Conditional Convergence in Harmonic-Type Series

Signed Convergence and Magnitude Convergence

A series whose terms alternate in sign can converge even when its terms, if replaced by their magnitudes, form a divergent series. The cancellation between positive and negative terms can make the signed partial sums settle toward a finite limit, although the total accumulated magnitude keeps growing.

This distinction is captured by the difference between conditional convergence and absolute convergence. The alternating harmonic series is the basic example: its signed series converges, but its series of magnitudes is the harmonic series, which diverges.

Definition: A series \(\sum_{n=1}^{\infty} a_n\) is absolutely convergent if \(\sum_{n=1}^{\infty}|a_n|\) converges. It is conditionally convergent if \(\sum_{n=1}^{\infty}a_n\) converges but \(\sum_{n=1}^{\infty}|a_n|\) diverges.

Absolute convergence is sufficient for convergence: if the sum of the magnitudes converges, then the signed series converges as well. Conditional convergence describes the distinct case in which the signed series converges, but the series of magnitudes does not. The key question is therefore not just whether the signed partial sums have a finite limit, but also what happens after the signs are removed.

The Alternating Harmonic Series

The alternating harmonic series is

\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} = 1-\frac12+\frac13-\frac14+\frac15-\cdots.

Its term magnitudes are \(b_n=1/n\). These are positive, decrease as \(n\) increases, and approach zero. The Alternating-Series Test says that a series of the form \(\sum_{n=1}^{\infty}(-1)^{n-1}b_n\) converges if \(b_n\) is nonincreasing and \(b_n\to0\). Both conditions hold here, so the signed series converges.

To decide whether that convergence is absolute, take the magnitude of each term:

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

The series on the right is the harmonic series. Its divergence can be seen by grouping consecutive terms into blocks whose lengths double. For each integer \(j\geq 0\), the block from \(n=2^j\) through \(n=2^{j+1}-1\) contains \(2^j\) terms. Every term in that block is at least \(1/2^{j+1}\), so its sum is at least \(1/2\). Consequently, the partial sum through \(2^m\) is at least \(m/2\), and these partial sums are unbounded as \(m\) increases. Thus the harmonic series diverges.

Worked Example: Classifying the Alternating Harmonic Series

Consider \(\sum_{n=1}^{\infty}(-1)^{n-1}/n\). First, set \(b_n=1/n\). For every \(n\geq1\), \(b_n>0\), and

b_{n+1}=\frac{1}{n+1}<\frac1n=b_n, \qquad \lim_{n\to\infty}b_n=0.

The Alternating-Series Test therefore gives convergence of the signed series. For example, its first four partial sums are \(S_1=1\), \(S_2=1-1/2=1/2\), \(S_3=1/2+1/3=5/6\), and \(S_4=5/6-1/4=7/12\). These values illustrate the alternating behavior; the convergence conclusion follows from the test, not from checking finitely many sums.

For absolute convergence, the magnitude series is \(\sum_{n=1}^{\infty}1/n\). The block argument above shows it diverges. Hence the signed series converges, while its magnitude series diverges: the alternating harmonic series is conditionally convergent. This classification is checked against both parts of the definition.

Alternating p-Series

The same distinction gives a full classification for the alternating p-series

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

For \(p>0\), the positive term magnitudes \(b_n=1/n^p\) decrease to zero. The Alternating-Series Test therefore gives convergence of the signed series. The magnitude series is the p-series \(\sum_{n=1}^{\infty}1/n^p\), which converges exactly when \(p>1\). Thus the alternating p-series is absolutely convergent for \(p>1\), and conditionally convergent for \(0<p\leq1\).

If \(p=0\), the magnitudes are all \(1\); if \(p<0\), they do not approach zero. In either case the terms of the signed series fail to approach zero, so the series diverges by the term test. The complete classification is:

Exponent	Signed alternating series	Series of magnitudes	Classification
\(p>1\)	Converges	Converges	Absolutely convergent
\(0<p\leq1\)	Converges	Diverges	Conditionally convergent
\(p\leq0\)	Diverges	Diverges	Not convergent

Worked Example: The Alternating Series with Exponent One-Half

Classify \(\sum_{n=1}^{\infty}(-1)^{n-1}/\sqrt n\). Its positive magnitudes are \(b_n=1/\sqrt n\). They decrease because \(n+1>n>0\), which implies \(1/\sqrt{n+1}<1/\sqrt n\), and they approach zero because \(\sqrt n\) grows without bound. The Alternating-Series Test proves convergence of the signed series.

Removing the signs gives

\sum_{n=1}^{\infty}\left|\frac{(-1)^{n-1}}{\sqrt n}\right| = \sum_{n=1}^{\infty}\frac1{n^{1/2}}.

