10.9 - Determining Absolute or Conditional Convergence

Three Types of Convergence Behavior

When you test a series for convergence, there are three possible outcomes.

1. Absolute Convergence

\text{If } \sum |a_n| \text{ converges} \implies \sum a_n \text{ converges (absolutely)}

When the series with absolute values converges, the original series converges too, and we say it converges absolutely.

2. Conditional Convergence

\text{If } \sum |a_n| \text{ diverges BUT } \sum a_n \text{ converges} \implies \text{conditional convergence}

When the absolute value series diverges but the original one converges, we say it converges conditionally.

3. Divergence

\text{If both } \sum |a_n| \text{ and } \sum a_n \text{ diverge} \implies \text{the series diverges}

When both diverge, the series just diverges. There’s no conditional or absolute label for it.

Testing Strategy: Always Test the Absolute Value First

Step 1: Test

\sum |a_n|

, the series with absolute values.

Step 2: Based on the result:

If $\sum |a_n|$ converges, you’re done. The series converges absolutely

If $\sum |a_n|$ diverges, test the original series $\sum a_n$

If $\sum a_n$ converges, it’s conditional convergence
If $\sum a_n$ diverges, the series diverges

Why test the absolute value first? Because if it converges, you’re immediately done and never have to touch the original.

Example 1: Absolute Convergence

Determine whether

\sum_{n=1}^{\infty} \dfrac{(-1)^n \cdot 3^n}{n!}

converges absolutely, converges conditionally, or diverges.

Step 1: Test the absolute value series.

\sum_{n=1}^{\infty} \left|\frac{(-1)^n \cdot 3^n}{n!}\right| = \sum_{n=1}^{\infty} \frac{3^n}{n!}

The absolute value strips off the alternating sign.

Step 2: Apply the ratio test.

L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{\frac{3^{n+1}}{(n+1)!}}{\frac{3^n}{n!}}

= \lim_{n \to \infty} \frac{3^{n+1}}{(n+1)!} \cdot \frac{n!}{3^n}

= \lim_{n \to \infty} \frac{3^n \cdot 3}{(n+1) \cdot n!} \cdot \frac{n!}{3^n} = \lim_{n \to \infty} \frac{3}{n+1} = 0

Step 3: Conclude.

Since

L = 0 < 1

, the series

\sum \frac{3^n}{n!}

converges by the ratio test.

\sum \frac{(-1)^n \cdot 3^n}{n!}

converges absolutely, and there’s no need to test the original series.

Example 2: Conditional Convergence

Determine whether

\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{\sqrt[3]{n}}

converges absolutely, converges conditionally, or diverges.

Step 1: Test the absolute value series.

\sum_{n=1}^{\infty} \left|\frac{(-1)^{n+1}}{\sqrt[3]{n}}\right| = \sum_{n=1}^{\infty} \frac{1}{n^{1/3}}

Step 2: Recognize it as a p-series.

This is a p-series with

p = \frac{1}{3}

. Since

p = \frac{1}{3} < 1

, it diverges.

Step 3: Test the original series.

Since the absolute value version diverged, we have to test the original alternating series:

\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt[3]{n}}

Apply the Alternating Series Test:

Condition 1:

\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{n^{1/3}} = 0

\checkmark

Condition 2: is

a_n = \frac{1}{n^{1/3}}

decreasing?

a_1 = 1, \quad a_2 = \frac{1}{\sqrt[3]{2}}, \quad a_3 = \frac{1}{\sqrt[3]{3}}, \ldots

Yes, the terms are decreasing.

\checkmark

Conclusion: the original series converges by the AST, but the absolute value series diverges. So the series converges conditionally.

Example 3: Divergence

Determine whether

\sum_{n=1}^{\infty} \dfrac{(-1)^n \cdot n}{n+2}

converges absolutely, converges conditionally, or diverges.

Step 1: Test the absolute value series.

\sum_{n=1}^{\infty} \left|\frac{(-1)^n \cdot n}{n+2}\right| = \sum_{n=1}^{\infty} \frac{n}{n+2}

Step 2: Apply the nth term test.

\lim_{n \to \infty} \frac{n}{n+2} = \lim_{n \to \infty} \frac{1}{1 + \frac{2}{n}} = 1 \neq 0

Since the limit isn’t 0, the series diverges by the nth term test.

