A conditionally convergent series is a warning sign. Its terms go to zero, its partial sums settle to a finite value, and yet the positive mass and the negative mass are both infinite when separated. The convergence is produced by cancellation rather than by small total size. Alternating series are the first systematic setting where this cancellation can be controlled. The simplest mystery is the harmonic series. The series $\sum_{n=1}^{\infty} 1/n$ diverges, so making the terms smaller only as $1/n$ is not enough for convergence. But inserting alternating signs changes the behaviour completely: \begin{align*} 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots. \end{align*} The terms still have the same absolute values as the harmonic series. The difference is that each positive term is followed by a slightly smaller negative correction, and each negative term is followed by a still smaller positive correction. The partial sums are trapped from above and below. [example: The Alternating Harmonic Series] Consider the partial sums \begin{align*} s_N=\sum_{n=1}^{N}\frac{(-1)^{n+1}}{n}. \end{align*} For $m\in\mathbb{N}$, the next two terms after $s_{2m}$ are positive and then negative, so \begin{align*} s_{2m+2}=s_{2m}+\frac{1}{2m+1}-\frac{1}{2m+2}. \end{align*} Hence \begin{align*} s_{2m+2}-s_{2m}=\frac{1}{2m+1}-\frac{1}{2m+2}=\frac{1}{(2m+1)(2m+2)}>0. \end{align*} Thus the even partial sums $(s_{2m})$ are increasing. Similarly, the next two terms after $s_{2m+1}$ are negative and then positive, so \begin{align*} s_{2m+3}=s_{2m+1}-\frac{1}{2m+2}+\frac{1}{2m+3}. \end{align*} Therefore \begin{align*} s_{2m+3}-s_{2m+1}=-\frac{1}{2m+2}+\frac{1}{2m+3}=-\frac{1}{(2m+2)(2m+3)}<0. \end{align*} Thus the odd partial sums $(s_{2m-1})$ are decreasing. The two subsequences are interlaced. Since \begin{align*} s_{2m+1}=s_{2m}+\frac{1}{2m+1}, \end{align*} we have \begin{align*} s_{2m}\le s_{2m+1}. \end{align*} Also \begin{align*} s_{2m-1}=s_{2m}+\frac{1}{2m}, \end{align*} so every even partial sum lies below the preceding odd partial sum. Hence the increasing sequence $(s_{2m})$ is bounded above, for instance by $s_1$, and the decreasing sequence $(s_{2m-1})$ is bounded below, for instance by $s_2$. By the monotone convergence principle, both subsequences have limits. Finally, their gap is exactly the last added positive term: \begin{align*} s_{2m+1}-s_{2m}=\frac{1}{2m+1}. \end{align*} Since $\frac{1}{2m+1}\to 0$, the two subsequential limits are equal. Therefore the alternating harmonic series converges, even though the associated positive harmonic series $\sum_{n=1}^{\infty}1/n$ diverges; the convergence comes from the ordered cancellation between successive positive and negative terms. [/example] This page studies alternating series as a child concept of [series](/page/Series). The parent theory asks when the partial sums of $\sum a_n$ converge. Here the special structure is that the signs alternate and the magnitudes usually decrease. The point is not only to prove convergence; it is to obtain a usable error estimate and to understand why conditional convergence is delicate. ## Definition A general series allows arbitrary signs, so there is no reason for cancellation to occur in an orderly way. Alternating series isolate the most regular possible sign pattern: plus, minus, plus, minus, or the same pattern shifted by one place. The remaining data are the non-negative magnitudes of the terms. [definition: Alternating Series] An alternating series is a real series of the form \begin{align*} \sum_{n=1}^{\infty} (-1)^{n+1} b_n \end{align*} or \begin{align*} \sum_{n=1}^{\infty} (-1)^n b_n, \end{align*} where $b_n \ge 0$ for every $n \in \mathbb{N}$. [/definition] The sign pattern still leaves a major question unanswered: when is the cancellation orderly enough to force convergence? If later corrections can be larger than earlier ones, the partial sums need not settle. The useful special case is therefore the one where the magnitudes shrink monotonically to zero. [definition: Decreasing Alternating Series] A decreasing alternating series is an alternating series \begin{align*} \sum_{n=1}^{\infty} (-1)^{n+1} b_n \end{align*} with $b_n \ge 0$ for every $n \in \mathbb{N}$, $b_{n+1} \le b_n$ for every $n \in \mathbb{N}$, and $b_n \to 0$ as $n \to \infty$. [/definition] The word decreasing is often used in the non-strict sense: equality between consecutive terms is allowed. What matters is that the later corrections never exceed the earlier errors they are meant to correct. To compare an infinite alternating sum with its proposed value, we need to watch the finite approximations separately according to parity. The odd approximations and even approximations usually lie on opposite sides of the sum, so naming them lets us state the bracketing mechanism precisely. [definition: Alternating Partial Sums] For an alternating series $\sum_{n=1}^{\infty} (-1)^{n+1}b_n$, the $N$-th partial sum is \begin{align*} s_N = \sum_{n=1}^{N} (-1)^{n+1}b_n. \end{align*} The even partial sums are $(s_{2m})_{m=1}^{\infty}$, and the odd partial sums are $(s_{2m-1})_{m=1}^{\infty}$. [/definition] The whole theory rests on the geometry of these two subsequences on the real line. Even partial sums approach the eventual answer from one side, odd partial sums approach from the other side, and the gap between them is the next omitted magnitude. ## Convergence by Trapping ### Monotone Subsequences The convergence test for alternating series is not a [comparison test](/theorems/173) in the usual sense. The absolute values may form a divergent positive series. Instead, the argument is an order argument: one monotone subsequence rises, another monotone subsequence falls, and the two are squeezed together. [quotetheorem:177] Once convergence is known, the next question is where the limit sits relative to the finite partial sums. Alternating series have a stronger structure than an arbitrary convergent series: the odd and even partial sums form a nested corridor around the limiting value. [remark: Bracketing of Alternating Partial Sums] Under the hypotheses of the [alternating series test](/theorems/177), the odd and even partial sums approach the same limit from opposite sides. With the convention \begin{align*} s_N=\sum_{n=1}^N(-1)^{n+1}b_n, \end{align*} one has \begin{align*} s_2\le s_4\le \cdots \le S \le \cdots \le s_5\le s_3\le s_1, \end{align*} where $S$ is the sum of the series. In particular, every partial sum lies on a known side of the limit, and the interval between two consecutive partial sums has length equal to the next magnitude. [/remark] This bracketing is what makes alternating series unusually useful for numerical approximation. A finite partial sum is not merely close in some eventual sense; it lies on a known side of the true value. [example: Bracketing the Alternating Harmonic Sum] Let \begin{align*} S = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}. \end{align*} Here $b_n=1/n$, and $b_{n+1}\le b_n$ because $n\le n+1$ with positive denominators, while $b_n\to 0$. Thus the bracketing result for decreasing alternating series applies. The fourth partial sum is \begin{align*} s_4 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4}. \end{align*} First, \begin{align*} 1-\frac{1}{2}=\frac{1}{2}. \end{align*} Then \begin{align*} \frac{1}{2}+\frac{1}{3}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6}. \end{align*} Therefore \begin{align*} s_4=\frac{5}{6}-\frac{1}{4}=\frac{10}{12}-\frac{3}{12}=\frac{7}{12}. \end{align*} The fifth partial sum adds the next positive term: \begin{align*} s_5=s_4+\frac{1}{5}. \end{align*} Substituting $s_4=7/12$ gives \begin{align*} s_5=\frac{7}{12}+\frac{1}{5}=\frac{35}{60}+\frac{12}{60}=\frac{47}{60}. \end{align*} By *Bracketing of Alternating Partial Sums*, the case $m=2$ gives $s_4\le S$, and the case $m=3$ gives $S\le s_5$. Hence \begin{align*} \frac{7}{12} \le S \le \frac{47}{60}. \end{align*} The interval has width \begin{align*} \frac{47}{60}-\frac{7}{12}=\frac{47}{60}-\frac{35}{60}=\frac{12}{60}=\frac{1}{5}, \end{align*} so this is a rough approximation, but it is certified by the alternating structure alone. The later endpoint discussion explains why this sum is equal to $\log 2$. [/example] ### Tails and Initial Terms