Divergence Test Slides Nth Term Reality Check
Lesson 1: The Divergence Test
The Paradox of Smallness
Consider the series: \[ \sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \dots \]
The terms get smaller and smaller. They approach zero.
"If the pieces get infinitely small, shouldn't the total stay finite?"
Spoiler:
The Logical Blueprint
Necessity
In order for a series \(\sum a_n\) to converge, the terms \(a_n\) must go to zero.
If \(\sum a_n\) converges, then \(\lim_{n \to \infty} a_n = 0\).
Sufficiency
Just because terms go to zero doesn't mean the sum is finite.
\(\lim_{n \to \infty} a_n = 0\) does NOT guarantee convergence.
The Divergence Test
If \(\lim_{n \to \infty} a_n \neq 0\), then the series \(\sum a_n\) diverges .
Crucial Warning:
If \(\lim_{n \to \infty} a_n = 0\), the test is inconclusive . We must look elsewhere!
Quickfire: Converge or Diverge?
\[ \sum_{n=1}^{\infty} \frac{2n}{n+1} \]
Check the limit...
\[ \sum_{n=1}^{\infty} \frac{n^2}{2^n} \]
Check the limit...
\[ \sum_{n=1}^{\infty} (-1)^n \]
Check the limit...
\[ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \]
Check the limit...
Divergence Test Worksheet Nth Term Reality Check
Divergence Test Practice
NAME:
DATE:
PART 1
The Logical Blueprint
Answer the following questions based on your understanding of the n-th term test.
1. True or False: If \(\lim_{n \to \infty} a_n = 0\), then the series \(\sum a_n\) must converge.
True
False
Explain your reasoning:
2. Under what single condition can you definitively state that a series diverges using only the n-th term test?
PART 2
Blueprint Analysis
For each series, evaluate the limit of the n-th term and state whether the Divergence Test guarantees divergence or is inconclusive.
3. \(\sum_{n=1}^{\infty} \frac{3n^2 - 4}{5n^2 + n}\)
Limit:
Show your limit calculation:
Conclusion: _________________________________________
4. \(\sum_{n=1}^{\infty} \cos\left(\frac{1}{n}\right)\)
Limit:
Show your limit calculation:
Conclusion: _________________________________________
5. \(\sum_{n=1}^{\infty} \frac{\ln(n)}{n}\)
Limit:
Show your limit calculation:
Conclusion: _________________________________________
6. \(\sum_{n=1}^{\infty} \frac{e^n}{n^{100}}\)
Limit:
Show your limit calculation:
Conclusion: _________________________________________
Critical Reflection
Sarah calculates \(\lim_{n \to \infty} a_n = 0\) for a specific series. She writes "Converges" on her paper and moves to the next question. Explain why Sarah might be wrong, and provide a counterexample series that has a limit of zero but does not converge.
Divergence Test Key Answer Key
Nth Term Reality Check
Teacher Resource
Part 1: Conceptual Check
1. True or False: If \(\lim_{n \to \infty} a_n = 0\), then the series \(\sum a_n\) must converge.
FALSE. Reasoning: A limit of zero is a necessary condition for convergence, but not a sufficient one. Some series (like the harmonic series) have terms that approach zero but the sum still grows to infinity.
2. Under what single condition can you definitively state that a series diverges using only the n-th term test?
When \(\lim_{n \to \infty} a_n \neq 0\) (or when the limit does not exist).
Part 2: Blueprint Analysis
3. \(\sum_{n=1}^{\infty} \frac{3n^2 - 4}{5n^2 + n}\)
Limit: \(3/5\)
Conclusion: DIVERGES (by Divergence Test because \(3/5 \neq 0\))
4. \(\sum_{n=1}^{\infty} \cos\left(\frac{1}{n}\right)\)
Limit: \(\cos(0) = 1\)
Conclusion: DIVERGES (by Divergence Test because \(1 \neq 0\))
5. \(\sum_{n=1}^{\infty} \frac{\ln(n)}{n}\)
Limit: \(0\) (via L'Hopital's Rule)
Conclusion: INCONCLUSIVE (The terms go to zero, so we need a different test—this series actually diverges via the Integral Test).
