The alternating series test

Theorem

The alternating series test

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Statement

Let \( (a_n)_{n \ge 1} \) be a sequence of real numbers satisfying: (i) \( a_n \ge 0 \) for all \( n \in \mathbb{N} \); (ii) \( (a_n) \) is monotone decreasing, i.e. \( a_{n+1} \le a_n \) for all \( n \); and (iii) \( a_n \to 0 \) as \( n \to \infty \). Then the alternating series \[ \sum_{n=1}^{\infty} (-1)^{n+1} a_n = a_1 - a_2 + a_3 - a_4 + \cdots \] converges to some \( s \in \mathbb{R} \). Moreover, for every \( N \in \mathbb{N} \) the partial sum \( s_N = \sum_{n=1}^{N} (-1)^{n+1} a_n \) satisfies the error bound \[ \lvert s - s_N \rvert \le a_{N+1}, \] and \( s \) always lies between any two consecutive partial sums \( s_N \) and \( s_{N+1} \).

Why it matters

This is the first convergence test in the analysis course that certifies convergence of a series whose terms do not converge absolutely: it produces genuine examples of conditional convergence, such as \( \sum (-1)^{n+1}/n \). Every comparison-type test (comparison, ratio, root, integral) can only ever establish absolute convergence, so without this theorem an entire class of series — the ones where convergence is caused by cancellation rather than smallness — would be invisible to us.

It also comes with something rare among convergence tests: a sharp, effective error estimate. The tail is controlled by the single next term \( a_{N+1} \), which makes alternating series a practical computational tool (for instance in evaluating \( \ln 2 \), \( \pi/4 \), or the values of many special functions) and a model case for the general phenomenon exploited by Dirichlet's test.

Hypotheses

Hypothesis (non-negativity and strict alternation): \( a_n \ge 0 \), so the signs of the terms \( (-1)^{n+1} a_n \) genuinely alternate.If the \( a_n \) are allowed arbitrary signs, the "alternating" structure is destroyed: take \( a_n = (-1)^{n+1}/n \), so that \( (-1)^{n+1} a_n = 1/n \) and the series is the harmonic series, which diverges. The cancellation mechanism the proof relies on is exactly the sign pattern \( +,-,+,-,\dots \)

Hypothesis (monotonicity): \( a_{n+1} \le a_n \) for all \( n \).Take \( a_{2k-1} = 1/k \) and \( a_{2k} = 1/k^2 \). Then \( a_n \to 0 \) and the signs alternate, but \( s_{2n} = \sum_{k=1}^{n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k^2} \to +\infty \), so the series diverges. Decreasingness is what forces consecutive terms to cancel almost completely.

Hypothesis (null sequence): \( a_n \to 0 \).Take \( a_n = 1 \) for all \( n \): the partial sums of \( 1 - 1 + 1 - 1 + \cdots \) oscillate between \( 1 \) and \( 0 \) and never converge. By the \( n \)-th term test this hypothesis is necessary for any convergent series, so it cannot be dropped.

Hypothesis (implicit): the terms are real numbers, and monotonicity is used through the order structure of \( \mathbb{R} \).The proof runs on the monotone convergence theorem, i.e. on order-completeness of \( \mathbb{R} \). In \( \mathbb{Q} \) the statement fails: one can build a decreasing rational null sequence whose alternating sum is irrational, so the partial sums are a Cauchy sequence in \( \mathbb{Q} \) with no limit in \( \mathbb{Q} \). Note also that only eventual monotonicity and eventual non-negativity are needed, since changing finitely many terms alters neither convergence nor the argument applied to the tail — but then the error bound \( \lvert s - s_N \rvert \le a_{N+1} \) is only valid once \( N \) is inside the monotone regime.

Proof

Write \( s_n = \sum_{k=1}^{n} (-1)^{k+1} a_k \) for the \( n \)-th partial sum. We must show \( (s_n) \) converges. We study the even-indexed and odd-indexed subsequences \( (s_{2n}) \) and \( (s_{2n-1}) \) separately.