The example above behaves well from the first term, but many useful series only become orderly later. Since adding or removing finitely many terms only changes finitely many partial sums, the convergence question should depend on the tail rather than on early noise. [definition: Eventually Decreasing Sequence] A real sequence $(b_n)_{n=1}^{\infty}$ is eventually decreasing if there exists $N_0 \in \mathbb{N}$ such that $b_{n+1} \le b_n$ for every $n \ge N_0$. [/definition] This tail condition raises the natural finite-prefix question: does the alternating test still work if the monotone behaviour begins only after some index? Since convergence is controlled by the tail of a series, the answer should preserve the same criterion after discarding finitely many early terms. The formal tool is still the ordinary alternating series test, but it is applied after reindexing the tail where monotonicity has begun. [quotetheorem:177] Indeed, if $b_n \ge 0$, $b_n\to 0$, and $(b_n)$ is eventually decreasing, choose $N_0$ so that $b_{n+1}\le b_n$ for every $n\ge N_0$. The tail \begin{align*} \sum_{n=N_0}^{\infty}(-1)^{n+1}b_n \end{align*} is, up to an overall sign and a reindexing, an alternating series whose magnitudes decrease from its first term onward and tend to zero. The quoted theorem therefore proves convergence of that tail. The finite initial segment changes the value of the sum, but it does not change whether the whole series converges. This is a recurring principle in series: convergence is a statement about eventual behaviour. ## Error Control For a positive series, a partial sum often gives little information unless a separate tail estimate is available. Alternating series package the tail estimate into the sign pattern itself. Once the magnitudes decrease to zero, the next term controls the entire remaining error. [definition: Remainder of a Series] Let $\sum_{n=1}^{\infty} a_n$ be a convergent real series with sum $s$, and let \begin{align*} s_N = \sum_{n=1}^{N} a_n. \end{align*} The $N$-th remainder is \begin{align*} R_N = s - s_N. \end{align*} [/definition] The remainder measures what the finite computation has missed. In an alternating series, the remainder has two special properties: its sign is the sign of the next term, and its magnitude is at most the next magnitude. [quotetheorem:9709] This estimate is the main practical payoff. To approximate the sum within a prescribed tolerance, it is enough to continue until the next omitted magnitude is below that tolerance. [example: How Many Terms Approximate $\log 2$?] For the alternating harmonic series, the endpoint identity gives \begin{align*} \log 2=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}. \end{align*} With $b_n=1/n$, the next omitted magnitude after the $N$-th partial sum is $b_{N+1}=1/(N+1)$, so the *[Alternating Series Remainder Estimate](/theorems/9709)* gives \begin{align*} \left|\log 2-\sum_{n=1}^{N}\frac{(-1)^{n+1}}{n}\right|\le \frac{1}{N+1}. \end{align*} To guarantee error at most $10^{-3}$, it is therefore enough to make the upper bound at most $10^{-3}$. Since $N+1>0$, the inequality \begin{align*} \frac{1}{N+1}\le 10^{-3} \end{align*} is equivalent to \begin{align*} 1\le 10^{-3}(N+1). \end{align*} Multiplying by $10^3$ gives \begin{align*} 10^3\le N+1. \end{align*} Since $10^3=1000$, this is \begin{align*} 1000\le N+1. \end{align*} Subtracting $1$ gives \begin{align*} 999\le N. \end{align*} Thus $N=999$ terms already guarantee error at most $10^{-3}$, because then the next omitted magnitude is \begin{align*} \frac{1}{999+1}=\frac{1}{1000}=10^{-3}. \end{align*} The convergence is slow, but the alternating structure gives a certified error bound from one visible next term. [/example] A useful refinement is that the true sum lies between consecutive partial sums. Taking their midpoint can improve the numerical approximation, although the basic theorem only guarantees the next-term error for a single partial sum. [quotetheorem:9710] This enclosure is often how alternating sums are used in computation: calculate two consecutive partial sums, and report an interval rather than a point