6. \(\sum_{n=1}^{\infty} \frac{e^n}{n^{100}}\)
Limit: \(\infty\) (Exponentials grow faster than polynomials)
Conclusion: DIVERGES (by Divergence Test because \(\infty \neq 0\))
Critical Reflection
Sarah is wrong because the Divergence Test only works in one direction. It can prove divergence, but it can never prove convergence.
Counterexample: The Harmonic Series \(\sum \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \dots\). The terms \(a_n = 1/n\) approach zero as \(n \to \infty\), but the sum is known to diverge.
Integral Test Slides Integral Area Insights
Lesson 2: Connecting Series to Calculus
The Series-Function Link
We want to know if \(\sum_{n=1}^{\infty} a_n\) converges.
What if we treated the terms \(a_n\) as heights of rectangles under a continuous function \(f(x)\)?
"If the area under the curve is finite, can the sum of the rectangles also be finite?"
Visualizing \(\sum a_n \approx \int f(x) dx\)
Prerequisites for the Test
For the function \(f(x)\) to represent our series, it must be:
Positive
\(f(x) > 0\)
Continuous
No gaps or jumps
Decreasing
For all \(x \ge k\)
The Integral Test
Suppose \(a_n = f(n)\). Then the series \(\sum_{n=1}^{\infty} a_n\) and the integral \(\int_{1}^{\infty} f(x) dx\) behave exactly the same way.
Case 1
If \(\int f(x) dx\) converges , then \(\sum a_n\) converges .
Case 2
If \(\int f(x) dx\) diverges , then \(\sum a_n\) diverges .
Test in Action
\[ \sum_{n=1}^{\infty} \frac{1}{n^2 + 1} \]
Define \(f(x) = \frac{1}{x^2 + 1}\)
Check conditions (Pos, Cont, Decr)
Set up integral: \(\int_{1}^{\infty} \frac{1}{x^2 + 1} dx\)
Evaluate: \(\lim_{b \to \infty} [\arctan(x)]_1^b\)
Calculation Results:
\[ = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]
The integral converges to a finite value, so the series converges .
Integral Test Investigation Integral Area Insights
Inquiry Activity: Series vs. Integrals
NAME:
PHASE 1
Geometric Bounds
The diagram below shows the function \(f(x) = 1/x\) and two different ways to represent the series \(\sum_{n=1}^{\infty} \frac{1}{n}\) using rectangles of width 1.
Left Riemann Sum (Overestimate): Rectangles have heights \(f(1), f(2), f(3) \dots\)
Right Riemann Sum (Underestimate): Rectangles have heights \(f(2), f(3), f(4) \dots\)
Task:
Write an inequality relating \(\sum_{n=2}^{\infty} \frac{1}{n}\), \(\int_1^{\infty} \frac{1}{x} dx\), and \(\sum_{n=1}^{\infty} \frac{1}{n}\).
____ < ____ < ____
1 2 3 f(x)
Sketch your interpretation of the inequality here.
PHASE 2
Testing Convergence
Problem A: The Logarithmic Test
\[ \sum_{n=2}^{\infty} \frac{1}{n \ln(n)} \]
Verify prerequisites (Is it positive, continuous, and decreasing for \(x \ge 2\)?):
Set up and evaluate the improper integral:
The series __________________________ because __________________________.
Problem B: Exponential Decay
\[ \sum_{n=1}^{\infty} n e^{-n^2} \]
Identify the substitution needed for integration:
Evaluate the integral:
The series __________________________ because __________________________.