Convergence of a series is by definition convergence of its sequence of partial sums. A

\[ s_{2n+2} - s_{2n} = a_{2n+1} - a_{2n+2} \ge 0, \] so \( (s_{2n})_{n \ge 1} \) is monotone increasing.

Direct computation: \( s_{2n+2} = s_{2n} + (-1)^{2n+2} a_{2n+1} + (-1)^{2n+3} a_{2n+2} = s_{2n} + a_{2n+1} - a_{2n+2} \). Non-negativity of the difference is precisely hypothesis (ii), \( a_{2n+2} \le a_{2n+1} \). B

\[ s_{2n+1} - s_{2n-1} = -a_{2n} + a_{2n+1} \le 0, \] so \( (s_{2n-1})_{n \ge 1} \) is monotone decreasing.

Same computation with odd indices: \( s_{2n+1} = s_{2n-1} - a_{2n} + a_{2n+1} \), and \( a_{2n+1} \le a_{2n} \) by hypothesis (ii). A

For every \( n \): \[ s_2 \le s_{2n} \le s_{2n} + a_{2n+1} = s_{2n+1} \le s_1 . \] Hence \( (s_{2n}) \) is bounded above by \( s_1 \) and \( (s_{2n-1}) \) is bounded below by \( s_2 \).

The identity \( s_{2n+1} = s_{2n} + a_{2n+1} \) and hypothesis (i), \( a_{2n+1} \ge 0 \), give \( s_{2n} \le s_{2n+1} \). Then \( s_{2n} \le s_{2n+1} \le s_1 \) since odd partial sums decrease (Step 3), and \( s_{2n+1} \ge s_{2n} \ge s_2 \) since even partial sums increase (Step 2). This interlacing is the key structural fact of the proof. B

\[ s_{2n} \to L \quad \text{and} \quad s_{2n-1} \to M \quad \text{for some } L, M \in \mathbb{R}. \]

Monotone convergence theorem (a bounded monotone sequence of reals converges — a consequence of the least upper bound axiom for \( \mathbb{R} \)): \( (s_{2n}) \) is increasing and bounded above (Steps 2, 4), \( (s_{2n-1}) \) is decreasing and bounded below (Steps 3, 4). B

\[ M - L = \lim_{n \to \infty} \left( s_{2n+1} - s_{2n} \right) = \lim_{n \to \infty} a_{2n+1} = 0, \quad \text{so } L = M =: s. \]

Algebra of limits applied to the convergent sequences \( (s_{2n+1}) \) and \( (s_{2n}) \); the last equality uses hypothesis (iii), \( a_n \to 0 \), together with the fact that every subsequence of a convergent sequence converges to the same limit (applied to the subsequence \( (a_{2n+1}) \)). C

Claim (interleaving lemma): if \( s_{2n} \to s \) and \( s_{2n-1} \to s \), then \( s_n \to s \). Proof: let \( \varepsilon \gt 0 \). Choose \( N_1 \) with \( \lvert s_{2n} - s \rvert \lt \varepsilon \) for all \( 2n \ge N_1 \), and \( N_2 \) with \( \lvert s_{2n-1} - s \rvert \lt \varepsilon \) for all \( 2n-1 \ge N_2 \). For \( n \ge \max(N_1, N_2) \), the index \( n \) is either even or odd, and in either case \( \lvert s_n - s \rvert \lt \varepsilon \). Hence \( s_n \to s \), i.e. \( \sum_{n=1}^{\infty} (-1)^{n+1} a_n \) converges.

