AP Calculus BC 10.5 Harmonic Series and p-Series Study Notes - New Syllabus
AP Calculus BC 10.5 Harmonic Series and p-Series Study Notes- New syllabus
AP Calculus BC 10.5 Harmonic Series and p-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:
- Harmonic Series and \( p \)-Series
Harmonic Series and \( p \)-Series
Harmonic Series and \( p \)-Series
In this topic, we study two important types of series: the Harmonic Series and the more general \( p \)-Series. The harmonic series is a special case of the \( p \)-series with \( p = 1 \).
Harmonic Series
The harmonic series is:
\( \displaystyle \sum_{n=1}^\infty \dfrac{1}{n} = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots \)
Key Result: The harmonic series diverges, even though its terms approach zero.
Reason: By grouping terms strategically, we can show that the partial sums grow without bound:
- \( 1 + \dfrac{1}{2} \) (already > 1.5)
- \( + \dfrac{1}{3} + \dfrac{1}{4} > \dfrac{1}{2} \)
- \( + \dfrac{1}{5} + \dots + \dfrac{1}{8} > \dfrac{1}{2} \)
- Each group adds at least \( \dfrac{1}{2} \), so the sum grows without bound.
This means that \( \lim_{n \to \infty} S_n = \infty \).
Example
Determine whether \( \displaystyle \sum_{n=1}^\infty \dfrac{1}{n} \) converges or diverges.
▶️ Answer/Explanation
This is the harmonic series with \( a_n = \dfrac{1}{n} \).
Using the Integral Test: \( \displaystyle \int_{1}^{\infty} \dfrac{1}{x} \, dx = \infty \).
Since the integral diverges, the series also diverges.
Final Answer: Divergent.
\( p \)-Series
A \( p \)-series has the form:
\( \displaystyle \sum_{n=1}^\infty \dfrac{1}{n^p} \)
where \( p \) is a constant.
Convergence Rule:
- If \( p > 1 \), the series converges.
- If \( p \leq 1 \), the series diverges.
This result can be proven using the Integral Test applied to \( f(x) = \dfrac{1}{x^p} \).
Example
Test \( \displaystyle \sum_{n=1}^\infty \dfrac{1}{n^3} \) for convergence.
▶️ Answer/Explanation
Here, \( p = 3 > 1 \).
By the \( p \)-series rule, the series converges.
Final Answer: Convergent.
Example
Test \( \displaystyle \sum_{n=1}^\infty \dfrac{1}{\sqrt{n}} \) for convergence.
▶️ Answer/Explanation
Here, \( p = \dfrac{1}{2} \leq 1 \).
By the \( p \)-series rule, the series diverges.
Final Answer: Divergent.
Harmonic Series vs \( p \)-Series
| Series Type | General Term | Condition for Convergence | Condition for Divergence |
|---|---|---|---|
| Harmonic Series | \( \displaystyle \dfrac{1}{n} \) | Never converges | Always diverges |
| \( p \)-Series | \( \displaystyle \dfrac{1}{n^p} \) | \( p > 1 \) | \( p \leq 1 \) |
Example
Determine whether the series \( \displaystyle \sum_{n=2}^\infty \dfrac{1}{n(\ln n)^2} \) converges or diverges.
▶️ Answer/Explanation
This is not a simple \( p \)-series because of the \( \ln n \) term in the denominator.
We can apply the Integral Test with \( f(x) = \dfrac{1}{x(\ln x)^2} \), for \( x \geq 2 \).
\( \displaystyle \int_{2}^{\infty} \dfrac{1}{x(\ln x)^2} \, dx \)
Let \( u = \ln x \), so \( du = \dfrac{1}{x} dx \).
The integral becomes \( \displaystyle \int_{\ln 2}^{\infty} \dfrac{1}{u^2} \, du \).
This integral evaluates to \( \left[ -\dfrac{1}{u} \right]_{\ln 2}^{\infty} = \dfrac{1}{\ln 2} \), which is finite.
Since the integral converges, the series also converges.
Final Answer: Convergent.