Integral Test Slides Convergence Lab
Project 01: The Integral Test
Calculus Series
The Core Question
"If the area under a curve is finite, must the sum of the rectangles under it be finite too?"
Discrete (Series)
\[ \sum_{n=1}^{\infty} a_n \]
A sum of distinct values.
Continuous (Integral)
\[ \int_{1}^{\infty} f(x) dx \]
The accumulation under a curve.
Lab Prerequisites
Before applying the test, the function \( f(x) \) must be:
1
Continuous
No breaks or holes on \([1, \infty)\)
2
Positive
\( f(x) > 0 \) for all \( x \ge 1 \)
3
Decreasing
\( f'(x) < 0 \) for all \( x \ge 1 \)
Protocol Check: Verify conditions before integrating!
The Integral Test
Let \( f \) be a continuous, positive, decreasing function on \( [1, \infty) \) and let \( a_n = f(n) \).
IF
\[ \int_{1}^{\infty} f(x) dx \]
CONVERGES
THEN
\[ \sum_{n=1}^{\infty} a_n \]
CONVERGES
...and if the integral diverges, the series diverges!
LAB TEST 1.1
Analysis of \( \sum \frac{1}{n^2+1} \)
Phase 1: Verify Conditions
• \( f(x) = \frac{1}{x^2+1} \) is continuous for \( x \ge 1 \)
• Always positive? Yes.
• Decreasing? Denominator grows, so yes.
Phase 2: Integrate
\[ \int_{1}^{\infty} \frac{1}{x^2+1} dx \]
\[ \lim_{t \to \infty} [ \arctan(x) ]_{1}^{t} \]
Result
\[ \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]
Since the integral is finite:
CONVERGES
Integral Test Lab Analysis Worksheet Convergence Lab Analysis
Protocol 01: The Integral Test
NAME:
DATE:
TEST PROTOCOL: THE INTEGRAL TEST
If \( f(x) \) is continuous, positive, and decreasing on \( [1, \infty) \), then \( \sum_{n=1}^{\infty} a_n \) and \( \int_{1}^{\infty} f(x)dx \) either both converge or both diverge.
Spec 01
Analyze the convergence of \( \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2+4} \)
Step A: Verify conditions (Continuous? Positive? Decreasing?)
Step B: Evaluate the improper integral
Conclusion:
Converges
Diverges
Spec 02
Analyze the convergence of \( \displaystyle \sum_{n=2}^{\infty} \frac{1}{n \ln n} \)
Step A: Verify conditions
Step B: Show work for integration
Conclusion:
Converges
Diverges
Spec 03
Consider the series \( \displaystyle \sum_{n=1}^{\infty} \frac{n}{e^{n^2}} \)
Use the Integral Test to determine if this "structure" is stable (convergent).
Calculation Zone
Protocol Debrief
"Wait! If the Integral Test tells me the integral converges to 5, does the series also sum to 5?"
Explain your reasoning here...
Integral Test Answer Key Answer Key
Confidential Instructor Resource: Protocol 01
Spec 01 Solution
Series: \( \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2+4} \)
Verification:
Continuous on \([1, \infty)\) (no zeros in denominator). Positive for all \(n \ge 1\). Decreasing since the denominator \(n^2+4\) is strictly increasing for positive \(n\).
Integration:
\[ \int_{1}^{\infty} \frac{1}{x^2+4} dx = \lim_{t \to \infty} \left[ \frac{1}{2} \arctan\left(\frac{x}{2}\right) \right]_{1}^{t} \] \[ = \lim_{t \to \infty} \left( \frac{1}{2} \arctan\left(\frac{t}{2}\right) - \frac{1}{2} \arctan\left(\frac{1}{2}\right) \right) \] \[ = \frac{1}{2} \left( \frac{\pi}{2} \right) - \frac{1}{2} \arctan\left(\frac{1}{2}\right) \approx 0.553 \]
Conclusion:
CONVERGES
Spec 02 Solution
Series: \( \displaystyle \sum_{n=2}^{\infty} \frac{1}{n \ln n} \)
Verification:
Continuous on \([2, \infty)\). Positive for \(n \ge 2\). Decreasing denominator grows.
