The Central Idea

Suppose

\[ \sum_{n=1}^{\infty} a_n \]

contains both positive and negative terms.

We can ask two different questions:

  1. Does the original signed series converge?
  2. Does the positive series of magnitudes \(\sum |a_n|\) converge?

Those two answers determine the convergence classification.

Absolute Convergence

A series

\[ \sum_{n=1}^{\infty} a_n \]

is absolutely convergent if

\[ \boxed{\sum_{n=1}^{\infty}|a_n|} \]

converges.

Conditional Convergence

A series is conditionally convergent if the original series converges but the series of absolute values diverges.

\[ \boxed{ \sum a_n \text{ converges} \quad\text{but}\quad \sum |a_n| \text{ diverges} } \]

Divergence Is a Third Possibility

Not every signed series is absolutely or conditionally convergent.

If

\[ \sum a_n \]

diverges, then the series receives neither convergence classification.

The Three Main Outcomes

Absolute Convergence Implies Convergence

One of the most important facts about infinite series is:

\[ \boxed{ \sum |a_n| \text{ converges} \Longrightarrow \sum a_n \text{ converges} }. \]

Therefore, once absolute convergence is established, no separate convergence proof for the signed series is required.

Why Absolute Convergence Is Stronger

Absolute convergence controls the total size of the terms without relying on cancellation.

If even

\[ |a_1|+|a_2|+|a_3|+\cdots \]

has a finite sum, then allowing signs can only produce a more controlled total.

A Useful Inequality

For every finite collection of terms,

\[ \left|\sum_{k=m}^{n} a_k\right| \le \sum_{k=m}^{n}|a_k|. \]

This triangle-inequality relationship is the basic reason absolute convergence forces ordinary convergence.

Cauchy View of the Proof

If

\[ \sum |a_n| \]

converges, then its tails can be made arbitrarily small:

\[ \sum_{k=m}^{n}|a_k|<\varepsilon \]

for sufficiently large \(m\) and \(n\).

Then

\[ \left|\sum_{k=m}^{n}a_k\right| \le \sum_{k=m}^{n}|a_k| <\varepsilon. \]

So the signed series satisfies the Cauchy criterion as well.

The Converse Is False

A convergent series does not have to converge absolutely.

The standard example is the alternating harmonic series.

The Alternating Harmonic Series

\[ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n = 1-\frac12+\frac13-\frac14+\cdots \]

converges by the Alternating Series Test.

Remove the Signs

Taking absolute values gives

\[ \sum_{n=1}^{\infty}\left|\frac{(-1)^{n-1}}n\right| = \sum_{n=1}^{\infty}\frac1n. \]

This is the harmonic series, which diverges.

Therefore the Alternating Harmonic Series Is Conditional

Since the signed series converges but the absolute-value series diverges,

\[ \boxed{ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n \text{ converges conditionally} }. \]

An Absolutely Convergent Example

Consider

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

Taking absolute values gives

\[ \sum_{n=1}^{\infty}\frac1{n^2}, \]

a convergent \(p\)-series with \(p=2\).

Therefore the original series converges absolutely.

Absolute Convergence Makes Alternation Secondary

If a signed series converges absolutely, its alternating structure is no longer needed to prove convergence.

The convergence already survives after all signs are removed.

Classification Workflow

For a signed series \(\sum a_n\), a reliable workflow is:

  1. Check the term condition \(a_n\to0\).
  2. Study the positive series \(\sum |a_n|\).
  3. If \(\sum |a_n|\) converges, classify the series as absolutely convergent.
  4. If \(\sum |a_n|\) diverges, determine whether the original signed series still converges.
  5. If it does, classify it as conditionally convergent.
  6. If it does not, classify it as divergent.

Why Check Absolute Convergence First?

Positive-term tests are often powerful and direct.

If they establish

\[ \sum |a_n| \]

as convergent, the classification is complete immediately.