estimate. ## Conditional and Absolute Convergence ### Two Kinds of Convergence Alternation can create convergence without making the total variation finite. This distinction is one of the first places where the order and signs of a series matter. It also explains why alternating series are powerful but fragile. [definition: Absolute Convergence] A real series $\sum_{n=1}^{\infty} a_n$ is absolutely convergent if the positive series \begin{align*} \sum_{n=1}^{\infty} |a_n| \end{align*} converges. [/definition] Absolute convergence means the series has finite total magnitude. Cancellation may still occur, but the convergence does not depend on it. The opposite behaviour also needs a name, because many alternating series converge only by balancing positive and negative contributions. Naming this case separates robust convergence from convergence that depends on the given order of the terms. [definition: Conditional Convergence] A real series $\sum_{n=1}^{\infty} a_n$ is conditionally convergent if $\sum_{n=1}^{\infty} a_n$ converges and \begin{align*} \sum_{n=1}^{\infty} |a_n| \end{align*} diverges. [/definition] The main examples should now measure how far alternation extends the usual positive-series tests. Power magnitudes $n^{-p}$ give a complete scale: the positive series has a sharp threshold at $p=1$, while the alternating version converges throughout the whole range $p>0$. [quotetheorem:9711] This theorem shows exactly what alternation adds. Without signs, the threshold for $\sum n^{-p}$ is $p>1$. With alternating signs, every exponent $p>0$ converges, because $n^{-p} \to 0$ and the magnitudes decrease. [example: Conditional Convergence at $p=1$] For $p=1$, the alternating $p$-series is \begin{align*} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}. \end{align*} Write $b_n=1/n$. For every $n\in\mathbb{N}$, $b_n\ge 0$. Also $n\le n+1$, and both denominators are positive, so \begin{align*} \frac{1}{n+1}\le \frac{1}{n}. \end{align*} Thus $b_{n+1}\le b_n$ for every $n$, and \begin{align*} \lim_{n\to\infty} b_n=\lim_{n\to\infty}\frac{1}{n}=0. \end{align*} Therefore the series \begin{align*} \sum_{n=1}^{\infty} (-1)^{n+1}b_n \end{align*} converges by the *Alternating Series Test*. Its absolute-value series is obtained term by term: \begin{align*} \left|\frac{(-1)^{n+1}}{n}\right|=\frac{|(-1)^{n+1}|}{n}=\frac{1}{n}. \end{align*} Hence \begin{align*} \sum_{n=1}^{\infty}\left|\frac{(-1)^{n+1}}{n}\right|=\sum_{n=1}^{\infty}\frac{1}{n}. \end{align*} The right-hand series is the harmonic series, which diverges by the *Divergence of the Harmonic Series*. The alternating harmonic series therefore converges but does not converge absolutely, so it is conditionally convergent. Its convergence depends on the alternating order of the positive and negative terms. [/example] ### Stability The distinction matters because absolute convergence is a form of convergence that survives many changes of viewpoint, while conditional convergence is tied to the particular cancellation pattern in which the terms are presented. One clean way to express this stability is to allow each term to depend on an input $x$. A numerical series is recovered by fixing $x$, so a theorem about pointwise convergence of series of functions includes the ordinary numerical case as the special case where the functions are constant. This formulation is useful here because it separates two ideas that are easy to confuse. Absolute convergence is strong enough to force convergence at every fixed input; conditional convergence, including the alternating harmonic series, shows that ordinary convergence can still occur without that stronger absolute control. [quotetheorem:275] This result gives a one-way guarantee: if the absolute-value series converges at a point, then the original signed series converges at that point. It does not say that every convergent series is absolutely convergent, nor does it give [uniform convergence](/page/Uniform%20Convergence) or rearrangement invariance by itself. For example, applying the theorem to constant functions recovers the familiar numerical principle that absolute convergence implies