Anatomy of an Alternating Series

Introduction

An infinite series is built from a sequence of terms, but the terms can have more structure than simply being positive or negative. In an alternating series, the signs follow a pattern: positive, negative, positive, negative, and so on, or the reverse.

To describe that pattern carefully, it helps to separate two pieces of information. A term has a sign, while its magnitude records how large it is without recording whether it is positive or negative. Keeping those roles distinct makes the notation for alternating series much easier to read.

Terms, Signs, and Magnitudes

Write a series as \(\sum_{n=1}^{\infty}u_n\), where \(u_n\) is its \(n\)th signed term. For a nonzero real number \(u_n\), its sign is positive if \(u_n>0\) and negative if \(u_n<0\). Its magnitude is its absolute value, \(|u_n|\), which is positive whether \(u_n\) is positive or negative.

Definition: A series is alternating when its nonzero terms follow alternating signs: successive nonzero terms have opposite signs. A common positive-first representation is \(u_n=(-1)^{n-1}a_n\), where \(a_n\geq 0\). Here \(u_n\) is the signed term and \(a_n\) is its nonnegative magnitude.

The factor \((-1)^{n-1}\) determines the sign according to the index: it is \(+1\) for odd \(n\) and \(-1\) for even \(n\). Thus, when \(a_n>0\), odd-indexed terms are positive and even-indexed terms are negative. If \(a_n=0\), then \(u_n=0\); zero has no positive or negative sign.

There is a useful convention to keep in mind. The formula \(u_n=(-1)^{n-1}a_n\) assigns a sign pattern by index wherever the terms are nonzero. But if some terms are zero, two successive nonzero terms need not have opposite signs: a zero can occur between them. In that case the formula still describes the indexed terms, but one should not claim that the nonzero terms alternate consecutively without checking them.

How the Standard Representation Works

The representation combines a sign factor and a magnitude. For a positive-first pattern, the sign factor at index \(n\) is \((-1)^{n-1}\), while the magnitude is \(a_n\). Multiplying them gives the signed term. In particular, the magnitude can be recovered from the signed term by taking its absolute value, provided the representation has the stated nonnegative \(a_n\).

u_n=(-1)^{n-1}a_n \qquad\Longrightarrow\qquad |u_n|=\bigl|(-1)^{n-1}\bigr|a_n=a_n, \quad a_n\geq 0.

The equality uses the fact that \(\bigl|(-1)^{n-1}\bigr|=1\). Conversely, suppose every term is nonzero, the first term is positive, and each successive term has the opposite sign. Their signs are then \(+1,-1,+1,-1,\ldots\), exactly the values of \((-1)^{n-1}\). Multiplying each sign by its magnitude \(|u_n|\) recovers the term.

u_n=(-1)^{n-1}|u_n|.

So in the positive-first case, setting \(a_n=|u_n|\) gives the standard representation. If the first nonzero term instead has negative sign and the terms alternate without zeros interrupting the pattern, the corresponding representation is \(u_n=-(-1)^{n-1}a_n\). The overall minus sign changes the starting sign; it does not change the magnitudes.

Worked Examples

Worked Example: Separating the Terms of the Alternating Harmonic Series

Consider the series whose \(n\)th term is \(u_n=\frac{(-1)^{n-1}}{n}\), for \(n\geq 1\). Identify the first six signed terms and their magnitudes.

The magnitude sequence is \(a_n=\frac{1}{n}\), which is positive for every positive integer \(n\). For each index, multiply this magnitude by the sign factor \((-1)^{n-1}\):

\begin{aligned} u_1&=(-1)^0\frac11=1, & a_1&=\frac11=1,\\ u_2&=(-1)^1\frac12=-\frac12, & a_2&=\frac12,\\ u_3&=(-1)^2\frac13=\frac13, & a_3&=\frac13,\\ u_4&=(-1)^3\frac14=-\frac14, & a_4&=\frac14,\\ u_5&=(-1)^4\frac15=\frac15, & a_5&=\frac15,\\ u_6&=(-1)^5\frac16=-\frac16, & a_6&=\frac16. \end{aligned}

The signed terms are \(1,-\frac12,\frac13,-\frac14,\frac15,-\frac16\), while the magnitudes are \(1,\frac12,\frac13,\frac14,\frac15,\frac16\). Check the representation: taking absolute values of the six signed terms gives exactly the listed \(a_n\) values, and the signs are positive at odd indices and negative at even indices.

Worked Example: Reading a Rational Term by Its Index

Let \(u_n=(-1)^{n-1}\frac{n+2}{n^2+1}\), for \(n\geq1\). Find the first four signed terms and write their magnitudes.

Here \(a_n=\frac{n+2}{n^2+1}\). Both numerator and denominator are positive for \(n\geq1\), so \(a_n>0\). Substitute the indices one at a time:

\begin{aligned} u_1&=(-1)^0\frac{1+2}{1^2+1} =1\cdot\frac32=\frac32, & a_1&=\frac32,\\ u_2&=(-1)^1\frac{2+2}{2^2+1} =-1\cdot\frac45=-\frac45, & a_2&=\frac45,\\ u_3&=(-1)^2\frac{3+2}{3^2+1} =1\cdot\frac5{10}=\frac12, & a_3&=\frac12,\\ u_4&=(-1)^3\frac{4+2}{4^2+1} =-1\cdot\frac6{17}=-\frac6{17}, & a_4&=\frac6{17}. \end{aligned}

Thus the first four signed terms are \(\frac32,-\frac45,\frac12,-\frac6{17}\), and their magnitudes are \(\frac32,\frac45,\frac12,\frac6{17}\). Check each pair: \(\left|\frac32\right|=\frac32\), \(\left|-\frac45\right|=\frac45\), \(\left|\frac12\right|=\frac12\), and \(\left|-\frac6{17}\right|=\frac6{17}\). The signs also match the odd-even index rule.

