The Alternating Series Test

Does the series $1 - \tfrac12 + \tfrac13 - \tfrac14 + \tfrac15 - \cdots$ converge? The terms $\tfrac1n$ do not decrease fast enough to be summable in absolute value. Indeed $\sum \tfrac1n$ diverges. Yet the alternation of signs produces cancellation that can rescue convergence. This tension between a slow decay and a sign flip is exactly what the Alternating Series Test resolves. It is also called the Leibniz test. It is one of the few convergence tests that gives you both an existence proof and a usable numerical error bound in the same breath. The result matters because most convergence tests in a first course are fundamentally tests for absolute convergence. This includes comparison, ratio, root, and integral tests. They succeed only when $\sum |a_n|$ converges. The alternating series $\sum(-1)^n \tfrac1n$ is the simplest example where every one of those tests is silent or gives a negative answer. Yet the series itself converges. The Leibniz test is the first tool a student meets that certifies conditional convergence. This means convergence of $\sum a_n$ without convergence of $\sum |a_n|$ . It does so with an elementary, self-contained argument built entirely from the Monotone Convergence Theorem. The big picture is that alternating signs force the partial sums to oscillate and squeeze toward a limit. This holds provided the magnitudes $|a_n|$ decrease monotonically to zero. Two interlaced monotone subsequences of partial sums trap the limit between them. One subsequence increases, the other decreases. The gap between consecutive partial sums shrinks to zero exactly because $|a_n|\to 0$ . This same trapping argument, refined slightly, yields a remarkably clean error estimate. Truncating the series after $n$ terms leaves an error. That error is no larger in absolute value than the very next term you dropped. This article gives the precise hypotheses and a complete proof by monotone subsequences. It also gives the error bound as a corollary, and two asymmetric worked examples. A counterexample shows why monotonicity cannot be dropped. Finally, it discusses where the dichotomy between absolute and conditional convergence becomes central later on. This test opens the door to that dichotomy.

Key takeaways

The test applies to series of the form $\sum (-1)^{n+1} b_n$ where $b_n \ge 0$ decreases monotonically to $0$ .
Convergence is proved by showing the even partial sums increase, the odd partial sums decrease, and the gap between them shrinks to $0$ .
The limit $S$ is trapped between any two consecutive partial sums: $S$ lies between $s_n$ and $s_{n+1}$ for every $n$ .
This gives a free error bound: $|S - s_n| \le b_{n+1}$ , the size of the first omitted term.
Monotonicity of $b_n$ is essential — without it the test can fail even if $b_n \to 0$ .
Key idea: alternating signs plus monotone decay to zero force two monotone, mutually trapping subsequences of partial sums to converge to a common limit.

Where this lives today

The Leibniz test is the standard first source of examples of conditionally convergent series, the phenomenon at the heart of Absolute vs Conditional Convergence: a series like $\sum (-1)^{n+1}/n$ converges but does not converge absolutely, and such series can be rearranged to sum to any real number (or diverge), a startling fact proved via the Riemann rearrangement theorem. This makes the alternating harmonic series a canonical cautionary example throughout analysis: term-by-term manipulations that are safe for absolutely convergent series can silently fail here. In numerical analysis, the error bound $|S - s_n|\le b_{n+1}$ is one of the few a priori, non-asymptotic error estimates available for a truncated infinite sum, and it underlies alternating-series acceleration techniques (e.g. Euler transform, Aitken's $\Delta^2$ process) used to speed up otherwise slow numerical series evaluation, such as computing $\ln 2$ , $\pi$ , or values of the Riemann zeta function from slowly convergent alternating representations. In probability and statistics, alternating series with monotone decreasing terms appear in tail bounds for discrete distributions and in inclusion–exclusion formulas, where the partial-sum trapping property (the theorem's part (a)) gives rigorous upper and lower bounds on probabilities computed via truncated inclusion–exclusion expansions — a technique known as the Bonferroni inequalities. In physics and engineering, alternating series arise in perturbation expansions and Fourier-type series (e.g., certain Fourier coefficients of periodic signals decay with alternating sign), where the Leibniz error bound offers quick, rigorous truncation guarantees without requiring full knowledge of the tail's exact behavior — useful in real-time or resource-constrained computation.