Integral Test Key Answer Key
Integral Area Insights
Teacher Resource
Phase 1: Geometric Bounds
The Inequality:
\(\sum_{n=2}^{\infty} \frac{1}{n} < \int_1^{\infty} \frac{1}{x} dx < \sum_{n=1}^{\infty} \frac{1}{n}\)
The integral is bounded by the left Riemann sum (overestimate starting at \(n=1\)) and the right Riemann sum (underestimate starting at \(n=2\)). If the integral is infinite, the sum on the right must be infinite. If the integral is finite, the sum on the left must be finite.
Phase 2: Testing Convergence
Problem A: \(\sum_{n=2}^{\infty} \frac{1}{n \ln(n)}\)
Prerequisites:
Yes. \(f(x) = \frac{1}{x \ln x}\) is positive for \(x > 1\), continuous for \(x > 1\), and the denominator increases as \(x\) increases, making \(f(x)\) decreasing.
Integral:
\(\int_2^{\infty} \frac{1}{x \ln x} dx \quad \text{Let } u = \ln x, du = \frac{1}{x} dx\)
CONCLUSION: DIVERGES because the corresponding improper integral diverges.
Problem B: \(\sum_{n=1}^{\infty} n e^{-n^2}\)
Substitution:
\(u = -x^2 \implies du = -2x dx \implies x dx = -\frac{1}{2} du\)
Integral:
\(\int_1^{\infty} x e^{-x^2} dx = -\frac{1}{2} \int_{-1}^{-\infty} e^u du = \frac{1}{2} \int_{-\infty}^{-1} e^u du\)
CONCLUSION: CONVERGES because the integral converges to a finite value (\(\frac{1}{2e}\)).
P Series Slides The P Series Powerhouse
Lesson 3: The Boundary of Infinity
The Infinite Overhang
If you stack identical blocks on the edge of a table, you can create an overhang.
How far can you go?
The max overhang is \(\frac{1}{2} \sum_{n=1}^{N} \frac{1}{n}\). Since the Harmonic Series diverges, the overhang can be infinitely long .
Harmonic Series: \(\frac{1}{2}(1 + \frac{1}{2} + \frac{1}{3} + \dots)\)
Blueprint of a P-Series
\[ \sum_{n=1}^{\infty} \frac{1}{n^p} \]
Where \(p\) is a positive constant.
The Convergence Rule
Converges
\(p > 1\)
The denominator grows "fast enough".
Diverges
\(p \le 1\)
The terms don't shrink fast enough.
Boundary Case: \(p=1\) is the Harmonic Series (Divergent).
Building Your Library
We will use these as standards of comparison. Categorize them quickly:
\( \sum \frac{1}{\sqrt{n}} \)
Diverges (p=1/2)
\( \sum \frac{1}{n^2} \)
Converges (p=2)
\( \sum \frac{1}{n^{1.0001}} \)
Converges (p>1)
\( \sum n^{-3} \)
Converges (p=3)
\( \sum \frac{1}{n} \)
Diverges (p=1)
Essential Mastery
Memorize the rule. It is your most used tool.
P Series Sorting Challenge P-Series Sorting Challenge
Building a Benchmark Library
Handout / Activity
Instructions:
In calculus, knowing your "benchmarks" is half the battle. Cut out the cards below. Sort them into two piles: CONVERGENT and DIVERGENT . For each card, identify the value of \(p\) and use it to justify your placement.
\[ \sum_{n=1}^{\infty} \frac{1}{n^5} \]
p = _____
\[ \sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n}} \]
p = _____
\[ \sum_{n=1}^{\infty} \frac{1}{n} \]
p = _____
\[ \sum_{n=1}^{\infty} n^{-1.1} \]
p = _____
\[ \sum_{n=1}^{\infty} \frac{5}{n^{\pi}} \]
p = _____
\[ \sum_{n=1}^{\infty} \frac{1}{n^{0.99}} \]
p = _____
\[ \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n} \]
p = _____
\[ \sum_{n=1}^{\infty} \frac{1}{n^e} \]
p = _____
\[ \sum_{n=1}^{\infty} \frac{n^2}{n^3} \]
p = _____
The P-Series Cheat Sheet
Series Form Condition Result \(\sum \frac{1}{n^p}\) \(p > 1\) Converges \(\sum \frac{1}{n^p}\) \(p \le 1\) Diverges \(\sum \frac{1}{n}\) Harmonic Series Diverges
Comparison Test Slides The Benchmarking Battle
Lesson 4: Direct and Limit Comparisons
The Weightlifter's Logic
In the Comparison Test, we don't look at the series alone. We compare it to a series we already know.
