When Decrease Begins After the First Term

Why the First Few Terms Need Not Decide Convergence

The alternating-series test is often introduced with a sequence of positive magnitudes that decreases from its first term. In applications, the magnitudes may not behave so neatly: they might rise or fall irregularly for a few terms, then decrease steadily toward zero.

That initial irregularity does not prevent us from using the test. Convergence is determined by the behavior of the tail, and a finite initial segment contributes only a fixed amount to every sufficiently late partial sum. The key is to identify where the required conditions begin and apply the test from that point onward.

Definition: A sequence \(b_n\) is eventually nonincreasing if there is an integer \(N\) such that \(b_{n+1}\le b_n\) for every \(n\ge N\). The condition need not hold for indices before \(N\).

The Alternating-Series Test for an Eventually Decreasing Tail

Consider a series with alternating signs and positive magnitudes:

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

The alternating-series test states that this series converges if \(b_n\) is nonincreasing and tends to zero. For the present situation, only the tail needs to satisfy those conditions.

Theorem: Suppose \(b_n\ge 0\) for every \(n\), and there is an integer \(N\) such that \(b_{n+1}\le b_n\) for every \(n\ge N\) and \(\lim_{n\to\infty}b_n=0\). Then

\sum_{n=1}^{\infty}(-1)^{n-1}b_n

converges. The terms before \(N\) do not have to decrease.

To see how the known test applies, split the series after the finite initial segment:

\sum_{n=1}^{\infty}(-1)^{n-1}b_n = \sum_{n=1}^{N-1}(-1)^{n-1}b_n + \sum_{n=N}^{\infty}(-1)^{n-1}b_n.

The first sum contains only finitely many terms, so it is a real number. In the tail, set \(c_k=b_{N+k-1}\). Then \(c_k\ge 0\), \(c_{k+1}\le c_k\), and \(c_k\to0\). Moreover,

\sum_{n=N}^{\infty}(-1)^{n-1}b_n = (-1)^{N-1}\sum_{k=1}^{\infty}(-1)^{k-1}c_k.

The factor \((-1)^{N-1}\) is either \(1\) or \(-1\); it only determines whether the tail begins with a positive or negative term. The alternating-series test applies to the series in \(c_k\), so the tail converges. Adding the finite initial sum preserves convergence.

Why a Finite Change Preserves Convergence

This reasoning is an instance of a general fact about infinite series. The partial sums of a series are the finite sums of its first terms. Convergence of the series means that this sequence of partial sums has a finite limit.

Suppose two series have the same terms from some index \(M\) onward, but may differ before \(M\). Let \(S_n\) be the \(n\)th partial sum of the first series, and let \(T_n\) be that of the second. For every \(n\ge M\), the terms after the initial segment are identical, so

$$ T_n=S_n+D, $$

where \(D\) is the fixed difference between the two finite initial sums. If \(S_n\to S\), then \(T_n=S_n+D\to S+D\). Conversely, if \(T_n\) converges, then \(S_n=T_n-D\) converges. Thus the two series either both converge or both fail to converge. Their sums, when they converge, can differ by \(D\).

Separate the initial segment.
Choose an index \(N\) from which the magnitudes are nonincreasing and write the series as a finite sum plus a tail.

Check the tail conditions.
Verify that the magnitudes remain nonnegative, do not increase from that index onward, and approach zero.

Apply the test to the tail.
The alternating-series test gives convergence of the tail, and adding the finite initial sum gives convergence of the full series.

Worked Examples

Worked Example: The Magnitudes Increase Once Before Decreasing

Consider \(\sum_{n=1}^{\infty}(-1)^{n-1}b_n\), where \(b_1=1\), \(b_2=\frac14\), and \(b_n=\frac1n\) for \(n\ge3\). The first three magnitudes are

b_1=1,\qquad b_2=\frac14,\qquad b_3=\frac13.

They are not decreasing from the beginning, since \(\frac14<\frac13\). But from \(n=3\) onward, \(b_n=\frac1n\), and

b_{n+1}=\frac1{n+1}<\frac1n=b_n\quad(n\ge3), \qquad \lim_{n\to\infty}\frac1n=0.

