Alternating p-Series Classification

Introduction

The exponent \(p\) controls how quickly the terms \(1/n^p\) shrink. For the alternating series, this raises two different questions: does the alternating series converge, and does it still converge when its signs are removed? The answers have different thresholds.

For every \(p>0\), the alternating signs allow the series to converge. But the series of absolute values converges only when \(p>1\). At \(p=1\), and for every \(p\) between zero and one, the series converges conditionally: it converges with its alternating signs, but not absolutely.

Definitions and the Alternating-Series Test

Definition: A series \(\sum_{n=1}^{\infty} u_n\) converges if its sequence of partial sums \(S_N=\sum_{n=1}^{N}u_n\) has a finite limit as \(N\) tends to infinity. It converges absolutely if \(\sum_{n=1}^{\infty}|u_n|\) converges. It converges conditionally if it converges but does not converge absolutely.

The series in this tutorial is

\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^p} = 1-\frac{1}{2^p}+\frac{1}{3^p}-\frac{1}{4^p}+\cdots, \qquad p>0.

The Alternating-Series Test states that \(\sum_{n=1}^{\infty}(-1)^{n-1}a_n\) converges if \(a_n\) is positive, nonincreasing, and tends to zero. Each condition matters: the terms must alternate in sign, their magnitudes must not increase, and those magnitudes must approach zero.

Here \(a_n=1/n^p\). For \(p>0\), these magnitudes are positive and decrease as \(n\) increases. Also \(n^p\) tends to infinity, so \(1/n^p\) tends to zero. Therefore the test establishes convergence for every \(p>0\). It does not, by itself, establish absolute convergence.

When Do the Absolute Values Sum to a Finite Amount?

Taking absolute values removes the alternating signs:

\left|\frac{(-1)^{n-1}}{n^p}\right|=\frac{1}{n^p}.

Thus the question of absolute convergence is exactly whether the positive series \(\sum_{n=1}^{\infty}1/n^p\) converges. We can determine its threshold directly by grouping its terms into blocks whose lengths double.

Theorem: For \(p>0\), the positive series \(\sum_{n=1}^{\infty}1/n^p\) converges if \(p>1\), and diverges if \(0<p\leq 1\).

A Dyadic-Block Proof of the Threshold

For each integer \(k\geq 0\), use the block of indices \(2^k\leq n<2^{k+1}\). There are \(2^k\) terms in this block. Because the terms \(1/n^p\) are positive, estimates for each block give estimates for the whole series.

Estimate a block from above.
For \(2^k\leq n<2^{k+1}\), we have \(1/n^p\leq 1/(2^k)^p=2^{-kp}\). There are \(2^k\) terms, so the sum in the block is at most \(2^k2^{-kp}=2^{k(1-p)}\).

Conclude convergence when \(p>1\).
The block sums are bounded above by \(2^{k(1-p)}\). When \(p>1\), these bounds form a geometric series with ratio \(2^{1-p}<1\), which has a finite sum. The original positive series, grouped into these blocks, therefore has a finite sum as well.

Estimate a block from below.
For \(2^k\leq n<2^{k+1}\), we have \(n<2^{k+1}\), so \(1/n^p>1/(2^{k+1})^p\). Summing over the \(2^k\) terms gives a block sum greater than \(2^k2^{-(k+1)p}=2^{-p}2^{k(1-p)}\).

Conclude divergence when \(0<p\leq1\).
If \(p\leq1\), then \(2^{k(1-p)}\geq1\). Every block therefore has sum greater than \(2^{-p}\), a fixed positive number. Infinitely many such blocks cannot have a finite total sum, so the positive series diverges.

This proves both directions of the threshold. The argument at \(p=1\) is included: each dyadic block has sum greater than \(1/2\), so the harmonic series diverges. No boundary case needs to be inferred from the cases on either side.

The Classification

Combining the alternating-series test with the positive-series result gives the full classification. The alternating series converges for all \(p>0\), while absolute convergence depends on whether \(p>1\).

Range of \(p\)	Alternating series	Series of absolute values	Classification
\(p>1\)	Converges	Converges	Absolutely convergent
\(p=1\)	Converges	Diverges	Conditionally convergent
\(0<p<1\)	Converges	Diverges	Conditionally convergent

Takeaway: For \(p>0\), \(\sum_{n=1}^{\infty}(-1)^{n-1}/n^p\) is always convergent. It is absolutely convergent exactly when \(p>1\), and conditionally convergent exactly when \(0<p\leq1\).