This is the standard lemma that a sequence whose even and odd subsequences converge to a common limit converges; the two subsequences exhaust all indices, so the \( \varepsilon\text{–}N \) definition of the limit is verified directly. C

Error bound: for every \( N \), \( s \) lies between \( s_N \) and \( s_{N+1} \), hence \[ \lvert s - s_N \rvert \le \lvert s_{N+1} - s_N \rvert = a_{N+1}. \]

By Steps 2–5, \( (s_{2n}) \) increases to \( s \) and \( (s_{2n-1}) \) decreases to \( s \) (a monotone sequence converges to its supremum resp. infimum), so \( s_{2n} \le s \le s_{2m-1} \) for all \( n, m \). In particular \( s \) lies in the interval with endpoints \( s_N \) and \( s_{N+1} \) (one endpoint is an even partial sum, the other odd), an interval of length exactly \( \lvert s_{N+1} - s_N \rvert = a_{N+1} \). B

Result

\[ a_n \ge 0, \quad a_{n+1} \le a_n, \quad a_n \to 0 \;\; \Longrightarrow \;\; \sum_{n=1}^{\infty} (-1)^{n+1} a_n \text{ converges, with } \lvert s - s_N \rvert \le a_{N+1}. \]

Reading. If you add and subtract alternately amounts that shrink steadily to zero, the running totals zigzag inward — each swing overshoots the limit by less than the previous one — and are trapped onto a single value. The error after stopping is never worse than the first term you ignored.

Scope. Applies to real series whose terms alternate strictly in sign with absolute values (eventually) decreasing to \( 0 \). It gives convergence only — typically conditional convergence; absolute convergence must be checked separately. It says nothing about series whose sign pattern is not \( +,-,+,-,\dots \) (for those, see Dirichlet's test), and the error bound requires the monotone regime to have started by index \( N \).

Corollaries & converses

Conditional convergence exists. \( \sum_{n=1}^{\infty} (-1)^{n+1}/n \) converges (to \( \ln 2 \)), yet \( \sum 1/n \) diverges: convergence and absolute convergence are genuinely different notions.
Free family of examples. \( \sum (-1)^{n+1} n^{-p} \) converges for every \( p \gt 0 \), absolutely iff \( p \gt 1 \).
Effective computation. To compute \( s \) within tolerance \( \varepsilon \), it suffices to take \( N \) with \( a_{N+1} \le \varepsilon \); moreover \( s - s_N \) has the same sign as the first omitted term \( (-1)^{N+2} a_{N+1} \).
Converse fails (monotonicity is not necessary). \( \sum (-1)^{n+1} b_n \) with \( b_n = \bigl(2 + (-1)^n\bigr)/n^2 \) converges (absolutely, by comparison with \( 3/n^2 \)) although \( (b_n) \) is not monotone. The test is sufficient, never necessary.
Converse fails in the other direction too. An alternating series with \( a_n \to 0 \) but \( (a_n) \) not decreasing may diverge (see "Fails without"), so \( a_n \to 0 \) alone characterises nothing for alternating series.
Special case of Dirichlet's test. Take \( b_n = (-1)^{n+1} \), whose partial sums are bounded by \( 1 \), and \( (a_n) \) decreasing to \( 0 \): Dirichlet's test recovers this theorem verbatim.

Fails without

Drop monotonicity. Let \( a_{2k-1} = \frac{1}{k} \), \( a_{2k} = \frac{1}{k^2} \). Then \( a_n \gt 0 \) and \( a_n \to 0 \), but \[ s_{2n} = \sum_{k=1}^{n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k^2} \ge \sum_{k=1}^{n} \frac{1}{k} - \frac{\pi^2}{6} \to +\infty, \] since the harmonic partial sums are unbounded while \( \sum 1/k^2 \) converges. The alternating series diverges: without decreasingness the positive terms can systematically dominate the negative ones.
Drop \( a_n \to 0 \). Let \( a_n = \frac{n+1}{n} \), which is positive and decreasing but with limit \( 1 \). Then \( (-1)^{n+1} a_n \not\to 0 \), so by the \( n \)-th term test the series \( \sum (-1)^{n+1} \frac{n+1}{n} \) diverges; concretely \( s_{2n+1} - s_{2n} = a_{2n+1} \to 1 \), so the partial sums cannot be Cauchy.
Drop the sign pattern. Let \( a_n = \frac{1 + (-1)^{n+1}}{n} \) placed so the "alternating" series becomes \( \sum_{n \text{ odd}} \frac{2}{n} \): grouping signs freely, without strict alternation the cancellation disappears and comparison with the harmonic series gives divergence. Alternation is structural, not cosmetic.