Step 3: Test the original series.

\lim_{n \to \infty} \frac{(-1)^n \cdot n}{n+2}

n \to \infty

, the fraction

\frac{n}{n+2} \to 1

, but

(-1)^n

keeps flipping between

-1

and

1

. So the limit does not exist, since it oscillates between

-1

and

1

Since

\lim_{n \to \infty} a_n \neq 0

, the original series diverges by the nth term test too.

Conclusion: both the absolute value series and the original diverge, so the series simply diverges.

Finding Intervals of Absolute Convergence

For series with a variable

x

in them, we can find the interval where the series converges absolutely.

Example 4: Finding the Interval of Absolute Convergence

For what values of

x

does

\sum_{n=1}^{\infty} \dfrac{(-1)^n \cdot n(x+4)^n}{6^n}

converge absolutely?

Step 1: Apply the ratio test with absolute values.

L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left|\frac{(n+1)(x+4)^{n+1}}{6^{n+1}} \cdot \frac{6^n}{n(x+4)^n}\right|

Step 2: Simplify.

= \lim_{n \to \infty} \left|\frac{n+1}{n} \cdot \frac{(x+4)^n \cdot (x+4)}{6^n \cdot 6} \cdot \frac{6^n}{(x+4)^n}\right| = \lim_{n \to \infty} \left|\frac{n+1}{n} \cdot \frac{x+4}{6}\right|

Step 3: Evaluate the limit.

= \lim_{n \to \infty} \frac{n+1}{n} \cdot \left|\frac{x+4}{6}\right| = 1 \cdot \left|\frac{x+4}{6}\right| = \left|\frac{x+4}{6}\right|

Step 4: For convergence, we need

L < 1

\left|\frac{x+4}{6}\right| < 1

Step 5: Solve the absolute value inequality.

-1 < \frac{x+4}{6} < 1

Multiply by 6:

-6 < x+4 < 6

Subtract 4:

-10 < x < 2

Answer: the series converges absolutely for

x \in (-10, 2)

Finding Points of Conditional Convergence

Once you’ve found the interval of absolute convergence, test the endpoints to check for conditional convergence.

Example 5: Absolute and Conditional Convergence Intervals

For what values of

x

does

\sum_{n=1}^{\infty} \dfrac{(-1)^n(x-1)^n}{n}

converge absolutely? Conditionally?

Step 1: Find the interval of absolute convergence.

L = \lim_{n \to \infty} \left|\frac{(x-1)^{n+1}}{n+1} \cdot \frac{n}{(x-1)^n}\right| = \lim_{n \to \infty} \left|\frac{n(x-1)}{n+1}\right| = |x-1| \lim_{n \to \infty} \frac{n}{n+1} = |x-1|

For convergence we need

|x-1| < 1

-1 < x-1 < 1 \quad \Rightarrow \quad 0 < x < 2

Absolute convergence on $(0, 2)$ .

Step 2: Test the endpoints for conditional convergence.

At $x = 0$ :

\sum_{n=1}^{\infty} \frac{(-1)^n(0-1)^n}{n} = \sum_{n=1}^{\infty} \frac{(-1)^n \cdot (-1)^n}{n} = \sum_{n=1}^{\infty} \frac{(-1)^{2n}}{n} = \sum_{n=1}^{\infty} \frac{1}{n}

That’s the harmonic series, which diverges. So no convergence at

x = 0

At $x = 2$ :

\sum_{n=1}^{\infty} \frac{(-1)^n(2-1)^n}{n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}

That’s the alternating harmonic series, which converges by the AST. It doesn’t converge absolutely there, because the absolute value series is

\sum \frac{1}{n}

, the harmonic series, which diverges. So it converges conditionally at $x = 2$ .

An endpoint can also converge absolutely (for example,

\sum \frac{(x-1)^n}{n^2}

x = 0

and

x = 2

, where it becomes

\sum \frac{(\pm 1)^n}{n^2}

). At each endpoint, test

\sum |a_n|

first, just like any other series.

