The Alternating Series Test

Why Alternating Signs Matter

For a series whose terms all have the same sign, the size of the terms often determines whether the partial sums keep growing without bound. Alternating signs can instead make successive partial sums move back and forth. When the magnitudes decrease to zero, those movements become smaller, and the partial sums settle toward a limit. The Alternating Series Test makes this idea precise.

The main conditions concern the magnitudes, not merely the signs: they must be nonincreasing and must tend to zero. A series with alternating signs but magnitudes that do not tend to zero fails the Necessary Condition for Series Convergence. Even when the test applies, it guarantees convergence, not necessarily absolute convergence.

Theorem (Alternating Series Test): Let $(b_n)_{n=1}^{\infty}$ be a sequence of nonnegative real numbers such that $b_{n+1}\leq b_n$ for every $n\geq1$, and $b_n\to0$. Then the alternating series $$ \sum_{n=1}^{\infty}(-1)^{n-1}b_n $$ converges. The same conclusion holds if the first sign is negative instead. If the series converges to $S$, and $S_N=\sum_{n=1}^{N}(-1)^{n-1}b_n$, then $$ |S-S_N|\leq b_{N+1}. $$

This tutorial introduces the test and develops estimates that help explain how it works. The full proof that the hypotheses guarantee convergence is the subject of the next tutorial. We will prove two supporting results here: a bound for every finite alternating block, and the remainder estimate under the assumption that the series converges.

A Bound for Finite Alternating Blocks

The basic estimate comes from grouping neighboring terms in pairs. If a block starts with a positive sign, the first term is at least as large as the second, the third is at least as large as the fourth, and so on. Pairing terms in the appropriate way shows that the block cannot be negative and cannot exceed its first term.

Proposition (Finite Alternating-Block Bound): Suppose $b_n\geq0$ and $b_{n+1}\leq b_n$. For integers $p\leq q$, set $$ T=\sum_{n=p}^{q}(-1)^{n-p}b_n. $$ Then $$ 0\leq T\leq b_p. $$ If the block begins with a negative sign instead, its sum lies between $-b_p$ and $0$.

Proof. Let the block have $L=q-p+1$ terms. If $L=1$, then $T=b_p$, so the conclusion holds. Suppose first that $L=2k$ is even, where $k\geq1$. Pair consecutive terms starting at $p$: $$ T=(b_p-b_{p+1})+(b_{p+2}-b_{p+3})+\cdots+(b_{p+2k-2}-b_{p+2k-1}). $$ Each difference is nonnegative because the sequence is nonincreasing, so $T\geq0$. For an upper bound, regroup the same terms as $$ T=b_p-(b_{p+1}-b_{p+2})-(b_{p+3}-b_{p+4})-\cdots-(b_{p+2k-3}-b_{p+2k-2})-b_{p+2k-1}. $$ When $k=1$, there are no paired differences to subtract, and this expression is $T=b_p-b_{p+1}$. In every case, all the subtracted quantities are nonnegative, so $T\leq b_p$.

Now suppose that $L=2k+1$ is odd. Pairing from the beginning and leaving the last term gives $$ T=(b_p-b_{p+1})+(b_{p+2}-b_{p+3})+\cdots+(b_{p+2k-2}-b_{p+2k-1})+b_{p+2k}\geq0, $$ where for $k=0$ this is simply $T=b_p$. For the upper bound, group instead as $$ T=b_p-(b_{p+1}-b_{p+2})-(b_{p+3}-b_{p+4})-\cdots-(b_{p+2k-1}-b_{p+2k}). $$ For $k=0$, there are no subtracted differences, and again $T=b_p$. Otherwise, each difference being subtracted is nonnegative. Thus $T\leq b_p$. This proves the result when the block starts with a positive sign. Multiplying by $-1$ proves the claim for a block that starts with a negative sign. $\square$

The pairings depend on the parity of the block length. In particular, for an even-length block, an upper-bound regrouping must leave the final term $b_{p+2k-1}$ unpaired, as shown above. Pairing terms inconsistently can change the sum and invalidate the estimate.

Remainder Estimates

Suppose an alternating series is known to converge to $S$. Its remainder after $N$ terms is $R_N=S-S_N$. A finite tail from $N+1$ through $m$ is an alternating block whose first magnitude is $b_{N+1}$. The finite-block bound therefore controls every such tail, regardless of how many terms it contains.

Theorem (Remainder Bound for a Convergent Alternating Series): Suppose $b_n\geq0$, $b_{n+1}\leq b_n$, and the series $\sum_{n=1}^{\infty}(-1)^{n-1}b_n$ converges to $S$. For every $N\geq1$, $$ |S-S_N|\leq b_{N+1}. $$

Proof. For $m>N$, the difference between two partial sums is $$ S_m-S_N=\sum_{n=N+1}^{m}(-1)^{n-1}b_n. $$ This is a finite alternating block. Its first sign may be positive or negative, so the Finite Alternating-Block Bound gives $$ |S_m-S_N|\leq b_{N+1}. $$ As $m\to\infty$, convergence gives $S_m\to S$, and therefore $|S_m-S_N|\to|S-S_N|$. Taking limits in the inequality yields $$ |S-S_N|\leq b_{N+1}. $$ This proves the estimate. $\square$

The estimate says that the error after $N$ terms is no larger than the first omitted magnitude. It does not say that the error equals $b_{N+1}$, or that the partial sum is always on the same side of $S$ for every $N$. The next tutorial proves convergence from the test’s hypotheses and develops the alternating partial sums’ position relative to the limit.