Common errors

Skipping the monotonicity check. Verifying only \( a_n \to 0 \) and citing the test. This is the single most common error: the counterexample \( a_{2k-1} = 1/k \), \( a_{2k} = 1/k^2 \) shows the conclusion can genuinely fail. For \( a_n = f(n) \) with differentiable \( f \), check \( f'(x) \le 0 \) eventually.
Off-by-one in the error bound. Writing \( \lvert s - s_N \rvert \le a_N \) instead of \( \le a_{N+1} \). The bound is the first omitted term.
Claiming absolute convergence. The test never yields absolute convergence; \( \sum (-1)^{n+1}/n \) converges while \( \sum 1/n \) does not. "Converges by the alternating series test" and "converges absolutely" must be kept separate.
Applying the test to a non-alternating series. Series like \( \sum \frac{\sin n}{n} \) have infinitely many sign changes but are not alternating; they need Dirichlet's test, not this one.
Concluding divergence when the test fails. If \( (a_n) \) is not monotone, the test is silent — it does not prove divergence (the test is not necessary; see the converse above).
Treating the sum as rearrangement-invariant. A conditionally convergent alternating series may be rearranged to converge to any real number (Riemann rearrangement theorem); manipulations that reorder terms are illegal without absolute convergence.

Discussion

The theorem is due to Leibniz, who stated the criterion in correspondence around 1682 in connection with the series \( \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots \); it is still often called the Leibniz criterion. Historically it predates any rigorous theory of convergence by well over a century — Leibniz's own justification was essentially the zigzag picture of the partial sums — and the modern proof via the monotone convergence theorem is a nineteenth-century repackaging, made possible once Bolzano, Cauchy and finally Dedekind and Cantor pinned down completeness of \( \mathbb{R} \).

Structurally, the proof is a nested-intervals argument in disguise: the intervals \( I_n = [\,s_{2n},\, s_{2n+1}\,] \) satisfy \( I_{n+1} \subseteq I_n \) and \( \operatorname{length}(I_n) = a_{2n+1} \to 0 \), so their intersection is a single point, which is the sum. This viewpoint explains why the error bound is automatic rather than an afterthought: the limit is trapped between consecutive partial sums at every stage. Few convergence tests localise the limit like this; the ratio and root tests, by contrast, give convergence with only indirect information about the tail.

The natural home of the result is the theory of series with mixed signs. The alternating series test is the simplest instance of Dirichlet's test (if \( (a_n) \) decreases to \( 0 \) and the partial sums of \( \sum b_n \) are bounded, then \( \sum a_n b_n \) converges), obtained by taking \( b_n = (-1)^{n+1} \). Dirichlet's test, in turn, is proved by Abel summation (summation by parts), the discrete analogue of integration by parts — so the cancellation exploited here is the same mechanism that makes \( \int_1^\infty \frac{\sin x}{x}\, dx \) converge. Conditional convergence is also the gateway to the Riemann rearrangement theorem: any series this test certifies as convergent but not absolutely convergent can be rearranged to sum to any prescribed real number, or to diverge.

Two finer points. First, the convergence \( s_N \to s \) can be arbitrarily slow: for \( \sum (-1)^{n+1}/n \) the error after \( N \) terms is \( \sim \frac{1}{2N} \), so raw partial sums are computationally poor even though the test is sharp; acceleration methods (Euler's transformation, or averaging \( \frac{1}{2}(s_N + s_{N+1}) \), which cancels the leading error term) exploit precisely the sign structure the proof reveals — indeed \( s - \frac{1}{2}(s_N + s_{N+1}) = \mathcal{O}(a_{N+1} - a_{N+2}) \) when \( (a_n) \) is convex. Second, the argument is irreducibly order-theoretic: in a general Banach space there is no monotone convergence theorem, and the correct generalisation is Dirichlet's test via Abel summation, whose proof uses completeness through the Cauchy criterion instead of suprema. Over \( \mathbb{Q} \) the statement is simply false, which makes this theorem a clean diagnostic for where completeness enters a first analysis course.