convergence. Alternating series sit exactly beyond its converse: their cancellation can create convergence even when the corresponding positive series diverges. That is why the later discussion of failure modes must keep track of which conclusions come from absolute control and which come only from ordered alternation. ## Failure Modes ### Non-Vanishing Terms The alternating series test has three hypotheses: non-negative magnitudes, monotone decrease, and convergence of the magnitudes to zero. The first is mostly a convention, because the signs can be separated from the magnitudes. The other two are mathematical content. If the terms do not tend to zero, convergence fails. If monotonicity fails completely, alternation alone gives no protection. A series cannot converge unless its terms tend to zero. Alternating signs do not change this basic necessary condition, so the first diagnostic test is to check the magnitudes before attempting any cancellation argument. [quotetheorem:3826] This necessary condition catches the most basic misuse of the alternating test. A sign pattern cannot repair terms that stay large. [example: Alternation Without Vanishing Terms] Consider the alternating series \begin{align*} \sum_{n=1}^{\infty} (-1)^{n+1}. \end{align*} Its $N$-th partial sum is \begin{align*} s_N=\sum_{n=1}^{N}(-1)^{n+1}. \end{align*} For an even index $N=2m$, the terms can be grouped into $m$ pairs: \begin{align*} s_{2m}=(1-1)+(1-1)+\cdots+(1-1). \end{align*} Each pair is $0$, so \begin{align*} s_{2m}=0. \end{align*} For an odd index $N=2m-1$, there are $m-1$ complete pairs and one remaining positive term: \begin{align*} s_{2m-1}=(1-1)+(1-1)+\cdots+(1-1)+1. \end{align*} The complete pairs contribute $0$, and the final term contributes $1$, so \begin{align*} s_{2m-1}=1. \end{align*} Thus the partial sums take the values $1,0,1,0,\ldots$ and do not approach a single real number. In the notation for alternating series, $b_n=1$ for every $n$, and therefore \begin{align*} \lim_{n\to\infty} b_n=\lim_{n\to\infty}1=1\ne 0. \end{align*} The signs alternate, but because the magnitudes do not vanish, the partial sums oscillate instead of converging. [/example] ### Irregular Magnitudes The monotonicity hypothesis is subtler. The terms can tend to zero, and the signs can alternate, but if large terms appear after much smaller ones, the bracketing argument breaks. The following example forces that failure into the open. [example: Vanishing Magnitudes Without Monotonicity] Define $b_{2m-1}=1/m$ and $b_{2m}=1/\sqrt{m}$ for $m \in \mathbb{N}$. Every $b_n$ is non-negative. The odd-indexed magnitudes satisfy $b_{2m-1}=1/m \to 0$, and the even-indexed magnitudes satisfy $b_{2m}=1/\sqrt{m}\to 0$, so $b_n\to 0$. However, the sequence is not eventually decreasing. For every $m\ge 2$, we have $\sqrt{m}<m$, and dividing by the positive number $m\sqrt{m}$ gives \begin{align*} \frac{1}{m}<\frac{1}{\sqrt{m}}. \end{align*} Thus \begin{align*} b_{2m-1}=\frac{1}{m}<\frac{1}{\sqrt{m}}=b_{2m}, \end{align*} so the magnitudes increase from index $2m-1$ to index $2m$ for infinitely many $m$. Now compute the even partial sums. Since $(-1)^{2m}=1$ and $(-1)^{2m+1}=-1$, the terms in positions $2m-1$ and $2m$ contribute \begin{align*} (-1)^{2m}b_{2m-1}+(-1)^{2m+1}b_{2m}=\frac{1}{m}-\frac{1}{\sqrt{m}}. \end{align*} Therefore \begin{align*} s_{2M}=\sum_{n=1}^{2M}(-1)^{n+1}b_n=\sum_{m=1}^{M}\left(\frac{1}{m}-\frac{1}{\sqrt{m}}\right). \end{align*} For $m\ge 4$, the inequality $\sqrt{m}\le m/2$ is equivalent to $2\sqrt{m}\le m$, and squaring both sides gives $4m\le m^2$, which holds exactly when $m\ge 4$. Hence, for $m\ge 4$, \begin{align*} \frac{1}{\sqrt{m}}\ge \frac{2}{m}. \end{align*} Subtracting from $1/m$ gives \begin{align*} \frac{1}{m}-\frac{1}{\sqrt{m}}\le \frac{1}{m}-\frac{2}{m}=-\frac{1}{m}. \end{align*} Thus, for $M\ge 4$, \begin{align*} s_{2M}\le \sum_{m=1}^{3}\left(\frac{1}{m}-\frac{1}{\sqrt{m}}\right)-\sum_{m=4}^{M}\frac{1}{m}. \end{align*} The tail $\sum_{m=4}^{M}1/m$ grows without bound as $M\to\infty$, so the right-hand side tends to $-\infty$. Hence $s_{2M}\to-\infty$, and the alternating series diverges. This