Determining Absolute or Conditional Convergence

3. Visual Connection

5. Concept Development

BC-only topic. A sign-changing series can converge because its positive and negative terms cancel, or it can converge even after every sign is made positive. Absolute and conditional convergence distinguish these two situations.

Three possible classifications

• Absolute convergence: \(\sum|a_n|\) converges.
• Conditional convergence: \(\sum a_n\) converges but \(\sum|a_n|\) diverges.
• Divergence: \(\sum a_n\) does not converge.

“Conditionally divergent” is not a valid classification.

Absolute convergence implies convergence

The converse is false: the alternating harmonic series converges, while the harmonic series formed from its absolute values diverges.

Why the theorem works

Separate each term into positive and negative parts:

Both \(a_n^+\) and \(a_n^-\) lie between \(0\) and \(|a_n|\). If \(\sum|a_n|\) converges, comparison shows that \(\sum a_n^+\) and \(\sum a_n^-\) converge, so their difference \(\sum a_n\) converges.

A reliable classification workflow

1. Check whether \(a_n\to0\). A nonzero or nonexistent term limit proves divergence immediately.
2. Form the magnitude series \(\sum|a_n|\).
3. Use a positive-term test on \(\sum|a_n|\): geometric, \(p\)-series, comparison, limit comparison, Integral Test, or Ratio Test.
4. If \(\sum|a_n|\) converges, stop: the original series converges absolutely.
5. If \(\sum|a_n|\) diverges, return to \(\sum a_n\). Its behavior is still undecided.
6. Use a sign-sensitive argument, often the Alternating Series Test, to decide whether the original series converges conditionally or diverges.

Why divergence of the absolute series is not enough

From \(\sum|a_n|\) diverging, one may conclude only that \(\sum a_n\) is not absolutely convergent. Cancellation may still make the signed series converge. A second argument about the original series is required.

Useful magnitude tests

• If \(|a_n|\) behaves like \(1/n^p\), use the \(p\)-series test or limit comparison.
• If factorials or fixed-base exponentials occur, try the Ratio Test.
• If a bounded factor appears, use a bound such as \(|\sin n|\le1\) or \(|\cos n|\le1\).
• If \(|a_n|\) is a positive continuous decreasing expression, the Integral Test may be efficient.

Alternating p-series map

For \(p>0\), the magnitudes \(1/n^p\) decrease to zero, so

If \(p\le0\), the terms do not approach zero and the series diverges.

Alternating does not mean conditional

An alternating series may be absolutely convergent, conditionally convergent, or divergent. Its sign pattern alone does not determine the classification.

Nonalternating sign changes

Absolute convergence can be proved even when signs do not alternate regularly. For example, a comparison bound on \(|\cos n|/n^2\) avoids any need to understand the irregular signs of \(\cos n\).

Rearranging terms

An absolutely convergent series keeps the same sum under every rearrangement or regrouping of its terms. Conditional convergence is more delicate: changing the order can change the behavior or value, so finite-sum algebra cannot be applied carelessly to such a series.

Finite changes

Adding, removing, or changing finitely many terms affects the numerical sum but not whether convergence is absolute, conditional, or divergent.

AP-style justification checklist

1. Name the series being tested: \(\sum a_n\) or \(\sum|a_n|\).
2. State the hypotheses required by the chosen test.
3. Give a conclusion for the magnitude series.
4. If the magnitude series diverges, separately justify convergence or divergence of the signed series.
5. Finish with exactly one classification: absolute, conditional, or divergent.

6. Detailed Worked Example and Error Check

Example 1: Absolute convergence of an alternating p-series

For \(\sum(-1)^{n+1}/n^2\),

The magnitude series is a convergent \(p\)-series with \(p=2\). Therefore the original series converges absolutely.

Example 2: The alternating harmonic series

The magnitudes \(1/n\) decrease to zero, so \(\sum(-1)^{n+1}/n\) converges by the Alternating Series Test. However, \(\sum1/n\) diverges. The original series converges conditionally.

Example 3: A fractional p-value

For \(\sum(-1)^n/n^{2/3}\), the signed series converges by the AST because \(1/n^{2/3}\) decreases to zero. Its magnitude series is a divergent \(p\)-series because \(p=2/3\le1\). Thus it converges conditionally.

Example 4: Limit comparison for the magnitude series

Consider \(\sum(-1)^n n/(n^2+1)\). The magnitudes eventually decrease to zero because \(f(x)=x/(x^2+1)\) has \(f'(x)=(1-x^2)/(x^2+1)^2<0\) for \(x>1\). The signed series converges by the AST. Also,

The magnitude series diverges by limit comparison with the harmonic series, so the original series converges conditionally.

Example 5: Ratio Test proves absolute convergence

For \(a_n=(-1)^n n/2^n\),

The series converges absolutely. There is no need to apply the AST afterward.

Example 6: Irregular signs but absolute convergence

Since \(|\cos n|\le1\),

The magnitude series converges by comparison with a \(p=2\) series. Hence \(\sum\cos n/n^2\) converges absolutely.

Example 7: Alternating but divergent

For \(\sum(-1)^n(n+1)/(n+2)\), the magnitudes approach \(1\). Thus the terms do not approach zero, and the series diverges by the nth term test.

Example 8: Shifted harmonic behavior

For \(\sum(-1)^{n+1}/(3n+1)\), the positive magnitudes decrease to zero, so the signed series converges by the AST. Meanwhile,

The magnitude series diverges by limit comparison with \(\sum1/n\). The original series therefore converges conditionally.

Example 9: A square-root denominator

For \(\sum(-1)^n/(\sqrt n+1)\), the magnitudes decrease to zero, giving signed convergence by the AST. Furthermore,

so the magnitude series diverges with the \(p=1/2\) series. The original series converges conditionally.

Example 10: Positive series terminology

The positive series \(\sum1/n^2\) is also absolutely convergent because \(|1/n^2|=1/n^2\). Absolute convergence is not restricted to alternating series.

Common errors

• Calling every alternating series conditionally convergent.
• Testing only \(\sum|a_n|\) and declaring the original series divergent when it diverges.
• Using the AST to claim absolute convergence.
• Forgetting that absolute convergence already proves convergence of the original series.
• Applying comparison directly to signed terms without first taking absolute values.
• Failing to verify decrease and zero limit before using the AST.
• Writing “not absolute” as though it automatically meant conditional.
• Using the term “conditionally divergent.”
• Ignoring the nth term test when the terms do not approach zero.
• Assuming an irregular sign-changing series must be divergent.
• Rearranging a conditionally convergent series as though it were a finite sum.