The tail therefore converges by the alternating-series test. The finite initial sum is \(b_1-b_2=1-\frac14=\frac34\), and the full series can be written as

\frac34+\left(\frac13-\frac14+\frac15-\frac16+\cdots\right).

The tail begins with a positive term and has decreasing magnitudes tending to zero, so the full series converges. As a check on the split, the third partial sum is \(1-\frac14+\frac13=\frac{13}{12}\); the displayed split gives \(\frac34+\frac13=\frac{13}{12}\) as well.

Worked Example: The Decreasing Tail Begins with a Negative Term

Define \(b_1=2\), \(b_2=\frac13\), \(b_3=\frac12\), and \(b_n=\frac1{n-3}\) for \(n\ge4\). The magnitudes rise from \(b_2=\frac13\) to \(b_3=\frac12\), and again from \(b_3=\frac12\) to \(b_4=1\). However, for \(n\ge4\),

b_n=\frac1{n-3}, \qquad b_{n+1}=\frac1{n-2}<\frac1{n-3}=b_n, \qquad \lim_{n\to\infty}b_n=0.

Thus the tail from \(n=4\) meets the test conditions. Because \(n=4\) has sign \((-1)^3=-1\), the series splits as

\left(2-\frac13+\frac12\right) + \left(-1+\frac12-\frac13+\frac14-\cdots\right).

The initial sum is \(2-\frac13+\frac12=\frac{12}{6}-\frac{2}{6}+\frac{3}{6}=\frac{13}{6}\). The tail is the negative of an alternating series with decreasing positive magnitudes \(1,\frac12,\frac13,\ldots\), so it converges. Therefore the full series converges. The split also checks the sign and starting index: its fourth partial sum is \(\frac{13}{6}-1=\frac76\), which equals \(2-\frac13+\frac12-1=\frac76\).

Worked Example: Replacing Two Terms Changes the Sum but Not Convergence

Let \(c_n=\frac{(-1)^{n-1}}{n}\), and define a second series with first terms \(d_1=7\), \(d_2=-4\), and \(d_n=c_n\) for \(n\ge3\). The series \(\sum c_n\) converges by the alternating-series test, since \(1/n\) decreases to zero. For every \(n\ge3\), the partial sums of the modified series satisfy

\sum_{k=1}^{n}d_k = \sum_{k=1}^{n}c_k +(7-1)+\left(-4+\frac12\right) = \sum_{k=1}^{n}c_k+\frac52.

The difference is constant because only the first two terms were replaced. Hence the modified partial sums converge whenever the original partial sums do. As a direct check at \(n=3\), the original partial sum is \(1-\frac12+\frac13=\frac56\), while the modified one is \(7-4+\frac13=\frac{10}{3}\); indeed, \(\frac56+\frac52=\frac{10}{3}\). The change preserves convergence but shifts the sum by \(\frac52\).

A Common Mistake

A common error is to treat a failure of decrease at an early index as proof that the alternating-series test cannot establish convergence. For the first example, one might write:

b_2=\frac14<\frac13=b_3, \quad\text{so the magnitudes are not decreasing; therefore the series diverges.}

The first statement is correct, but the conclusion does not follow. The test’s conditions hold from \(n=3\) onward, and that is enough: the tail converges, and the first two terms add only the fixed amount \(\frac34\). A condition failing at one early step means only that the test cannot be applied to the whole sequence starting at its first term. It does not establish divergence.

There is a related distinction worth keeping clear. Removing or changing finitely many terms does not affect whether a series converges, but it generally does affect the sum. The constant difference in partial sums records that change.

Check Your Understanding

Use the tail conditions and the effect of a finite initial segment to answer these questions.

If \(b_1,b_2,b_3\) are irregular but \(b_n\) is nonnegative, nonincreasing for \(n\ge4\), and tends to zero, what can the alternating-series test establish?
For an alternating series whose decreasing tail starts at an even index \(N\), what is the sign of the first tail term?
Why does adding a finite initial sum to a convergent tail preserve convergence?
If two series differ only in their first three terms, how are their sufficiently late partial sums related?
Does changing finitely many terms necessarily preserve the sum, or only convergence?