example has $b_n\to 0$, but the magnitudes are not eventually decreasing; alternating signs alone do not create the bracketing needed for convergence. [/example] This example also explains why the theorem is stated with decreasing magnitudes rather than with a vague idea of cancellation. Regular cancellation is a structural condition, not a hope that positives and negatives balance eventually. ## Rearrangement and Cancellation Conditional convergence is not just weaker than absolute convergence; it behaves differently under rearrangement. In a finite sum, changing the order of the terms changes nothing. In an infinite series, changing the order changes the path taken by the partial sums, and conditionally convergent series have enough positive and negative mass to steer that path. [definition: Rearrangement of a Series] Let $\sum_{n=1}^{\infty} a_n$ be a real series. A rearrangement of this series is a series \begin{align*} \sum_{k=1}^{\infty} a_{\pi(k)}, \end{align*} where $\pi: \mathbb{N} \to \mathbb{N}$ is a bijection. [/definition] The definition creates a direct question: which convergence results survive after the terms are reordered? The stable benchmark is absolute convergence, where the total magnitude is finite and the order of summation cannot change the value. [quotetheorem:178] This result gives one reason absolute convergence is the preferred notion in many arguments. It allows the terms to be treated as a collection rather than as a fragile ordered process. For conditionally convergent series, both the positive terms and the negative terms carry infinite total mass. This creates enough freedom to aim the partial sums toward a prescribed target by taking positive terms until the sum passes the target, then negative terms until it falls below, and repeating with smaller and smaller terms. [quotetheorem:3831] The theorem reveals the cost of cancellation. A conditionally convergent series has infinitely much positive mass and infinitely much negative mass; the original order makes these masses cancel in one particular way, but other orders can force different limiting behaviour. [example: Rearranging the Alternating Harmonic Series Toward a Larger Sum] Start with the alternating harmonic series \begin{align*} 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots. \end{align*} Now rearrange it by taking two unused positive terms, then one unused negative term: \begin{align*} 1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \frac{1}{9} + \frac{1}{11} - \frac{1}{6} + \cdots. \end{align*} In the $m$-th block, the two positive terms are \begin{align*} \frac{1}{4m-3}+\frac{1}{4m-1}, \end{align*} and the negative term is \begin{align*} -\frac{1}{2m}. \end{align*} Thus after $M$ complete blocks the partial sum is \begin{align*} B_M=\sum_{m=1}^{M}\left(\frac{1}{4m-3}+\frac{1}{4m-1}-\frac{1}{2m}\right). \end{align*} The positive denominators appearing here are exactly the first $2M$ odd positive integers, so \begin{align*} \sum_{m=1}^{M}\left(\frac{1}{4m-3}+\frac{1}{4m-1}\right)=\sum_{k=1}^{2M}\frac{1}{2k-1}. \end{align*} Also \begin{align*} \sum_{m=1}^{M}\frac{1}{2m}=\frac{1}{2}\sum_{m=1}^{M}\frac{1}{m}=\frac{1}{2}H_M, \end{align*} where $H_M=\sum_{m=1}^{M}1/m$. Since \begin{align*} H_{4M}=\sum_{k=1}^{2M}\frac{1}{2k-1}+\sum_{k=1}^{2M}\frac{1}{2k}, \end{align*} and \begin{align*} \sum_{k=1}^{2M}\frac{1}{2k}=\frac{1}{2}\sum_{k=1}^{2M}\frac{1}{k}=\frac{1}{2}H_{2M}, \end{align*} we get \begin{align*} \sum_{k=1}^{2M}\frac{1}{2k-1}=H_{4M}-\frac{1}{2}H_{2M}. \end{align*} Therefore \begin{align*} B_M=H_{4M}-\frac{1}{2}H_{2M}-\frac{1}{2}H_M. \end{align*} Use the standard asymptotic fact $H_N-\log N\to \gamma$. Write $\varepsilon_N=H_N-\log N$, so $\varepsilon_N\to\gamma$. Then \begin{align*} B_M=(\log(4M)+\varepsilon_{4M})-\frac{1}{2}(\log(2M)+\varepsilon_{2M})-\frac{1}{2}(\log M+\varepsilon_M). \end{align*} The logarithmic part is \begin{align*} \log(4M)-\frac{1}{2}\log(2M)-\frac{1}{2}\log M=\log 4+\log M-\frac{1}{2}\log 2-\frac{1}{2}\log M-\frac{1}{2}\log M. \end{align*} The $\log M$ terms cancel, leaving \begin{align*} \log 4-\frac{1}{2}\log 2=2\log 2-\frac{1}{2}\log 