Common misconceptions. The test does not say "alternating series with terms tending to zero converge" — monotonicity is essential, not decorative. It is a one-way implication: failure of its hypotheses proves nothing. And it never establishes absolute convergence, so theorems requiring absolute convergence (rearrangement, products of series) remain unavailable after applying it.

Worked examples

Example 1. Show that the alternating harmonic series \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \) converges, and find \( N \) such that \( s_N \) approximates the sum with error at most \( 10^{-2} \).

Set \( a_n = \frac{1}{n} \). Then \( a_n \gt 0 \) for all \( n \in \mathbb{N} \).

Hypothesis (i) holds by inspection. A

\[ a_{n+1} = \frac{1}{n+1} \lt \frac{1}{n} = a_n \quad \text{for all } n, \] so \( (a_n) \) is (strictly) decreasing.

\( n \lt n+1 \) and reciprocals of positive reals reverse the order. Hypothesis (ii) holds. A

\[ \lim_{n \to \infty} \frac{1}{n} = 0. \]

Standard limit (Archimedean property of \( \mathbb{R} \)). Hypothesis (iii) holds, so the alternating series test applies and the series converges to some \( s \). A

Error requirement: \( \lvert s - s_N \rvert \le a_{N+1} = \frac{1}{N+1} \le 10^{-2} \iff N + 1 \ge 100 \iff N \ge 99. \)

Direct application of the error bound from the theorem; solve the inequality for \( N \). B

\[ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \text{ converges (in fact to } \ln 2\text{)}, \quad \text{and } \lvert s - s_{99} \rvert \le \tfrac{1}{100} = 10^{-2}. \]

Reading. All three hypotheses are one-line checks here; the payoff is a convergent series that is not absolutely convergent, plus an explicit stopping rule: 99 terms guarantee two decimal places.

Scope. The identification of the sum as \( \ln 2 \) needs the Taylor series of \( \ln(1+x) \) at \( x=1 \) (Abel's theorem); the alternating series test alone gives convergence and the error bound.

Example 2. Determine whether \( \sum_{n=2}^{\infty} (-1)^{n} \frac{\ln n}{n} \) converges, and classify the convergence as absolute or conditional.

Set \( a_n = \frac{\ln n}{n} \) for \( n \ge 2 \). Then \( a_n \gt 0 \) for \( n \ge 2 \) since \( \ln n \gt 0 \) there.

Hypothesis (i), on the tail \( n \ge 2 \); finitely many initial terms never affect convergence. A

Let \( f(x) = \frac{\ln x}{x} \). Then \[ f'(x) = \frac{1 - \ln x}{x^2} \lt 0 \quad \text{for } x \gt e, \] so \( (a_n) \) is decreasing for \( n \ge 3 \).

Quotient rule; \( 1 - \ln x \lt 0 \iff x \gt e \), and a function with negative derivative on \( (e, \infty) \) is decreasing there (mean value theorem). Eventual monotonicity suffices, as noted in Hypotheses. B

\[ \lim_{n \to \infty} \frac{\ln n}{n} = 0. \]

Standard growth comparison: \( \ln n \le \sqrt{n} \) for large \( n \) (e.g. since \( \ln x / \sqrt{x} \to 0 \) by L'Hôpital or by \( \ln x \le 2(\sqrt{x} - 1) \)), so \( 0 \le \frac{\ln n}{n} \le \frac{1}{\sqrt{n}} \to 0 \); squeeze theorem. Hypothesis (iii) holds. B

By the alternating series test applied to the tail \( n \ge 3 \), \( \sum_{n \ge 3} (-1)^n \frac{\ln n}{n} \) converges; adding the finitely many terms \( n = 2 \) back, the full series converges.