Ask the tutor: How does the Leibniz error bound compare to error bounds from the integral test or ratio test for series that satisfy both sets of hypotheses?

\sum_{n=1}^\infty a_n

Intuition

Picture walking along the number line: you step right by $b_1$ , then left by $b_2$ , then right by $b_3$ , and so on, with each step shorter than the last. Because each step is smaller than the previous one, you never retrace past where you were two steps ago — you keep overshooting less and less, homing in on a resting point. The even-indexed positions $s_2, s_4, s_6, \dots$ creep upward (each even step corrects less than the odd step that undoes it), while the odd-indexed positions $s_1, s_3, s_5,\dots$ creep downward, and the two staircases converge toward each other because the step sizes $b_n \to 0$ . This is precisely a monotone-sequence argument in disguise: one bounded monotone increasing sequence and one bounded monotone decreasing sequence, squeezed together by a gap that vanishes.

Figure

Partial sums zigzagging toward a limit, trapped between odd and even subsequences

Central theorem

(b_n)_{n\ge1}

s_n = \sum_{k=1}^n (-1)^{k+1}b_k

S - s_n

n=2k

Worked examples

Example 1 — an alternating $p$ -series with a shifted index and estimating a sum to three decimal places. Consider

\sum_{n=1}^\infty \frac{(-1)^{n+1}}{(n+2)^{3/2}} = \frac{1}{3^{3/2}} - \frac{1}{4^{3/2}} + \frac{1}{5^{3/2}} - \frac{1}{6^{3/2}} + \cdots

Here $b_n = (n+2)^{-3/2}$ . Checking the hypotheses. Certainly $b_n = (n+2)^{-3/2} > 0$ for all $n\ge1$ , so (1) holds. For monotonicity, the function $f(x) = (x+2)^{-3/2}$ is strictly decreasing on $x>0$ since $f'(x) = -\tfrac32(x+2)^{-5/2} < 0$ ; hence $b_{n+1} = f(n+1) < f(n) = b_n$ , so (2) holds (strictly, in fact). For the vanishing condition, $b_n = (n+2)^{-3/2}\to 0$ as $n\to\infty$ because $(n+2)^{3/2}\to\infty$ , so (3) holds. By the theorem, the series converges to some sum $S$ . Estimating $S$ . Compute partial sums numerically:

b_1 = 3^{-3/2} = \frac{1}{3\sqrt3} \approx 0.192450,\quad b_2 = 4^{-3/2} = \frac18 = 0.125000,

b_3 = 5^{-3/2}\approx 0.089443,\quad b_4 = 6^{-3/2}\approx 0.068041,\quad b_5 = 7^{-3/2}\approx 0.053995.

s_1 = 0.192450,\ \ s_2 = 0.067450,\ \ s_3 = 0.156893,\ \ s_4 = 0.088852,\ \ s_5 = 0.142847.

By the error bound (b), $|S - s_5| \le b_6 = 8^{-3/2} = \frac{1}{16\sqrt2}\approx 0.044194$ . That is a loose bound; to guarantee three-decimal accuracy ( $|S-s_n|<0.0005$ ) we need $b_{n+1} < 0.0005$ , i.e. $(n+3)^{3/2} > 2000$ , i.e. $n+3 > 2000^{2/3}\approx 158.7$ , so $n \ge 156$ suffices. This illustrates a genuine feature of the bound: it is rigorous but can require many terms for high precision when $b_n$ decays slowly (here like $n^{-3/2}$ ). Using the corollary, since we stopped at $n=5$ (odd), we know $S \le s_5 = 0.142847$ ; and since $s_4$ used an even index, $S \ge s_4 = 0.088852$ . So without any further computation we already know

0.088852 \le S \le 0.142847,

consistent with the sharper bound $|S-s_5|\le b_6\approx 0.0442$ , giving $S \in [0.0986, 0.1870]$ intersected with the trapping interval — the trapping interval from Step 5 (using consecutive terms $s_4, s_5$ directly) is actually tighter here than the generic bound (b) applied at $n=5$ , since (b) is a consequence of (a) and can be loose when $b_{n+1}$ is not tiny yet.