Integration (u-sub):
Let \(u = \ln x\), \(du = \frac{1}{x} dx\). \[ \int_{2}^{\infty} \frac{1}{x \ln x} dx = \int_{\ln 2}^{\infty} \frac{1}{u} du = \lim_{t \to \infty} [\ln u]_{\ln 2}^{t} = \infty \]
Conclusion:
DIVERGES
Spec 03 Solution
Series: \( \displaystyle \sum_{n=1}^{\infty} n e^{-n^2} \)
Integral: \( \displaystyle \int_{1}^{\infty} x e^{-x^2} dx \)
Debrief Guidance
"No. The sum of the series is not necessarily equal to the value of the integral. The integral provides a bound or an indicator of behavior, but since the series represents rectangles and the integral represents the smooth area, they differ. However, they share the same 'fate' (convergence or divergence)."
P-Series Slides P-Series Sensitivity
Benchmarking the Boundary
Project 02
Identifying the Structure
A p-series is an infinite series of the form:
LAB SPEC
\[ \sum_{n=1}^{\infty} \frac{1}{n^p} = \frac{1}{1^p} + \frac{1}{2^p} + \frac{1}{3^p} + \dots \]
p = 1
Harmonic
p = 2
Square Series
p = 0.5
Root Series
Proof of Stability
Let's apply Protocol 01 (Integral Test) to the general p-series:
\[ \int_{1}^{\infty} \frac{1}{x^p} dx \]
System Logs:
Case 1: \( p > 1 \)
Result: Integral Finite (Convergent)
Case 2: \( p \le 1 \)
Result: Integral Infinite (Divergent)
The P-Series Protocol
CONVERGES
p > 1
DIVERGES
p ≤ 1
Crucial Case: The Harmonic Series
\[ \sum \frac{1}{n} \quad (p=1) \implies \text{Divergent} \]
Sensitivity Analysis
\[ \sum \frac{1}{n^{1.0001}} \]
Stable
\[ \sum \frac{1}{n^{0.9999}} \]
Unstable
Expert Insight
"The series behaves like an elastic band. At \( p=1 \), it snaps. Even the smallest push beyond 1 keeps it bound to a finite sum."
Benchmark Series Reference Sheet Lab Notebook: Benchmark Series
Reference Sheet 02-B
P-Series
\[ \sum_{n=1}^{\infty} \frac{1}{n^p} \]
Converges p > 1
Diverges p ≤ 1
Key Example: The Harmonic Series \( \sum \frac{1}{n} \) diverges (\(p=1\)).
Geometric Series
\[ \sum_{n=1}^{\infty} ar^{n-1} \]
Converges |r| < 1
Diverges |r| ≥ 1
Sum = \( \frac{a}{1-r} \)
Thinking like a Lab Tech
When we see a complex series, we look for its "inner skeleton"—the dominant part that looks like a benchmark series.
Complex Form
\[ \frac{n^2}{n^3+5} \]
Skeleton
\( \frac{1}{n} \)
Complex Form
\[ \frac{3}{\sqrt{n^5+n}} \]
Skeleton
\( \frac{1}{n^{2.5}} \)
Complex Form
\[ \frac{4^n}{5^n+1} \]
Skeleton
\( (4/5)^n \)
PRO-TIP: Look at the highest powers!
In rational expressions, the highest powers of \( n \) in the numerator and denominator determine the series' "DNA". Subtract them to find the benchmark \( p \).
P-Series Sensitivity Worksheet P-Series Sensitivity Test
Protocol 02: Benchmark Identification
NAME:
DATE:
Phase 1: Rapid Identification
Identify the value of \( p \) and circle the convergence status for each series.
\( \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^5} \)
p = ________
Conv Div
\( \displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n}} \)
p = ________
Conv Div
\( \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{\pi}} \)
p = ________
Conv Div
\( \displaystyle \sum_{n=1}^{\infty} \frac{5}{n^{0.99}} \)
p = ________
Conv Div
Phase 2: Structural Derivation
Use the Integral Test to prove the convergence status of the Harmonic Series \( \displaystyle \sum_{n=1}^{\infty} \frac{1}{n} \).
Step 1: Set up the improper integral
Step 2: Evaluate the limit
Expert Reflection
If a series is given by \( a_n = \frac{1}{n^p} \), why does the value \( p=1 \) act as such a definitive "tipping point" for divergence?
Direct Comparison Slides Direct Comparison
The Scaling Protocol
Project 03
Logical Constraints
Scenario A
"If you can lift a box, you can definitely lift a lighter box."
Mathematical Parallel:
Smaller than Convergent is Convergent.
Scenario B
"If you can't lift a box, you definitely can't lift a heavier box."