"If you can't lift a 50lb weight (diverges), you definitely can't lift a 100lb weight."
"If you can easily carry a 100lb box (converges), you can definitely carry a 50lb box."
Unknown
Benchmark
Comparison
Direct Comparison Test (DCT)
The Small Side
If \(0 \le a_n \le b_n\) and \(\sum b_n\) converges , then \(\sum a_n\) also converges .
Benchmarking a smaller series against a larger convergent one.
The Large Side
If \(0 \le b_n \le a_n\) and \(\sum b_n\) diverges , then \(\sum a_n\) also diverges .
Benchmarking a larger series against a smaller divergent one.
The Problem with Precision
Consider \(\sum \frac{1}{n^2 - 1}\).
We want to compare to \(\sum \frac{1}{n^2}\), but...
\(\frac{1}{n^2 - 1} > \frac{1}{n^2}\)
DCT fails! We found something bigger than a convergent series. We know nothing.
The Solution: LCT
Limit Comparison Test: If the limit of their ratio is a finite positive number \(L\), they have the same fate.
\(\lim_{n \to \infty} \frac{a_n}{b_n} = L\)
Choosing Your Champion
How to pick \(b_n\)? Look for the dominant terms .
Target Series
\[ \sum \frac{n^2 + 1}{n^4 + n} \]
Dominant: \(n^2\) on top, \(n^4\) on bottom.
Champion: \( \sum \frac{n^2}{n^4} = \sum \frac{1}{n^2} \)
Target Series
\[ \sum \frac{1}{3^n + 2} \]
Dominant: \(3^n\) in the denominator.
Champion: \( \sum \frac{1}{3^n} \)
Benchmarking Battle Worksheet The Benchmarking Battle
Direct & Limit Comparison Practice
NAME:
Direct Comparison (DCT)
\(a_n \le b_n\) and \(\sum b_n\) conv \(\implies \sum a_n\) conv
\(a_n \ge b_n\) and \(\sum b_n\) div \(\implies \sum a_n\) div
Limit Comparison (LCT)
If \(\lim_{n \to \infty} \frac{a_n}{b_n} = L\) (where \(0 < L < \infty\)),
then \(\sum a_n\) and \(\sum b_n\) share the same result.
01
\(\sum_{n=1}^{\infty} \frac{1}{2n^2 + n + 1}\)
Comparison Series (\(b_n\)): __________
Method: Direct Comparison Test
Workspace (Construct Inequality)
Result: __________________________________________________
02
\(\sum_{n=1}^{\infty} \frac{n + 5}{n^2 - 4}\)
Comparison Series (\(b_n\)): __________
Method: Limit Comparison Test
Workspace (Evaluate Limit)
Result: __________________________________________________
03
\(\sum_{n=1}^{\infty} \frac{\sin^2(n)}{n^3}\)
Comparison Series (\(b_n\)): __________
Method: Direct Comparison Test
Workspace (Analyze Trigonometric Bounds)
Result: __________________________________________________
The Strategist's Choice
Which test (DCT or LCT) would be more efficient for the series \(\sum \frac{2^n}{3^n - 1}\)? Explain your reasoning, identify your comparison champion, and state the final result.
Comparison Test Key Answer Key
The Benchmarking Battle
Teacher Resource
Problem 1: \(\sum \frac{1}{2n^2 + n + 1}\)
Champion: \(\sum \frac{1}{n^2}\) (Convergent P-Series, \(p=2\))
\(2n^2 + n + 1 > n^2\) for all \(n \ge 1\).