The theorem, plus the fact that prepending finitely many terms preserves convergence of a series. A

Absolute convergence fails: for \( n \ge 3 \), \[ \frac{\ln n}{n} \ge \frac{1}{n}, \] and \( \sum \frac{1}{n} \) diverges, so \( \sum \frac{\ln n}{n} \) diverges by comparison.

\( \ln n \ge 1 \) for \( n \ge 3 \); comparison test with the harmonic series. B

\[ \sum_{n=2}^{\infty} (-1)^{n} \frac{\ln n}{n} \text{ converges conditionally.} \]

Reading. A typical exam-grade application: monotonicity is not obvious and must be earned with calculus, and it only holds eventually — which is exactly enough.

Scope. The conclusion is conditional convergence only; any manipulation requiring absolute convergence (rearrangement, Cauchy products) is forbidden for this series.

Problems

Show that \( \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n-1} \) converges. Is the convergence absolute?
Solution

Let \( a_n = \frac{1}{2n-1} \gt 0 \). Since \( 2(n+1) - 1 = 2n+1 \gt 2n-1 \), taking reciprocals gives \( a_{n+1} \lt a_n \), so \( (a_n) \) is decreasing; and \( 0 \lt a_n \lt \frac{1}{n} \to 0 \) gives \( a_n \to 0 \) by the squeeze theorem. All three hypotheses of the alternating series test hold, so the series converges. It is not absolutely convergent: \( \frac{1}{2n-1} \ge \frac{1}{2n} \) and \( \sum \frac{1}{2n} = \frac{1}{2} \sum \frac{1}{n} \) diverges, so by comparison \( \sum \frac{1}{2n-1} \) diverges. The convergence is conditional. (The sum is \( \pi/4 \), Leibniz's series, though that identification needs more than this test.)
Determine the convergence behaviour of \( \displaystyle\sum_{n=1}^{\infty} (-1)^{n} \frac{n}{n^2 + 1} \).
Solution

Let \( a_n = \frac{n}{n^2+1} \gt 0 \). Monotonicity: with \( f(x) = \frac{x}{x^2+1} \), \[ f'(x) = \frac{(x^2+1) - x \cdot 2x}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2} \lt 0 \quad \text{for } x \gt 1, \] so \( (a_n) \) is decreasing for \( n \ge 1 \): the derivative handles \( n \ge 2 \), and \( a_1 = \tfrac{1}{2} \gt \tfrac{2}{5} = a_2 \) directly. Also \( 0 \lt a_n \le \frac{n}{n^2} = \frac{1}{n} \to 0 \). The alternating series test gives convergence. Absolute convergence fails: \( \frac{n}{n^2+1} \ge \frac{n}{n^2 + n^2} = \frac{1}{2n} \) for \( n \ge 1 \), and \( \sum \frac{1}{2n} \) diverges, so by comparison the series of absolute values diverges. Conclusion: conditionally convergent.
How many terms of \( \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} \) suffice to compute its sum with error at most \( 10^{-4} \)? Show also that the error \( s - s_N \) has the sign of the first omitted term \( (-1)^{N+2} a_{N+1} \).
Solution