Using the \(p\)-Series Benchmark

Recall:

\[ \sum_{n=1}^{\infty}\frac1{n^p} \]

converges when

\[ p>1 \]

and diverges when

\[ p\le1. \]

Alternating \(p\)-Series Classification

Consider

\[ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^p}. \]

The classification depends on \(p\).

Case 1: \(p>1\)

If \(p>1\), then

\[ \sum\frac1{n^p} \]

converges.

Therefore

\[ \sum\frac{(-1)^{n-1}}{n^p} \]

converges absolutely.

Case 2: \(0

If \(0 \[ \sum\frac1{n^p} \]

diverges, but the alternating magnitudes decrease to zero.

Therefore the signed series converges conditionally.

Case 3: \(p\le0\)

If \(p\le0\), then

\[ \frac1{n^p} \]

does not approach zero.

Therefore the signed series diverges by the term test.

A Complete \(p\)-Series Classification

\[ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^p} \]

is:

Using Direct Comparison

To prove absolute convergence, apply a positive-term comparison to

\[ |a_n|. \]

If

\[ 0\le |a_n|\le b_n \]

and \(\sum b_n\) converges, then \(\sum |a_n|\) converges.

Example: Direct Comparison for Absolute Convergence

Consider

\[ \sum_{n=1}^{\infty}(-1)^n\frac1{n^2+3}. \]

Since

\[ 0\le\frac1{n^2+3}\le\frac1{n^2}, \]

and \(\sum 1/n^2\) converges, the original series converges absolutely.

Using Limit Comparison

If the magnitude \(|a_n|\) behaves like a familiar positive benchmark \(b_n\), the limit comparison test can classify absolute convergence.

Example: Limit Comparison

Consider

\[ \sum_{n=1}^{\infty}(-1)^n\frac{3n+1}{n^3+4}. \]

For large \(n\),

\[ \frac{3n+1}{n^3+4}\sim\frac3{n^2}. \]

Since \(\sum 1/n^2\) converges, the series converges absolutely.

Using the Ratio Test

For a signed series, the ratio test naturally analyzes absolute values:

\[ L= \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|. \]

If \(L<1\), the series converges absolutely.

Ratio Test Outcomes

If

\[ L= \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|, \]

then:

Example: Factorial Terms

Consider

\[ \sum_{n=1}^{\infty}(-1)^n\frac{n!}{3^n n^2}. \]

The ratio test gives

\[ \left| \frac{a_{n+1}}{a_n} \right| = \frac{n+1}{3}\cdot\frac{n^2}{(n+1)^2} = \frac{n^2}{3(n+1)}. \]

This tends to infinity, so the terms do not produce a convergent series.

Example: Factorial in the Denominator

Consider

\[ \sum_{n=1}^{\infty}(-1)^n\frac{2^n}{n!}. \]

Then

\[ \left| \frac{a_{n+1}}{a_n} \right| = \frac{2}{n+1}\to0. \]

Therefore the series converges absolutely.

Using the Root Test

The root test studies

\[ L= \lim_{n\to\infty}\sqrt[n]{|a_n|}. \]

If \(L<1\), the series converges absolutely.

Root Test Outcomes

Example: Powers to the \(n\)-th Degree

Consider

\[ \sum_{n=1}^{\infty}(-1)^n \left(\frac{2n}{3n+1}\right)^n. \]

The root test gives

\[ \sqrt[n]{|a_n|} = \frac{2n}{3n+1} \to\frac23. \]

Since \(2/3<1\), the series converges absolutely.

Absolute Convergence of Geometric-Type Series

Consider

\[ \sum_{n=0}^{\infty}(-r)^n. \]

Taking absolute values gives

\[ \sum_{n=0}^{\infty}|r|^n. \]

Therefore the series converges absolutely whenever

\[ |r|<1. \]

Signed Geometric Series Are Usually Absolute

A convergent geometric series with alternating signs still has

\[ |r|<1, \]

so its absolute-value series is another convergent geometric series.