This is a p-series with \(p=1/2\), and \(1/2\leq1\), so it diverges. For a numerical check of the first terms, \(S_1=1\), \(S_2=1-1/\sqrt2\), and \(S_3=1-1/\sqrt2+1/\sqrt3\); these partial sums do not establish convergence by themselves. The test establishes signed convergence, while the p-series classification establishes magnitude divergence. Therefore the series is conditionally convergent.

A Rational Harmonic-Type Example

Not every term magnitude is exactly \(1/n^p\). A useful approach is to verify the alternating-series conditions directly, then compare the magnitude series with a familiar divergent series. The comparison must be made in the correct direction: to prove divergence, it is enough to show that the terms are at least as large as the terms of a divergent positive series.

Worked Example: A Rational Series with Harmonic-Size Terms

Consider

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

Set \(b_n=n/(n^2+1)\). Each \(b_n\) is positive, and \(b_n\to0\), since dividing numerator and denominator by \(n^2\) gives

\frac{n}{n^2+1} = \frac{1/n}{1+1/n^2} \longrightarrow 0.

To check that the magnitudes decrease, compare successive terms. All denominators are positive, so cross-multiplication preserves the inequality:

\begin{aligned} \frac{n+1}{(n+1)^2+1}<\frac{n}{n^2+1} &\Longleftrightarrow (n+1)(n^2+1)<n(n^2+2n+2)\\ &\Longleftrightarrow n^3+n^2+n+1<n^3+2n^2+2n\\ &\Longleftrightarrow 0<n^2+n-1. \end{aligned}

For every integer \(n\geq1\), \(n^2+n-1\geq1>0\). Thus \(b_n\) decreases, and the Alternating-Series Test proves convergence of the signed series. Its first terms are \(1/2-2/5+3/10-4/17+\cdots\); the test applies to the whole sequence, including its first term.

Now examine the magnitude series. For \(n\geq1\), \(n^2\geq1\), so \(n^2+1\leq2n^2\). Taking reciprocals of positive quantities and multiplying by \(n>0\) yields

\frac{n}{n^2+1}\geq\frac{n}{2n^2}=\frac1{2n}.

The comparison series \(\sum_{n=1}^{\infty}1/(2n)\) diverges because it is one-half of the harmonic series. Since every magnitude is at least the corresponding term of this divergent positive series, the magnitude series diverges by comparison. The signed series converges and the magnitude series diverges, so this series is conditionally convergent.

Why the Two Tests Answer Different Questions

The Alternating-Series Test uses the sign pattern and the behavior of the positive magnitudes to establish convergence of the signed series. It does not establish absolute convergence. Absolute convergence is a separate question: replace each term by its magnitude, then classify the resulting positive series.

This difference matters whenever cancellation is doing essential work. If the magnitude series diverges, the original series cannot be called absolutely convergent, even if its signed partial sums approach a finite limit. Conversely, if the magnitude series converges, absolute convergence already guarantees convergence of the original signed series.

Check the signed series.
For an alternating series, verify that its positive magnitudes decrease (at least from some index onward) and tend to zero. A finite initial segment does not affect convergence.

Form the magnitude series.
Replace each term by its absolute value, so every term is nonnegative.

Classify the result.
If both series converge, convergence is absolute. If the signed series converges but the magnitude series diverges, convergence is conditional.

A Common Mistake

A common error is to apply the Alternating-Series Test and then conclude that the series converges absolutely. The test proves convergence of the signed series only. For the alternating harmonic series, the mistaken reasoning would be:

\frac1n\searrow0 \quad\Longrightarrow\quad \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n \text{ converges absolutely.}

The first implication establishes signed convergence, but it does not justify the word “absolutely.” The correction is to check the magnitudes separately:

\sum_{n=1}^{\infty} \left|\frac{(-1)^{n-1}}n\right| = \sum_{n=1}^{\infty}\frac1n \quad\text{diverges}.

Therefore the series converges conditionally, not absolutely. Whenever a test uses cancellation, state exactly which series it proves convergent before drawing a conclusion.

Check Your Understanding

For each question, distinguish the signed series from its series of magnitudes.

What two facts are required to classify a convergent series as conditionally convergent?
Why does the Alternating-Series Test prove convergence of \(\sum_{n=1}^{\infty}(-1)^{n-1}/n\)?
What series results when the signs are removed from the alternating harmonic series, and why does it diverge?
Classify \(\sum_{n=1}^{\infty}(-1)^{n-1}/n^{3/2}\) as absolutely convergent, conditionally convergent, or divergent. Give the reason for each relevant test.
For the series \(\sum_{n=1}^{\infty}(-1)^{n-1}n/(n^2+1)\), what inequality compares its magnitudes with harmonic terms, and what does that inequality establish?