Example 2 — a series where the terms are not simply $1/n^p$ , requiring a monotonicity argument via calculus. Consider

\sum_{n=1}^\infty (-1)^{n+1}\, \frac{n}{n^2+5}.

Here $b_n = \dfrac{n}{n^2+5}$ , and monotonicity is not obvious from the formula alone — we must prove it. Checking the hypotheses. Nonnegativity is clear: $b_n = n/(n^2+5) > 0$ for $n\ge1$ . For the vanishing condition, $b_n = \dfrac{n}{n^2+5} = \dfrac{1/n}{1+5/n^2} \to \dfrac{0}{1} = 0$ as $n\to\infty$ , using the algebraic limit laws applied to $1/n\to0$ and $5/n^2\to0$ . For monotonicity, consider $f(x) = \dfrac{x}{x^2+5}$ on $x>0$ . By the quotient rule,

f'(x) = \frac{(x^2+5)\cdot 1 - x\cdot 2x}{(x^2+5)^2} = \frac{5 - x^2}{(x^2+5)^2}.

This is negative exactly when $x^2 > 5$ , i.e. $x > \sqrt5 \approx 2.236$ . So $f$ is increasing on $(0,\sqrt5)$ and decreasing on $(\sqrt5,\infty)$ . In particular $f$ is not monotone on all of $x>0$ : $b_1 = \tfrac16\approx0.1667$ , $b_2=\tfrac29\approx0.2222$ , $b_3=\tfrac{3}{14}\approx0.2143$ , and indeed $b_1 < b_2$ — the sequence increases at first! This means the Leibniz test in its stated form does not directly apply to the whole series, since hypothesis (2) fails for $n=1$ ( $b_2 > b_1$ ). However, the theorem only needs monotonicity eventually, together with the tail argument — and indeed a standard strengthening (whose proof is a trivial modification of the one above, applied to the tail $\sum_{n\ge3}(-1)^{n+1}b_n$ , and using that finitely many terms cannot affect convergence) shows convergence still holds provided $b_n$ is eventually monotone decreasing to $0$ . Here, for $n \ge 3 > \sqrt5$ , $f$ is strictly decreasing, so $(b_n)_{n\ge3}$ is monotone decreasing, and $b_n\to0$ . Applying the theorem to the tail series $\sum_{n=3}^\infty (-1)^{n+1}b_n$ shows it converges to some $T$ ; adding back the finite sum $b_1 - b_2$ shows the original series converges to $S = b_1 - b_2 + T$ (finitely many terms never affect convergence, only the value of the sum, by the Cauchy Criterion for Series applied to finite truncations). The error bound now applies only to the tail: for $n \ge 3$ , $|T - (\text{tail partial sum through } n)| \le b_{n+1}$ . Concretely with $n=4$ : $b_4 = 4/21\approx0.19048$ , $b_5=5/30=\tfrac16\approx0.16667$ ; the tail partial sum $t_4 = b_3-b_4 = 0.21429-0.19048=0.02381$ (started at $n=3$ ) satisfies $|T-t_4|\le b_5\approx0.1667$ . This example demonstrates the practically important point: always check monotonicity by an honest computation (e.g. via the derivative of the continuous extension), and if it fails for small $n$ , isolate the finitely many bad terms before invoking the test on the tail.