Here \( a_n = \frac{1}{n^2} \) is positive, decreasing, and null, so the test applies and \( \lvert s - s_N \rvert \le a_{N+1} = \frac{1}{(N+1)^2} \). We need \( \frac{1}{(N+1)^2} \le 10^{-4} \iff (N+1)^2 \ge 10^4 \iff N + 1 \ge 100 \iff N \ge 99 \). So \( 99 \) terms suffice. For the sign: from the proof, the even partial sums increase to \( s \) and the odd partial sums decrease to \( s \), so \( s_{2m} \le s \le s_{2m-1} \) for all \( m \). Hence if \( N \) is even, \( s - s_N \ge 0 \), matching the positive first omitted term \( (-1)^{N+2} a_{N+1} = + a_{N+1} \); if \( N \) is odd, \( s - s_N \le 0 \), matching the negative omitted term. In both cases \( s - s_N \) has the sign of the first omitted term. (The sum is \( \pi^2/12 \), from \( \sum 1/n^2 = \pi^2/6 \) by splitting even and odd terms.)
Define \( a_{2k-1} = \frac{1}{k} \) and \( a_{2k} = \frac{1}{k^2} \) for \( k \ge 1 \). Show that \( a_n \gt 0 \) and \( a_n \to 0 \), but that \( \displaystyle\sum_{n=1}^{\infty} (-1)^{n+1} a_n \) diverges. Which hypothesis of the alternating series test fails?
Solution

Positivity is clear. For the limit: given \( \varepsilon \gt 0 \), choose \( K \gt 1/\varepsilon \); for all \( n \gt 2K \), either \( n = 2k-1 \) with \( k \gt K \) and \( a_n = 1/k \lt \varepsilon \), or \( n = 2k \) with \( k \gt K \) and \( a_n = 1/k^2 \le 1/k \lt \varepsilon \). So \( a_n \to 0 \). Now compute the even partial sums: \[ s_{2n} = \sum_{k=1}^{n} \left( a_{2k-1} - a_{2k} \right) = \sum_{k=1}^{n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k^2}. \] The second sum is bounded above by \( \sum_{k=1}^{\infty} 1/k^2 = \pi^2/6 \) (any convergent bound suffices, e.g. \( 1 + \sum_{k \ge 2} \frac{1}{k(k-1)} = 2 \) by telescoping), while \( \sum_{k=1}^n 1/k \to \infty \) (harmonic series). Hence \( s_{2n} \ge \sum_{k=1}^{n} \frac{1}{k} - 2 \to +\infty \), so \( (s_n) \) is unbounded and the series diverges. The failed hypothesis is monotonicity: for every \( k \ge 2 \) we have \( k^2 \gt k+1 \), hence \( a_{2k} = \frac{1}{k^2} \lt \frac{1}{k+1} = a_{2k+1} \), so the sequence increases at infinitely many steps and \( (a_n) \) is not even eventually decreasing.
Let \( (a_n) \) be decreasing with \( a_n \to L \gt 0 \). Prove that the partial sums \( s_n = \sum_{k=1}^n (-1)^{k+1} a_k \) do not converge, but that the even and odd partial sums each converge, and compute the gap between their limits. Deduce a second proof that hypothesis (iii) of the alternating series test cannot be dropped.
Solution

Apply the test's own machinery to \( b_n = a_n - L \): \( (b_n) \) is decreasing (differences unchanged), \( b_n \ge 0 \) (a decreasing sequence lies above its limit: if some \( a_m \lt L \), then \( a_n \le a_m \lt L \) for \( n \ge m \), contradicting \( a_n \to L \)), and \( b_n \to 0 \). By the alternating series test, \( \sum (-1)^{k+1} b_k \) converges; call its sum \( \sigma \) and its partial sums \( t_n \). Now \[ s_n = t_n + L \sum_{k=1}^{n} (-1)^{k+1} = t_n + L \cdot \chi_n, \quad \text{where } \chi_n = 1 \text{ if } n \text{ odd}, \; \chi_n = 0 \text{ if } n \text{ even}. \] Therefore \( s_{2n} = t_{2n} \to \sigma \) and \( s_{2n-1} = t_{2n-1} + L \to \sigma + L \): both subsequential limits exist, and they differ by exactly \( L \gt 0 \). A sequence with two distinct subsequential limits does not converge (limits of sequences are unique, and every subsequence of a convergent sequence shares its limit), so \( (s_n) \) diverges. Since every hypothesis of the alternating series test except \( a_n \to 0 \) is satisfied here, hypothesis (iii) is indispensable — and the computation shows precisely how failure occurs: the partial sums split into two parallel convergent tracks separated by the gap \( L \).