Mathematical Parallel:
Larger than Divergent is Divergent.
Direct Comparison Test (DCT)
Suppose \( 0 \le a_n \le b_n \) for all \( n \)...
Condition 1
If \( \sum b_n \) converges,
then \( \sum a_n \) also converges.
Proof: Finite ceiling exists.
Condition 2
If \( \sum a_n \) diverges,
then \( \sum b_n \) also diverges.
Proof: Infinite floor exists.
Lab Test 3.1
Does \( \sum \frac{1}{n^2+5} \) converge?
1. Identify Benchmark
The "skeleton" is \( \sum \frac{1}{n^2} \) (a convergent p-series).
2. Build Inequality
Since \( n^2+5 > n^2 \), we know:
\[ \frac{1}{n^2+5} < \frac{1}{n^2} \]
Conclusion
The series is smaller than a convergent series.
Convergent
The Protocol Fails IF...
Inequality Direction Error
"I am larger than something finite."
"I am smaller than something infinite."
Result: INCONCLUSIVE
Direct Comparison Worksheet Direct Comparison Analysis
Protocol 03: The Inequality Test
NAME:
DATE:
Goal: Prove Convergence
Find a convergent benchmark series \( \sum b_n \) such that \( a_n \le b_n \).
Goal: Prove Divergence
Find a divergent benchmark series \( \sum b_n \) such that \( a_n \ge b_n \).
Spec 01
Determine the convergence of \( \displaystyle \sum_{n=1}^{\infty} \frac{1}{3^n+2} \)
1. Benchmark Selection (\( \sum b_n \))
Target: Geometric? P-series?
2. Benchmark Behavior
Convergent or Divergent? Why?
3. Build and Justify the Inequality
Show that \( a_n \le b_n \) or \( a_n \ge b_n \)...
Spec 02
Determine the convergence of \( \displaystyle \sum_{n=2}^{\infty} \frac{1}{\sqrt{n}-1} \)
1. Benchmark Selection (\( \sum b_n \))
2. Benchmark Behavior
3. Build and Justify the Inequality
Security Alert
Identify the "Direction Flaw"
A student attempts to use DCT on \( \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2-5} \) by comparing it to \( \sum \frac{1}{n^2} \).
Student's Logic:
\( n^2-5 < n^2 \)
Why is this inconclusive?
Limit Comparison Slides Limit Comparison
The Growth Rate Protocol
Project 04
System Failure
In Lesson 3, we saw that Direct Comparison fails if the inequality points the "wrong way."
\[ \sum_{n=1}^{\infty} \frac{1}{n^2-5} \quad \text{vs} \quad \sum_{n=1}^{\infty} \frac{1}{n^2} \]
"I am larger than a convergent series... so I could be anything."
We need a test that looks at growth rates instead of absolute sizes.
Limit Comparison Test (LCT)
Suppose \( a_n > 0 \) and \( b_n > 0 \). If the limit of their ratio exists:
\[ \lim_{n \to \infty} \frac{a_n}{b_n} = L \]
"If \( 0 < L < \infty \), then both series share the exact same fate."
Dominance Analysis
To use LCT effectively, find the Dominant Terms.
Example Series:
\[ \frac{2n^2 + 3n}{5n^3 - n + 7} \]
Benchmark Choice:
\[ \frac{n^2}{n^3} = \frac{1}{n} \]
Lab Tip
In the limit as \( n \) goes to infinity, the constants and lower-power terms become insignificant "noise."
LCT ignores the noise and looks at the structure.
Lab Test 4.1
Analysis of \( \sum \frac{1}{n^2-5} \)
1. Setup: Compare \( a_n = \frac{1}{n^2-5} \) with \( b_n = \frac{1}{n^2} \)
\[ \lim_{n \to \infty} \frac{\frac{1}{n^2-5}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^2}{n^2-5} = 1 \]
Since L = 1 (finite and positive), they share the same fate.
Status:
CONVERGES
Limit Comparison Worksheet Dominance & Limit Analysis
Protocol 04: The Growth Rate Test
NAME:
DATE:
LCT Summary
Pick \( b_n \) by taking the highest power in the numerator and denominator of \( a_n \).
Spec 01
Analyze \( \displaystyle \sum_{n=1}^{\infty} \frac{2n^2+n}{n^4-3} \)
1. Identify Dominant Terms
Target Benchmark (\( b_n \)):
2. Behavior of \( b_n \)
Conv/Div? Rule?