\sum_{n=1}^\infty (-1)^{n+1} c_n, \qquad \text{where } c_n = \begin{cases} \dfrac{1}{n} & n \text{ odd} \\[4pt] \dfrac{1}{n^2} & n \text{ even}\end{cases}.

Scope, edge cases & common traps

Situation	What happens	Why
$b_n \downarrow 0$ eventually (not from $n=1$ )	Test still gives convergence	Finitely many terms don't affect convergence; apply theorem to the tail
$b_n \to 0$ but not monotone	Test is silent; series may converge or diverge	See counterexample above
$b_n$ monotone but $b_n \to L \ne 0$	Series diverges	Fails the Divergence Test: terms $(-1)^{n+1}b_n$ don't tend to $0$
$b_n$ monotone $\downarrow 0$ , but only weakly ( $b_{n+1}\le b_n$ , equalities allowed)	Test still applies	Proof only used $\le$ , never strict inequality
Want absolute convergence	Leibniz test is silent on this	It proves convergence of $\sum a_n$ , not of $\sum\|a_n\|$ ; see Absolute vs Conditional Convergence

Common trapThe classic error: forgetting to verify monotonicity, or assuming

b_n\to 0

alone suffices

A very common mistake is to check only that $b_n \to 0$ and conclude the alternating series converges, silently skipping the monotonicity hypothesis — or worse, to assume monotonicity because the formula for $b_n$ "looks decreasing" without any justification (derivative sign check, ratio $b_{n+1}/b_n \le 1$ , or explicit algebra). As the counterexample shows, $b_n\to 0$ with $b_n\ge0$ is not sufficient: monotonicity is doing essential work in forcing both subsequences of partial sums to be individually monotone, which is what lets the Monotone Convergence Theorem apply. Always state and verify monotonicity as its own step, exactly as in the second worked example, where naive inspection would have missed the failure at $n=1$ .

The mathematician behind it

Origins · 1 minGottfried Wilhelm Leibniz and how this was discovered

Gottfried Wilhelm Leibniz

1646–1716 · Co-founder of calculus, philosopher, and polymath in Hanover

Leibniz developed the differential and integral calculus independently of Newton, introduced much of the modern notation still used today ( $dx$ , $\int$ , and the term "function"), and made foundational contributions to logic, combinatorics, and infinite series. His 1682 result on the series $\pi/4 = 1 - \tfrac13+\tfrac15-\tfrac17+\cdots$ (the Leibniz series for $\pi$ ) is a direct instance of the theorem that bears his name, though the general test for arbitrary alternating series with monotone decreasing terms was made fully rigorous only later, in the 19th-century arithmetization of analysis associated with Cauchy and his successors.

History

- 1673–76: Leibniz derives the series

\pi/4 = 1-\tfrac13+\tfrac15-\tfrac17+\cdots

while in Paris, studying quadratures of the circle.

- 1682: Leibniz publishes this alternating series result, establishing convergence by an early cancellation argument.

- 1821: Cauchy's Cours d'Analyse rigorizes the notion of convergence via what is now the Cauchy Criterion for Series, setting the stage for a fully modern proof of the general test.

- Late 1800s: The general Leibniz/alternating series test, with explicit monotone-tail hypotheses and the error bound, becomes a standard fixture of analysis textbooks following the arithmetization of analysis.

- 20th century: The test becomes a standard early example distinguishing conditional from absolute convergence, foreshadowing rearrangement phenomena (the Riemann rearrangement theorem) for conditionally convergent series.

Does the Alternating Series Test prove absolute convergence?

No. It only proves that the series $\sum(-1)^{n+1}b_n$ itself converges — that its partial sums $s_n$ have a limit. It says nothing about $\sum |a_n| = \sum b_n$ , which may very well diverge, as in the classic example $\sum (-1)^{n+1}/n$ : the alternating harmonic series converges by the Leibniz test, but $\sum 1/n$ is the divergent harmonic series. When a series converges but not absolutely, it is called conditionally convergent, and this distinction is developed fully under Absolute vs Conditional Convergence. If you need absolute convergence, you must test $\sum|a_n|$ separately using tools like comparison, ratio, or integral tests.

