Absolute Convergence for MATH 141
Exam Relevance for MATH 141
Absolute vs conditional convergence is tested on MATH 141 finals. Late-course topic with good exam weight.
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Understanding Absolute Convergence
At the end of the Alternating Series Test, we briefly mentioned "absolute" vs "conditional" convergence. Now it's time to dig deeper. Absolute convergence is one of the most important concepts in series — it tells us not just whether a series converges, but how strongly it converges. And here's the key insight: if the absolute values converge, the original series must also converge.
What is Absolute Convergence?
Let's break down the terminology:
- Absolute = taking the absolute value (removing signs)
- Convergence = the series adds up to a finite number
A series $\sum a_n$ is absolutely convergent if the series of absolute values converges:
$$\sum_{n=1}^{\infty} |a_n| \text{ converges}$$
The big idea: Taking absolute values makes a series "harder" to converge (you're adding instead of sometimes subtracting). So if the harder version converges, the original definitely converges too.
The Absolute Convergence Theorem
If $\sum |a_n|$ converges, then $\sum a_n$ converges.
This is incredibly useful! It means we can use all our tests for positive series (p-test, comparison test, etc.) on the absolute values, and if that converges, we're done.
Important: The converse is NOT true. $\sum a_n$ can converge even when $\sum |a_n|$ diverges.
Absolute vs Conditional Convergence
There are exactly three possibilities for any series:
| Type | What happens | Example |
|---|---|---|
| Absolutely convergent | $\sum \|a_n\|$ converges (so $\sum a_n$ converges too) | $\displaystyle\sum \frac{(-1)^n}{n^2}$ |
| Conditionally convergent | $\sum a_n$ converges BUT $\sum \|a_n\|$ diverges | $\displaystyle\sum \frac{(-1)^{n+1}}{n}$ |
| Divergent | $\sum a_n$ diverges | $\displaystyle\sum (-1)^n$ |
Think of it this way:
- Absolute convergence = "strong" convergence (survives even without the sign cancellation)
- Conditional convergence = "fragile" convergence (only works because of sign cancellation)
Why Does Absolute Convergence Imply Convergence?
Here's the intuition: If you're adding up $|a_1| + |a_2| + |a_3| + \cdots$ and getting a finite sum $M$, then when you switch back to $a_1 + a_2 + a_3 + \cdots$ (with some negatives), you can't possibly get more than $M$ — the negatives only help!
More precisely, the partial sums of $\sum a_n$ are "trapped" between $-\sum |a_n|$ and $+\sum |a_n|$, so they must converge.
How to Test for Absolute Convergence
Step 1: Take absolute values of all terms: $|a_n|$
Step 2: Test whether $\sum |a_n|$ converges using any test you know:
- p-test
- Comparison test
- Limit comparison test
- Geometric series test
- etc.
Step 3: Conclude:
- If $\sum |a_n|$ converges → original series converges absolutely
- If $\sum |a_n|$ diverges → need to test $\sum a_n$ directly (might be conditional)
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$ converges absolutely.
Step 1: Take absolute values
$$|a_n| = \left|\frac{(-1)^n}{n^2}\right| = \frac{1}{n^2}$$
Step 2: Test the absolute series
$$\sum_{n=1}^{\infty} \frac{1}{n^2}$$
This is a p-series with $p = 2 > 1$, so it converges.
Step 3: Conclude
Since $\sum |a_n|$ converges, the original series $\sum \frac{(-1)^n}{n^2}$ converges absolutely.
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots$ converges absolutely or conditionally.
Step 1: Take absolute values
$$|a_n| = \frac{1}{n}$$
Step 2: Test the absolute series
$$\sum_{n=1}^{\infty} \frac{1}{n}$$
This is the harmonic series (p-series with $p = 1$), which diverges.
Step 3: Does the original converge?
By the Alternating Series Test (from the previous lesson), we know $\sum \frac{(-1)^{n+1}}{n}$ converges.
Conclusion: The series converges conditionally — it converges, but not absolutely.
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{\cos(n)}{n^2}$ converges.
Step 1: Take absolute values
$$|a_n| = \left|\frac{\cos(n)}{n^2}\right| = \frac{|\cos(n)|}{n^2}$$
Step 2: Use comparison
Since $|\cos(n)| \leq 1$ for all $n$:
$$\frac{|\cos(n)|}{n^2} \leq \frac{1}{n^2}$$
Since $\sum \frac{1}{n^2}$ converges (p-series, $p = 2 > 1$), by the Comparison Test, $\sum |a_n|$ converges.
Step 3: Conclude
The series converges absolutely, which means $\sum \frac{\cos(n)}{n^2}$ converges.
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{\sin(n)}{n^{3/2}}$ converges.
Step 1: Take absolute values
$$|a_n| = \frac{|\sin(n)|}{n^{3/2}}$$
Step 2: Use comparison
Since $|\sin(n)| \leq 1$:
$$\frac{|\sin(n)|}{n^{3/2}} \leq \frac{1}{n^{3/2}}$$
Since $\sum \frac{1}{n^{3/2}}$ converges (p-series, $p = 3/2 > 1$), by Comparison Test, $\sum |a_n|$ converges.
Step 3: Conclude
The series converges absolutely.
Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}}$ converges absolutely or conditionally.
Step 1: Take absolute values
$$|a_n| = \frac{1}{\sqrt{n}} = \frac{1}{n^{1/2}}$$
Step 2: Test the absolute series
$$\sum_{n=1}^{\infty} \frac{1}{n^{1/2}}$$
This is a p-series with $p = 1/2 < 1$, so it diverges.
Step 3: Test the original
Check the Alternating Series Test:
- $b_n = \frac{1}{\sqrt{n}}$ is decreasing ✓
- $\lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0$ ✓
So the original series converges.
Conclusion: The series converges conditionally.
Key Strategy: When to Use Absolute Convergence
Use absolute convergence when:
- The series has irregular signs (not alternating), like $\cos(n)$ or $\sin(n)$
- The series has alternating signs and you want a quick test
- You want to use comparison-type tests
If absolute test fails (absolute series diverges):
- Try the Alternating Series Test if signs alternate
- The series might converge conditionally
Why Does This Matter?
Absolutely convergent series are "well-behaved":
- You can rearrange the terms in any order and get the same sum
- They're easier to work with in calculations
Conditionally convergent series are "fragile":
- Rearranging terms can change the sum (even to any number you want!)
- The famous Riemann Rearrangement Theorem says you can rearrange a conditionally convergent series to sum to ANY value
Common Mistakes and Misunderstandings
❌ Mistake: Thinking conditional convergence means divergence
Wrong: "$\sum |a_n|$ diverges, so $\sum a_n$ diverges."
Why it's wrong: Absolute divergence doesn't tell you anything about the original series — it might still converge conditionally.
Correct: If $\sum |a_n|$ diverges, test $\sum a_n$ separately (e.g., with Alternating Series Test).
❌ Mistake: Forgetting to take absolute values
Wrong: Testing $\sum \frac{(-1)^n}{n^2}$ directly with the p-test.
Why it's wrong: The p-test only applies to positive series. You need to take absolute values first.
Correct: Test $\sum \frac{1}{n^2}$ (the absolute version), conclude absolute convergence.
❌ Mistake: Confusing the implication direction
Wrong: "$\sum a_n$ converges, so $\sum |a_n|$ must converge."
Why it's wrong: Convergence does NOT imply absolute convergence. The alternating harmonic series is the classic counterexample.
Correct: Absolute convergence implies convergence, but NOT vice versa.
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