3. Execute the Limit
LIMIT CALCULATION
Conclusion:
CONVERGES
DIVERGES
Spec 02
Analyze \( \displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt{4n^2+n}} \)
1. Dominant Terms & Benchmark
2. Limit Result (\( L \))
L =
3. Full Analysis Space
Note: LCT is most effective when the series term is a rational or radical expression involving powers of \( n \).
Convergence Decision Tree Handout Strategic Decision Matrix
Mastering the Convergence Lab Procedures
1. Quick Check
Does \( \displaystyle \lim_{n \to \infty} a_n \ne 0 \)?
YES
Divergence Test:
NO (Inconclusive)
2. Form Check
Is it a rational or radical expression of \( n \)?
YES
Comparison Tests
DCT Use if inequality is simple and points the right way.
LCT Most powerful. Use if constants (+ or -) make DCT messy.
NO
Functional Tests
Integral Test Use if \( f(x) \) is easily integrable (look for u-sub).
Geometric/P-Series Does it perfectly match a benchmark form?
The "Cheat Sheet" of Benchmarks
Constants: Ignore them when picking benchmarks (\( +5 \), \( -1 \)).
Trig: Use DCT to bound \( \sin(n) \) and \( \cos(n) \) between -1 and 1.
Exponentials: These are geometric benchmarks.
Logarithms: Slow growth. Often requires Integral Test.
Series Sort Activity Cards The Series Sort
Strategic Test Workshop Activity
Series Alpha
\( \displaystyle \sum \frac{n}{n^2+1} \)
Best Test: _________
Series Beta
\( \displaystyle \sum \frac{1}{n^3+n} \)
Best Test: _________
Series Gamma
\( \displaystyle \sum \frac{1}{\sqrt{n}} \)
Best Test: _________
Series Delta
\( \displaystyle \sum \frac{\sin^2 n}{n^2} \)
Best Test: _________
Series Epsilon
\( \displaystyle \sum \frac{1}{n \ln n} \)
Best Test: _________
Series Zeta
\( \displaystyle \sum \frac{e^n}{1+e^{2n}} \)
Best Test: _________
Series Eta
\( \displaystyle \sum \frac{1}{3^n-1} \)
Best Test: _________
Series Theta
\( \displaystyle \sum \frac{n^2+1}{n^2-1} \)
Best Test: _________
Series Iota
\( \displaystyle \sum \frac{\sqrt{n}}{n^2+5} \)
Best Test: _________
Instructions:
1. Cut out the cards above.
2. Sort them into categories based on the most efficient convergence test.
3. Be prepared to justify your "sorting logic" to the Lab Manager (your teacher).
Series Sort Answer Key Workshop Key: Series Sort
Instructor Reference
Alpha: \( \sum \frac{n}{n^2+1} \)
LCT with \( 1/n \) (Divergent) or Integral Test.
Beta: \( \sum \frac{1}{n^3+n} \)
DCT with \( 1/n^3 \) (Convergent).
Gamma: \( \sum \frac{1}{\sqrt{n}} \)
P-Series (\(p = 1/2 \le 1\)) Divergent.
Delta: \( \sum \frac{\sin^2 n}{n^2} \)
DCT: \( 0 \le \frac{\sin^2 n}{n^2} \le \frac{1}{n^2} \) (Convergent).
Epsilon: \( \sum \frac{1}{n \ln n} \)
Integral Test (Divergent).
Zeta: \( \sum \frac{e^n}{1+e^{2n}} \)
Integral Test (\(u = e^x\)) or LCT with \( e^{-n} \).
Eta: \( \sum \frac{1}{3^n-1} \)
LCT with \( (1/3)^n \) (Convergent Geometric).
Theta: \( \sum \frac{n^2+1}{n^2-1} \)
Divergence Test: \( \lim a_n = 1 \ne 0 \).
Iota: \( \sum \frac{\sqrt{n}}{n^2+5} \)
LCT with \( 1/n^{1.5} \) (Convergent p-series).
Facilitator Tips:
• Encourage debate: Some series (like Zeta) can be solved multiple ways. Ask: "Which is faster?"
• The Trap: Theta looks like a comparison candidate, but the Divergence Test catches it instantly. Watch for students doing unnecessary work.
• The Direction: For Eta, DCT is messy because \( 3^n-1 < 3^n \), making the fraction larger. LCT is the "safer" choice here.