Worked Applications

Worked Example: The Alternating Harmonic Series

Consider $$ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}. $$ Here $b_n=1/n$. These magnitudes are positive, satisfy $b_{n+1}=1/(n+1)\leq1/n=b_n$, and tend to zero. The Alternating Series Test therefore gives convergence. The remainder after $N$ terms satisfies $$ |S-S_N|\leq\frac{1}{N+1}. $$ For instance, using the first $99$ terms gives an error at most $1/100$.

This convergence is not absolute. The series of absolute values is $\sum_{n=1}^{\infty}1/n$, the p-series with $p=1$, which diverges by the p-Series Convergence Criterion. Thus the alternating harmonic series converges conditionally: it converges, but its series of absolute values does not.

Worked Example: Alternating Terms with Square-Root Magnitudes

Consider $$ \sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n+4}}. $$ Set $b_n=1/\sqrt{n+4}$. Since $n+5>n+4>0$, taking square roots and reciprocals gives $$ 0<b_{n+1}=\frac{1}{\sqrt{n+5}}<\frac{1}{\sqrt{n+4}}=b_n. $$ Also $b_n\to0$, so the Alternating Series Test proves that the series converges. After $N$ terms its remainder satisfies $$ |S-S_N|\leq\frac{1}{\sqrt{N+5}}. $$

It is not absolutely convergent. For $n\geq1$, $n+4\leq5n$, so $$ \frac{1}{\sqrt{n+4}}\geq\frac{1}{\sqrt{5n}}=\frac{1}{\sqrt5}\frac{1}{\sqrt n}. $$ The p-series $\sum_{n=1}^{\infty}1/\sqrt n$ diverges because $p=1/2$. The Comparison Test for Nonnegative Series then shows that the series of absolute values diverges as well.

Worked Example: Alternating Signs Without Vanishing Terms

Consider $$ \sum_{n=1}^{\infty}(-1)^{n-1}\left(1+\frac1n\right). $$ The magnitudes $b_n=1+1/n$ decrease, since $$ b_n-b_{n+1} =\left(1+\frac1n\right)-\left(1+\frac{1}{n+1}\right) =\frac{1}{n(n+1)}>0. $$ But they tend to $1$, not to zero. In fact, the terms themselves do not tend to zero: their absolute values tend to $1$. The Necessary Condition for Series Convergence therefore proves that this series diverges. Alternating signs and decreasing magnitudes alone are not enough; the limit-zero condition is essential.

Worked Example: Absolute Convergence Can Also Occur

For $$ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(n+1)^2}, $$ the magnitudes $b_n=1/(n+1)^2$ decrease and tend to zero, so the Alternating Series Test applies. In this case the series is also absolutely convergent, because $$ \sum_{n=1}^{\infty}\left|\frac{(-1)^{n-1}}{(n+1)^2}\right| =\sum_{n=1}^{\infty}\frac{1}{(n+1)^2} \leq\sum_{n=1}^{\infty}\frac{1}{n^2}, $$ and the p-series on the right converges for $p=2$. Alternation is sufficient for convergence here, but it is not needed to establish absolute convergence.

How to Apply the Test Carefully

When using the test, separate the sign pattern from the magnitudes. Write the terms in the form $(-1)^{n-1}b_n$, or its negative, and then check the two conditions on $b_n$ independently. Nonincreasing allows equality; strict decrease is not required. The limit must be zero, not merely a finite number.

If the conditions hold, the conclusion is convergence. To decide whether that convergence is absolute, examine $\sum b_n$ separately. The alternating harmonic series and the square-root example converge conditionally, while the squared-denominator example converges absolutely. If the magnitudes fail to decrease or do not tend to zero, the test itself gives no convergence conclusion; in the latter case, the Necessary Condition may establish divergence directly.

Takeaway: For an alternating series with nonnegative, nonincreasing magnitudes tending to zero, the Alternating Series Test guarantees convergence. Once convergence is established, the error after $N$ terms is at most the first omitted magnitude $b_{N+1}$.

Check Your Understanding

Use the conditions, finite-block estimate, and examples to answer the following questions.

Which two conditions must the magnitudes $b_n$ satisfy for the Alternating Series Test?
Why does a decreasing sequence of alternating term magnitudes that tends to $1$ still fail the test?
For a convergent alternating series, what bound does the theorem give for the remainder after $N$ terms?
Why does convergence of an alternating series not by itself establish absolute convergence?
In the finite-block proof, how does the pairing used for an even number of terms differ between the lower and upper bounds?

What You'll Learn