Worked Examples

Worked Example: The Exponent \(p=\frac12\)

Consider

\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt n} = 1-\frac{1}{\sqrt2}+\frac{1}{\sqrt3}-\frac{1}{\sqrt4}+\cdots.

The magnitudes are \(a_n=1/\sqrt n\), which are positive. Their successive ratio is

\frac{a_{n+1}}{a_n} = \frac{1/\sqrt{n+1}}{1/\sqrt n} = \sqrt{\frac{n}{n+1}} <1,

so they decrease. Also \(a_n\to0\), since \(\sqrt n\to\infty\). The Alternating-Series Test therefore proves that the signed series converges. For absolute convergence, the dyadic block with \(2^k\leq n<2^{k+1}\) has sum greater than

2^{-1/2}2^{k(1-1/2)} = \frac{2^{k/2}}{\sqrt2} \geq \frac{1}{\sqrt2}.

The absolute-value series diverges because infinitely many blocks each have sum greater than \(1/\sqrt2\). Thus the alternating series is conditionally convergent. The checks agree with the classification: \(0<p<1\).

Worked Example: The Boundary Case \(p=1\)

The series is

1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots.

Its magnitudes \(a_n=1/n\) are positive, decrease because \(1/(n+1)<1/n\), and tend to zero. The Alternating-Series Test proves convergence. But the magnitudes do not have a finite sum: in each block \(2^k\leq n<2^{k+1}\), every term is greater than \(1/2^{k+1}\), and there are \(2^k\) terms. Hence

\sum_{n=2^k}^{2^{k+1}-1}\frac1n > 2^k\frac{1}{2^{k+1}} = \frac12.

For example, the block from \(n=2\) through \(n=3\) is \(1/2+1/3=5/6>1/2\), and the block from \(n=4\) through \(n=7\) is \(1/4+1/5+1/6+1/7>4(1/8)=1/2\). The same lower bound applies to every block, so the absolute-value series diverges. The signed series converges, but not absolutely: it is conditionally convergent.

Worked Example: The Exponent \(p=2\)

Consider

\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2} = 1-\frac14+\frac19-\frac1{16}+\cdots.

The magnitudes \(a_n=1/n^2\) are positive and decrease, since

\frac{a_{n+1}}{a_n} = \frac{n^2}{(n+1)^2} <1.

They tend to zero, so the Alternating-Series Test proves convergence. The absolute-value series also converges: for each dyadic block its sum is at most

2^{k(1-2)}=2^{-k}.

The geometric series of these bounds converges because its ratio is \(1/2<1\). Therefore \(\sum 1/n^2\) converges, and the alternating series is absolutely convergent. This check matches \(p>1\).

A Common Misclassification

A frequent error is to apply the Alternating-Series Test and conclude that the series is absolutely convergent. The test establishes convergence of the alternating series, not convergence after the signs are removed.

For example, at \(p=1\), the valid reasoning is that \(1/n\) decreases to zero, so \(1-1/2+1/3-\cdots\) converges. The incorrect next step would be to claim that \(\sum 1/n\) converges because the alternating series does. Removing the signs changes the series: the dyadic blocks of \(\sum 1/n\) each have sum greater than \(1/2\), so that positive series diverges. The correct classification is conditional convergence.

Why the Threshold Matters

The exponent \(p=1\) separates two different behaviors of the magnitudes. When \(p>1\), the terms shrink quickly enough that even their positive sum is finite. When \(0<p\leq1\), their positive sum is infinite, although alternating signs still produce convergence. This distinction is why a convergence classification should always specify whether it concerns the signed series or the series of absolute values.

For this family, the decision can be made in order: first verify the alternating-series conditions, then examine the absolute-value series and compare \(p\) with one. The first check gives convergence for all positive \(p\); the second separates absolute from conditional convergence.

Check Your Understanding

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

For which positive values of \(p\) is the alternating p-series absolutely convergent?
Why does the Alternating-Series Test apply when \(p=1/3\)?
What lower bound does a dyadic block give for the absolute-value series when \(p=1\)?
Classify the series with \(p=3/4\) as absolutely convergent, conditionally convergent, or divergent.
Explain why convergence of the alternating harmonic series does not imply convergence of the harmonic series.