Worked Example: Zero Terms and the Sign Pattern

Define \(a_n\) for \(n=1,\ldots,6\) by \(a_1=2\), \(a_2=0\), \(a_3=3\), \(a_4=4\), \(a_5=0\), and \(a_6=6\). Let \(u_n=(-1)^{n-1}a_n\). Compute each term and interpret the zeros.

\begin{aligned} u_1&=(-1)^0(2)=2,\\ u_2&=(-1)^1(0)=0,\\ u_3&=(-1)^2(3)=3,\\ u_4&=(-1)^3(4)=-4,\\ u_5&=(-1)^4(0)=0,\\ u_6&=(-1)^5(6)=-6. \end{aligned}

The adjacent pair \(u_3,u_4\) consists of nonzero terms with opposite signs. The other adjacent pairs each include a zero term. Since zero has no sign, those pairs cannot be described as having opposite signs. Also, the nonzero terms \(u_1=2\) and \(u_3=3\) have the same sign, with \(u_2=0\) between them. This example shows why a zero term must not be treated as if it were positive or negative.

The check is direct: the absolute values of the six signed terms are \(2,0,3,4,0,6\), precisely the specified magnitudes \(a_1,\ldots,a_6\). The formula correctly assigns the index-based sign factor even at indices where the product is zero.

Starting with a Negative Term

The displayed convention \(u_n=(-1)^{n-1}a_n\) starts with a positive sign when \(a_1>0\). Not every alternating series is written that way: a series may begin negative. A leading minus sign accounts for that reversal.

Worked Example: Representing a Negative-First Pattern

Consider \(u_n=\frac{(-1)^n}{n+1}\), for \(n\geq1\). Rewrite it to show a nonnegative magnitude and find its first four terms.

Since \((-1)^n=-(-1)^{n-1}\), the term can be written as

u_n=-(-1)^{n-1}a_n, \qquad a_n=\frac{1}{n+1}\geq0.

Now evaluate:

\begin{aligned} u_1&=\frac{(-1)^1}{1+1}=-\frac12, & a_1&=\frac12,\\ u_2&=\frac{(-1)^2}{2+1}=\frac13, & a_2&=\frac13,\\ u_3&=\frac{(-1)^3}{3+1}=-\frac14, & a_3&=\frac14,\\ u_4&=\frac{(-1)^4}{4+1}=\frac15, & a_4&=\frac15. \end{aligned}

The terms are \(-\frac12,\frac13,-\frac14,\frac15\), and their magnitudes are \(\frac12,\frac13,\frac14,\frac15\). As a check, taking absolute values recovers the stated \(a_n\) at every index, while the negative leading factor makes odd-indexed terms negative and even-indexed terms positive.

A Common Sign Error

A frequent mistake is to shift the exponent by one without changing the starting sign. For example, a student might claim that \(\frac{(-1)^n}{n}\) has positive first term because it is being read using the positive-first pattern \((-1)^{n-1}a_n\). Direct substitution shows otherwise.

\frac{(-1)^1}{1}=-1, \qquad \text{not }+1.

The correction is to preserve the sign produced by the exponent. Since \((-1)^n=-(-1)^{n-1}\), the accurate positive-magnitude representation is

\frac{(-1)^n}{n} = -(-1)^{n-1}\frac1n.

A reliable check is to substitute \(n=1\) before describing the starting sign. This catches exponent shifts immediately. It is also important not to call \(a_n\) the signed term: in the representation \(u_n=(-1)^{n-1}a_n\), \(a_n\) is nonnegative and \(u_n\) includes the sign.

Why the Distinction Matters

The sign and magnitude carry different information. The sign tells which direction a term contributes when added; the magnitude tells the size of that contribution without its direction. For example, \(u_n=-\frac47\) has magnitude \(\frac47\), not \(-\frac47\). Confusing the two can lead to incorrect statements about the terms or incorrect use of a formula.

This distinction also makes an expression easier to inspect. In \(u_n=(-1)^{n-1}a_n\), one can check the sign pattern using the exponent and examine the magnitudes using \(a_n\). These are separate tasks. The notation alone does not establish any claim about whether an infinite series converges; such claims require additional conditions and results.

Key takeaway: In a positive-first alternating representation, \(u_n\) is the signed term, \(a_n\geq0\) is its magnitude, and \((-1)^{n-1}\) supplies the index-based sign. A zero magnitude produces a zero term, which has no sign.

Check Your Understanding

Answer the following by identifying the sign factor, the signed term, and the magnitude carefully.

For \(u_n=(-1)^{n-1}\frac{2}{n+3}\), what are \(u_1,u_2,u_3\) and their magnitudes?
In \(u_n=(-1)^{n-1}a_n\) with \(a_n\geq0\), which part determines the sign pattern, and which part gives the magnitude?
Rewrite \(\frac{(-1)^n}{n^2}\) using a leading minus sign and a nonnegative magnitude.
If \(a_4=0\), does \(u_4=(-1)^3a_4\) have a positive or negative sign? Explain.
Why should the first term be checked before deciding whether a formula starts positive or negative?