Conditional Convergence Requires Cancellation

A conditionally convergent series cannot survive after all signs are removed.

Its convergence depends essentially on the arrangement of positive and negative contributions.

Positive and Negative Parts

Any real term \(a_n\) can be decomposed into positive and negative parts:

\[ a_n=a_n^+-a_n^-, \]

where

\[ a_n^+=\max(a_n,0) \]

and

\[ a_n^-=\max(-a_n,0). \]

Absolute Value in Terms of Positive and Negative Parts

\[ |a_n|=a_n^++a_n^-. \]

This decomposition helps explain why absolute convergence is stronger than ordinary convergence.

Absolute Convergence Controls Both Sides

If

\[ \sum |a_n| \]

converges, then both

\[ \sum a_n^+ \]

and

\[ \sum a_n^- \]

converge separately.

Conditional Convergence Has a Different Balance

For a conditionally convergent real series, the positive contributions and negative contributions do not each have finite total magnitude.

Their interaction through cancellation is essential.

Rearrangement Becomes Important

Absolute convergence is stable under rearrangement: reordering the terms does not change the sum.

Conditional convergence is much more delicate.

Rearrangement of Absolutely Convergent Series

If

\[ \sum a_n \]

converges absolutely, then any rearrangement of its terms converges to the same sum.

Why Rearrangement Is Safe Under Absolute Convergence

Since the total magnitude

\[ \sum |a_n| \]

is finite, moving terms changes the order of accumulation but cannot create an uncontrolled amount of additional positive or negative mass.

Conditional Rearrangement Is Not Safe

A conditionally convergent series may change its sum when its terms are rearranged.

This is one reason conditional convergence is viewed as weaker.

Riemann Rearrangement Phenomenon

For a conditionally convergent real series, appropriate rearrangements can produce dramatically different behavior.

A later analysis course may prove the full theorem. Here, the practical lesson is: term order matters for conditional convergence.

Grouping and Rearrangement Are Different

Grouping consecutive terms without changing their order is not the same as arbitrarily rearranging terms.

Keep this distinction clear when manipulating infinite series.

Finite Rearrangements Do Not Matter

Changing the position of only finitely many terms does not affect convergence or the total sum.

The delicate issue concerns infinite rearrangements.

Absolute Convergence and Linear Combinations

If

\[ \sum a_n \quad\text{and}\quad \sum b_n \]

converge absolutely, then for constants \(c\) and \(d\),

\[ \sum(ca_n+db_n) \]

also converges absolutely.

Why Linear Combinations Stay Absolute

By the triangle inequality,

\[ |ca_n+db_n| \le |c||a_n|+|d||b_n|. \]

The right side belongs to a convergent positive-term series.

Term Limits Still Come First

Before classifying a signed series, always check whether

\[ a_n\to0. \]

If not, the series diverges immediately.

Example: Divergence Before Classification

Consider

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

Since

\[ \frac{n+1}{n}\to1, \]

the terms do not approach zero. The series diverges.

Example: Conditional by Alternating Series Test

Consider

\[ \sum_{n=2}^{\infty}\frac{(-1)^n}{\ln n}. \]

The magnitude \(1/\ln n\) decreases to zero, so the signed series converges.

The positive series

\[ \sum_{n=2}^{\infty}\frac1{\ln n} \]

diverges, so the original series is conditionally convergent.

Example: Absolute by Comparison

Consider

\[ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{n+1}{n^3+2}. \]

For large \(n\), the magnitude behaves like

\[ \frac1{n^2}. \]

A comparison or limit comparison with \(\sum1/n^2\) proves absolute convergence.

Example: Conditional with a Radical

Consider

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

The signed series converges by the Alternating Series Test.

But

\[ \sum\frac1{\sqrt n} \]

is a \(p\)-series with \(p=1/2\), so it diverges.