Problem 2: \(\sum \frac{n + 5}{n^2 - 4}\)
Champion: \(\sum \frac{n}{n^2} = \sum \frac{1}{n}\) (Divergent Harmonic Series)
\(\lim_{n \to \infty} \frac{(n+5)/(n^2-4)}{1/n} = \lim_{n \to \infty} \frac{n^2 + 5n}{n^2 - 4} = 1\).
Since \(1\) is a finite positive constant, both series share the same fate.
Problem 3: \(\sum \frac{\sin^2(n)}{n^3}\)
Champion: \(\sum \frac{1}{n^3}\) (Convergent P-Series, \(p=3\))
Since \(0 \le \sin^2(n) \le 1\), it follows that \(\frac{\sin^2(n)}{n^3} \le \frac{1}{n^3}\).
The Strategist's Choice: \(\sum \frac{2^n}{3^n - 1}\)
Choice: LCT is more efficient here because the denominator is smaller than the benchmark, making DCT messy/inconclusive if comparing directly to \(\sum (2/3)^n\).
Alternating Series Slides Alternating Sign Signals
Lesson 5: Oscillation and Approximation
The Seesaw Paradox
The Harmonic Series \(\sum \frac{1}{n}\) diverges.
But what if every other term was subtracted?
\[ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \]
By constantly pulling back, we prevent the sum from escaping to infinity.
Convergence through Cancellation
The Alternating Series Test
For a series \(\sum (-1)^n b_n\) to converge, it must satisfy TWO conditions:
1. Decreasing
The positive parts \(b_n\) must be non-increasing for all \(n\).
\(b_{n+1} \le b_n\)
2. Zero Limit
The limit of the positive terms must be zero.
\(\lim_{n \to \infty} b_n = 0\)
How Close Are We?
Alternating Series Estimation Theorem
If you sum up to term \(n\), the error (the difference between your partial sum and the actual infinite sum) is less than the absolute value of the next term.
\[ |R_n| = |S - S_n| \le b_{n+1} \]
Approximation accuracy is built right into the series!
The Convergence Hierarchy
Absolute Convergence
The series converges even if you make all the terms positive.
Example: \(\sum \frac{(-1)^n}{n^2}\)
Conditional Convergence
The series converges as written, but diverges if you make all terms positive.
Example: \(\sum \frac{(-1)^n}{n}\)
Convergence Quest Blueprint Convergence Quest Blueprint
Decision-Making Flowchart Activity
NAME:
You are an architect of infinity. To determine if a series converges or diverges, you need a systematic plan. Complete the flowchart below by filling in the missing tests and conditions we have covered in this sequence.
START: SERIES \(\sum a_n\)?
Primary Check
Nth Term Test
Does \(\lim_{n \to \infty} a_n = 0\)?
NO
DIVERGES
YES
Pick a Strategy...
P-Series? \(\frac{1}{n^p}\)
Geometric? \(ar^n\)
TEST TYPE A
Use when integrable...
TEST TYPE B
Compare to benchmarks...
TEST TYPE C
If terms swap signs...
The Ultimate Challenge
Identify which test you would use for the following series and WHY .
\(\sum \frac{(-1)^n}{n^2 + 1}\)
\(\sum \frac{\sqrt{n}}{n^2 + n}\)
Alternating Series Exit Ticket Alternating Sign Signals
Exit Ticket: Error and Convergence
NAME:
1. Criteria Check
Consider the alternating series \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n + 1}\). State the two conditions of the Alternating Series Test and show that this series satisfies them.
Condition A:
Condition B:
2. The Margin of Error
If we approximate the sum of \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}\) using the first 5 terms (\(S_5\)), what is the maximum possible error of our approximation? Write your answer as a simplified fraction.
Hint: Use the Alternating Series Estimation Theorem.
3. Absolute vs. Conditional
Explain the difference between absolute convergence and conditional convergence. Use the series \(\sum \frac{(-1)^n}{\sqrt{n}}\) as an example to illustrate your point.
