 1 + 1 + 1 + \dots \) converges or diverges.
▶️ Answer/Explanation
The \( n \)-th partial sum is \( S_n = n \).
As \( n \to \infty \), \( S_n \to \infty \).
Since the partial sums grow without bound, the series is divergent.
Final Answer: Divergent, sum does not exist.
Example
Determine whether the series \( 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dots + 5 + 5 + 5 + \dots \) converges or diverges.
▶️ Answer/Explanation
The series can be thought of in two parts:
- Part 1: \( 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dots \) This is a geometric series with \( a = 1 \), \( r = \dfrac{1}{2} \). Since \( |r| < 1 \), it converges with sum \( S = \dfrac{1}{1 – \dfrac{1}{2}} = 2 \).
- Part 2: \( 5 + 5 + 5 + \dots \) — The \( n \)-th partial sum of this part is \( 5n \), which grows without bound as \( n \to \infty \), so it diverges.
Even though Part 1 converges, adding a divergent part causes the entire series to diverge.
Final Answer: Divergent, because the second part grows without bound.