Therefore the original series converges conditionally.

Example: Absolute with a Radical

Consider

\[ \sum_{n=1}^{\infty}\frac{(-1)^n}{n^{3/2}}. \]

The absolute-value series

\[ \sum\frac1{n^{3/2}} \]

converges because \(3/2>1\). Therefore the original series converges absolutely.

Example: Rational Expression Near \(1/n\)

Consider

\[ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{n}{n^2+1}. \]

The magnitude behaves like \(1/n\), so the absolute-value series diverges.

The magnitudes decrease eventually and approach zero, so the signed series converges by the Alternating Series Test.

Therefore it converges conditionally.

Example: Rational Expression Near \(1/n^2\)

Consider

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

The magnitude behaves like \(1/n^2\).

Therefore the absolute-value series converges, and the original series converges absolutely.

Asymptotic Thinking Helps Classification

When

\[ |a_n|\sim\frac{C}{n^p}, \]

the exponent \(p\) often predicts whether absolute convergence is possible.

Conditional Convergence Needs a Second Step

Showing

\[ \sum |a_n| \]

diverges does not prove conditional convergence.

You must still prove that

\[ \sum a_n \]

converges.

Common Error: Calling Every Alternating Series Conditional

Alternating signs do not imply conditional convergence.

For example,

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

is alternating but converges absolutely.

Common Error: Stopping After Absolute Divergence

If

\[ \sum |a_n| \]

diverges, the original signed series may still converge conditionally.

Continue the analysis.

Common Error: Assuming Absolute Divergence Means Original Divergence

The implication

\[ \sum |a_n| \text{ diverges} \Longrightarrow \sum a_n \text{ diverges} \]

is false.

Common Error: Reversing the Main Theorem

It is true that

\[ \sum |a_n|\text{ convergent} \Longrightarrow \sum a_n\text{ convergent}. \]

The reverse implication is false.

Common Error: Ignoring Absolute Values in the Ratio Test

For signed terms, use

\[ \left|\frac{a_{n+1}}{a_n}\right|. \]

The ratio test is fundamentally a test of magnitude.

Common Error: Calling a Divergent Series Conditional

Conditional convergence requires actual convergence of the original series.

A divergent alternating series is simply divergent.

Common Error: Forgetting Eventual Behavior

A finite number of early irregular terms does not alter absolute or conditional convergence.

Classification is governed by the infinite tail.

A Reliable Classification Procedure

  1. Check whether \(a_n\to0\).
  2. Form the positive series \(\sum |a_n|\).
  3. Use a positive-term test on \(\sum |a_n|\).
  4. If it converges, stop: the original series converges absolutely.
  5. If it diverges, analyze the original signed series.
  6. If the original converges, classify it as conditional.
  7. If the original diverges, classify it as divergent.

Worked Example: Absolute Convergence

Classify

\[ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n^4+1}. \]

Since

\[ 0\le\frac1{n^4+1}\le\frac1{n^4}, \]

and \(\sum1/n^4\) converges, the series converges absolutely.

Worked Example: Conditional Convergence

Classify

\[ \sum_{n=1}^{\infty}(-1)^{n-1}\frac1{\sqrt n}. \]

The signed series converges by the Alternating Series Test.

The absolute-value series is

\[ \sum\frac1{n^{1/2}}, \]

which diverges.

Therefore the original series converges conditionally.

Worked Example: Divergence

Classify

\[ \sum_{n=1}^{\infty}(-1)^n\frac{n+2}{n+1}. \]

Since

\[ \frac{n+2}{n+1}\to1, \]

the terms do not approach zero. The series diverges.

Worked Example: Ratio Test Classification

Classify

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

The ratio of magnitudes is

\[ \frac{3^{n+1}/(n+1)!}{3^n/n!} = \frac3{n+1}\to0. \]

Therefore the series converges absolutely.