Can the monotonicity hypothesis be weakened to "eventually monotone"?

Yes, and this is used routinely. If $b_n$ is monotone decreasing to $0$ only for $n \ge N_0$ , apply the theorem to the tail series $\sum_{n\ge N_0}(-1)^{n+1}b_n$ , which converges by the theorem as stated. Since altering or removing finitely many terms from a series changes at most the value of the sum, not whether it converges (a direct consequence of the Cauchy Criterion for Series applied to tails), the full series $\sum_{n\ge1}(-1)^{n+1}b_n$ also converges. This is exactly the technique used in the second worked example, where monotonicity failed only at the very first step.

Is the error bound

|S-s_n|\le b_{n+1}

tight, or can it be improved?

The bound is tight in general — it can be attained (in the limiting sense) by suitable choices of $b_n$ — but for specific series it is often loose. The trapping property (part (a) of the theorem) is actually sharper: it tells you $S$ lies strictly between $s_n$ and $s_{n+1}$ , so you can sometimes get better bounds by directly comparing $s_n$ and $s_{n+1}$ rather than invoking the generic corollary. In the first worked example, comparing consecutive partial sums directly gave information consistent with, but computed independently of, the blanket bound $b_{n+1}$ .

What happens if

b_n

is monotone and nonnegative but does not tend to

0

Then the test's hypotheses are not met, and in fact the series necessarily diverges, by the Divergence Test (the $n$ -th term test): if $b_n \to L \ne 0$ , the terms $(-1)^{n+1}b_n$ do not tend to $0$ , since they alternate between values near $L$ and near $-L$ . A necessary condition for any series $\sum a_n$ to converge is $a_n \to 0$ , so nonvanishing terms immediately rule out convergence, alternating or not.

Does the test work for complex-valued or vector-valued alternating series?

Not directly in this form, because the proof relies essentially on order properties of $\mathbb{R}$ (monotone sequences, the Monotone Convergence Theorem), which have no direct analogue in $\mathbb{C}$ or $\mathbb{R}^k$ without extra structure. However, one can sometimes apply the real-valued Leibniz test to the real and imaginary parts (or components) separately, if each happens to be alternating with monotone decreasing magnitude, and then combine the results using the fact that a complex (or vector) sequence converges iff each component sequence converges.

How does this test foreshadow the notion of conditional convergence?

The alternating harmonic series is the prototypical example of a series that converges only because of the arrangement and cancellation of its terms, not because its terms are "absolutely small enough." This is the seed of the entire subject of Absolute vs Conditional Convergence: absolutely convergent series behave like finite sums (any rearrangement gives the same sum, by a theorem sometimes attributed to Dirichlet), while conditionally convergent series can be rearranged to converge to any prescribed real number, or to diverge, by the Riemann rearrangement theorem. The Leibniz test is usually the first place a student sees a convergent-but-not-absolutely-convergent series constructed explicitly.

Can I use the Leibniz test to determine convergence of

\sum (-1)^n \sin(1/n)

Yes. Let $b_n = \sin(1/n)$ for $n \ge 1$ . Since $1/n \in (0,1]\subset(0,\pi/2)$ , $b_n>0$ ; since $\sin$ is increasing on $(0,\pi/2)$ and $1/n$ is decreasing, $b_n=\sin(1/n)$ is decreasing in $n$ ; and $b_n \to \sin(0)=0$ by continuity of $\sin$ at $0$ combined with $1/n\to 0$ . All three Leibniz hypotheses hold, so $\sum(-1)^n\sin(1/n)$ converges. Since $\sin(1/n)\sim 1/n$ for large $n$ (comparison test with the divergent harmonic series, via $\lim_{n\to\infty}\frac{\sin(1/n)}{1/n}=1$ ), the convergence is only conditional, not absolute.