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Absolute vs. Conditional
Convergence

Learn to classify infinite series by testing whether they converge absolutely, conditionally, or diverge: a critical skill for the AP Calculus BC exam.

AP Calculus BC

What Is Series Convergence?

A series \(\displaystyle\sum_{n=1}^{\infty}a_n\) converges if the sequence of partial sums \(S_N = a_1 + a_2 + \cdots + a_N\) approaches a finite limit as \(N \to \infty\). If no such limit exists, the series diverges.

Defining Absolute Convergence

A series \(\displaystyle\sum a_n\) is said to converge absolutely if the series of absolute values also converges:

Think of it as the strongest form of convergence: removing all sign cancellation still leaves a convergent series.

Defining Conditional Convergence

A series \(\displaystyle\sum a_n\) converges conditionally if it converges, but not absolutely:

Conditional convergence relies on cancellation between positive and negative terms. The classic example is the alternating harmonic series: \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\).

Three Possible Outcomes

Test: does \(\sum a_n\) converge?
DIVERGES No convergence No Yes
Now test: does \(\sum |a_n|\) converge?
ABSOLUTE Yes CONDITIONAL No

The Absolute Convergence Theorem

The converse is false: a series can converge without converging absolutely. That gap is precisely what conditional convergence fills.

Why Does the Distinction Matter?

Absolute
You can rearrange the terms in any order and always get the same sum. Safe for algebraic manipulation.
Conditional
Rearranging terms can change the sum to any value, or even make it diverge (Riemann Rearrangement Theorem).

Absolute vs. Conditional: Side by Side

Absolute Conditional
\(\sum a_n\) Converges Converges
\(\sum |a_n|\) Converges Diverges
Rearrangement safe? Yes No
Classic example \(\sum \frac{(-1)^n}{n^2}\) \(\sum \frac{(-1)^{n+1}}{n}\)

Step-by-Step Decision Process

1
Take absolute values. Form the series \(\displaystyle\sum |a_n|\).
2
Test \(\sum |a_n|\) for convergence using the Ratio Test, Root Test, Comparison, Limit Comparison, or Integral Test.
3
If \(\sum |a_n|\) converges: conclude absolute convergence. Done.
4
If \(\sum |a_n|\) diverges: go back and test the original series \(\sum a_n\) (often via the Alternating Series Test).
5
If \(\sum a_n\) converges: conclude conditional convergence. If it diverges: divergent.

Example 1: \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\)

1
Absolute values: \(\displaystyle\sum \left|\frac{(-1)^{n+1}}{n^2}\right| = \sum \frac{1}{n^2}\).
2
Test: \(\sum \frac{1}{n^2}\) is a \(p\)-series with \(p = 2 > 1\), so it converges.
3
Conclusion: The original series converges absolutely.

Example 2: \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\)

1
Absolute values: \(\displaystyle\sum \frac{1}{n}\) is the harmonic series, which diverges.
2
Test original: Apply the Alternating Series Test. Let \(b_n = \frac{1}{n}\): (i) \(b_n > 0\), (ii) \(b_{n+1} \le b_n\), (iii) \(\lim_{n\to\infty} b_n = 0\). All conditions met.
3
Conclusion: The series converges conditionally.

Example 3: Ratio Test Approach

Determine convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n \cdot n}{3^n}\).

1
Ratio Test on \(|a_n|\): \(\displaystyle L = \lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|} = \lim_{n\to\infty}\frac{n+1}{3^{n+1}} \cdot \frac{3^n}{n} = \frac{1}{3}\lim_{n\to\infty}\frac{n+1}{n} = \frac{1}{3}\).
2
Since \(L = \frac{1}{3} < 1\), the series \(\sum |a_n|\) converges.
3
Conclusion: The series converges absolutely.

Example 4: \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{\sqrt{n}}\)

1
Absolute values: \(\displaystyle\sum \frac{1}{\sqrt{n}} = \sum \frac{1}{n^{1/2}}\) is a \(p\)-series with \(p = \frac{1}{2} \le 1\), so it diverges.
2
Alternating Series Test: \(b_n = \frac{1}{\sqrt{n}}\) is positive, decreasing, and \(\to 0\). So \(\sum a_n\) converges.
3
Verdict: Converges conditionally.

Your Convergence Toolbox

Ratio Test
Best for factorials, exponentials: compute \(\lim |a_{n+1}/a_n|\).
\(p\)-Series
\(\sum 1/n^p\) converges iff \(p > 1\). Quick comparison target.
AST
Alternating Series Test: checks the original series when \(\sum|a_n|\) diverges.

Common Mistakes

✕ Stopping at AST
Showing the Alternating Series Test passes, then writing "converges" without testing \(\sum|a_n|\). You must determine absolute vs. conditional.
✕ Ratio Test = 1
When the Ratio Test gives \(L = 1\), it is inconclusive. Choosing "converges" or "diverges" from \(L=1\) alone is a common error.
✕ Wrong direction
Saying "it diverges absolutely, so it diverges." Wrong: \(\sum|a_n|\) diverging does NOT mean \(\sum a_n\) diverges. You still must check the original.

Question 1 of 5

The series \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^3}\) is:

Question 2 of 5

The alternating harmonic series \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\) is:

Question 3 of 5

If the Ratio Test gives \(\displaystyle L = \lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|} = 1\), what can you conclude?

Question 4 of 5

Determine the convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n \cdot n!}{2^n}\).

Question 5 of 5

Which statement is always true?