Final answer:

Absolute convergence: $0 < x < 2$

Conditional convergence: $x = 2$

Diverges: $x \leq 0$ and $x > 2$

Important Theorems

Absolute Convergence Implies Convergence

\text{If } \sum |a_n| \text{ converges} \implies \sum a_n \text{ converges}

This is always true. Absolute convergence is a stronger condition than regular convergence.

The Converse is NOT True

\sum a_n \text{ converges} \not\Rightarrow \sum |a_n| \text{ converges}

Counterexample: the alternating harmonic series converges, but the plain harmonic series diverges.

Rearranging the Terms

If a series converges absolutely, you can regroup or rearrange its terms in any order, and the sum stays the same. That’s how finite sums work, and absolutely convergent series behave the same way.

Conditionally convergent series don’t get this guarantee. Their positive terms add up to

\infty

and their negative terms add up to

-\infty

. The finite sum only comes from the way they cancel in the given order, so changing the order can change the sum.

Example: the alternating harmonic series converges to

\ln 2

. Rearrange it so each positive term is followed by two negative terms:

1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} - \frac{1}{8} + \frac{1}{5} - \frac{1}{10} - \frac{1}{12} + \cdots

Combine each positive term with the negative term right after it:

1 - \frac{1}{2} = \frac{1}{2}

\frac{1}{3} - \frac{1}{6} = \frac{1}{6}

\frac{1}{5} - \frac{1}{10} = \frac{1}{10}

, and so on. The series becomes

\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \frac{1}{8} + \frac{1}{10} - \frac{1}{12} + \cdots = \frac{1}{2}\left(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots\right) = \frac{1}{2}\ln 2

Same terms, different order, and the sum dropped from

\ln 2

\frac{1}{2}\ln 2

. That’s why the word absolutely matters.

Summary of All the Tests You Might Use

Test	When to Use	Result
nth term test	Always try it first	Can only prove divergence
p-series test	$\sum \frac{1}{n^p}$	Gives a direct answer
Geometric series	$\sum ar^n$	Gives a direct answer
Ratio test	Factorials, exponentials	$L<1$ converges, $L>1$ diverges
Alternating series test	Alternating signs	Proves convergence
Comparison tests	You can bound it by a known series	Proves convergence or divergence
Integral test	The function is easy to integrate	Both behave the same way

Strategy Flowchart


Start with the series Σ aₙ
         ↓
Test Σ |aₙ| (absolute value)
         ↓
    Does it converge?
    /              \
  YES               NO
   ↓                 ↓
CONVERGES      Test the original Σ aₙ
ABSOLUTELY          ↓
(DONE)         Does it converge?
              /              \
            YES               NO
             ↓                 ↓
         CONVERGES        DIVERGES
       CONDITIONALLY      (neither)

Common Mistakes to Avoid

Testing the original series first. Always test the absolute value first, because it saves time when it converges.

Forgetting that absolute value kills the alternation.

\sum |(-1)^n a_n| = \sum a_n

, meaning just the positive terms.

Assuming convergence means absolute convergence. You have to test the absolute value series separately.

Not checking the endpoints for conditional convergence. The endpoints of an interval can converge conditionally.

Confusing the definitions. Absolute means

\sum |a_n|

converges. Conditional means

\sum |a_n|

diverges but

\sum a_n

converges.

Key Takeaways

Test the absolute value first. If $\sum |a_n|$ converges, you’re done, and it’s absolute convergence

Absolute implies convergence. If a series converges absolutely, it converges

Rearranging is safe only with absolute convergence. Reorder or regroup an absolutely convergent series and the sum stays the same. A conditionally convergent series can change its sum

Conditional convergence is when $\sum |a_n|$ diverges but $\sum a_n$ converges

Alternating series are often conditional. Strip off the alternation and they may diverge

The classic example: the alternating harmonic series converges conditionally

For intervals: find the interval of absolute convergence first, then test the endpoints

Endpoints matter. They may converge conditionally at the boundary

All the previous tests still apply: ratio, comparison, AST, p-series, and so on

Absolute value removes signs: $|(-1)^n a_n| = a_n$

Only three outcomes exist: absolute convergence, conditional convergence, or divergence

Strategy hierarchy: absolute first, then the original if you need it

There’s no such thing as “conditional divergence.” It either converges (absolutely or conditionally) or it diverges

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