Worked Example: Limit Comparison Classification

Classify

\[ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{2n+5}{n^2+3}. \]

The magnitude behaves like

\[ \frac2n. \]

Thus the absolute-value series diverges by limit comparison with the harmonic series.

The magnitude approaches zero and is eventually decreasing, so the signed series converges by the Alternating Series Test.

Therefore it converges conditionally.

Practice 1: Absolute or Conditional?

Classify

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

Practice 2: Harmonic-Type Behavior

Classify

\[ \sum_{n=1}^{\infty}\frac{(-1)^n}{n}. \]

Practice 3: Radical Denominator

Classify

\[ \sum_{n=1}^{\infty}\frac{(-1)^n}{n^{3/4}}. \]

Practice 4: Rational Comparison

Classify

\[ \sum_{n=1}^{\infty}(-1)^{n-1}\frac1{n^2+7}. \]

Practice 5: Rational Harmonic Behavior

Classify

\[ \sum_{n=1}^{\infty}(-1)^n\frac{n+1}{n^2+1}. \]

Practice 6: Term Test

Classify

\[ \sum_{n=1}^{\infty}(-1)^n\frac{2n+1}{n+4}. \]

Practice 7: Factorial Series

Classify

\[ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{4^n}{n!}. \]

Practice 8: Root-Test Structure

Classify

\[ \sum_{n=1}^{\infty}(-1)^n \left(\frac{n}{2n+1}\right)^n. \]

Practice 9: Determine the Classification Range

For what values of \(p\) does

\[ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^p} \]

converge absolutely, converge conditionally, or diverge?

Practice 10: Explain the Logic

A student proves that

\[ \sum |a_n| \]

diverges and immediately concludes that

\[ \sum a_n \]

diverges. Explain the error.

Practice Answers

Practice 1

The absolute-value series is

\[ \sum\frac1{n^3}, \]

which converges. Therefore the series converges absolutely.

Practice 2

The alternating harmonic series converges, while

\[ \sum\frac1n \]

diverges. Therefore the series converges conditionally.

Practice 3

The signed series converges by the Alternating Series Test because

\[ \frac1{n^{3/4}}\to0 \]

and decreases. The absolute-value series is a \(p\)-series with \(p=3/4\), so it diverges. Therefore the series converges conditionally.

Practice 4

Since

\[ 0\le\frac1{n^2+7}\le\frac1{n^2}, \]

the absolute-value series converges. Therefore the series converges absolutely.

Practice 5

The magnitude

\[ \frac{n+1}{n^2+1} \]

behaves like \(1/n\), so the absolute-value series diverges. The magnitudes approach zero and are eventually decreasing, so the signed series converges by the Alternating Series Test. Therefore it converges conditionally.

Practice 6

Since

\[ \frac{2n+1}{n+4}\to2, \]

the terms do not approach zero. The series diverges.

Practice 7

The ratio of absolute values is

\[ \frac{4}{n+1}\to0. \]

Therefore the series converges absolutely.

Practice 8

The root test gives

\[ \sqrt[n]{|a_n|} = \frac{n}{2n+1} \to\frac12. \]

Since \(1/2<1\), the series converges absolutely.

Practice 9

Practice 10

Divergence of

\[ \sum |a_n| \]

means only that absolute convergence fails. The original signed series may still converge through cancellation. A separate convergence test must be applied to

\[ \sum a_n. \]

What to Remember

The main hierarchy is:

\[ \boxed{ \text{absolute convergence} \Longrightarrow \text{convergence} }. \]

But the reverse implication is false.

A practical classification strategy is to analyze the absolute-value series first, then return to the signed series only if absolute convergence fails.

Where We Go Next

Convergence tests tell us whether a numerical series has a finite sum. The next step is to let the terms depend on a variable.

The next lesson, Power Series, studies infinite polynomial-like expressions of the form

\[ \sum_{n=0}^{\infty}c_n(x-a)^n, \]

where convergence can change as \(x\) changes.