2=\frac{3}{2}\log 2. \end{align*} The error part is \begin{align*} \varepsilon_{4M}-\frac{1}{2}\varepsilon_{2M}-\frac{1}{2}\varepsilon_M\to \gamma-\frac{1}{2}\gamma-\frac{1}{2}\gamma=0. \end{align*} Hence \begin{align*} B_M\to \frac{3}{2}\log 2. \end{align*} The partial sums inside the next incomplete block differ from some $B_M$ by either one or two positive unused odd reciprocals, and these added reciprocals tend to $0$. Thus the whole rearranged series, not only the block-end subsequence, converges to $\frac{3}{2}\log 2$. Since the original alternating harmonic series has sum $\log 2$, this rearrangement converges to a larger value, showing concretely how conditional convergence depends on the order of the terms. [/example] The rearrangement phenomenon does not contradict the alternating series test. The test applies to a specific ordered series with decreasing magnitudes. A rearranged series may no longer have that alternating decreasing form. ## Power Series and Endpoint Tests Alternating series often appear at the edge of a [power series](/page/Power%20Series) interval of convergence. Inside the interval, absolute convergence may be available. At an endpoint, absolute convergence can fail, and alternating cancellation becomes the decisive tool. [definition: Power Series] A real power series centred at $0$ is a series whose $n$-th term is the function $f_n: \mathbb{R} \to \mathbb{R}$ defined by \begin{align*} f_n(x) &= c_n x^n, \end{align*} where $(c_n)_{n=0}^{\infty}$ is a sequence of real coefficients. Its $N$-th partial sum is the real-valued function $s_N: \mathbb{R} \to \mathbb{R}$ defined by \begin{align*} s_N(x) &= \sum_{n=0}^{N} c_n x^n. \end{align*} [/definition] Power series connect alternating series to functions. Substituting a negative endpoint often inserts the factor $(-1)^n$, transforming an ordinary coefficient series into an alternating one. To know where endpoint questions begin, we first need to know the open interval where the power series behaves absolutely. The [radius of convergence](/theorems/262) records that absolute-convergence zone and leaves the boundary points as separate problems. [definition: Radius of Convergence] For a real power series $\sum_{n=0}^{\infty} c_n x^n$, the [radius of convergence](/theorems/265) is the number $R \in [0,\infty]$ such that the power series converges absolutely for $|x|<R$ and diverges for $|x|>R$. [/definition] The radius does not decide what happens at $x=R$ or $x=-R$. Endpoint behaviour must be checked separately, and alternating series are one of the standard tools. [example: Endpoint Convergence for $\log(1+x)$] For $|x|<1$, the power series is given by \begin{align*} \log(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}. \end{align*} At the endpoint $x=1$, the $n$-th term becomes \begin{align*} \frac{(-1)^{n+1}1^n}{n}=\frac{(-1)^{n+1}}{n}, \end{align*} because $1^n=1$ for every $n\in\mathbb{N}$. Thus the endpoint series at $x=1$ is \begin{align*} \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}. \end{align*} Writing $b_n=1/n$, we have $b_n\ge 0$ for every $n$, and $n\le n+1$ gives \begin{align*} \frac{1}{n+1}\le \frac{1}{n}. \end{align*} Also \begin{align*} \lim_{n\to\infty}\frac{1}{n}=0. \end{align*} Therefore the endpoint series at $x=1$ converges by the *Alternating Series Test*. This proves convergence of the endpoint numerical series; *Abel Limit Theorem for Power Series* is the separate result that identifies its sum as $\log 2$. At the endpoint $x=-1$, the $n$-th term becomes \begin{align*} \frac{(-1)^{n+1}(-1)^n}{n}=\frac{(-1)^{2n+1}}{n}. \end{align*} Since $2n+1$ is odd, \begin{align*} (-1)^{2n+1}=-1. \end{align*} Hence \begin{align*} \frac{(-1)^{n+1}(-1)^n}{n}=-\frac{1}{n}, \end{align*} so the endpoint series at $x=-1$ is \begin{align*} -\sum_{n=1}^{\infty}\frac{1}{n}. \end{align*} The harmonic series diverges by *Divergence of the Harmonic Series*, so multiplying its partial sums by $-1$ gives divergence to $-\infty$. Thus the same power series converges at $x=1$ but diverges at $x=-1$, showing that endpoints must be tested separately. [/example] Endpoint analysis is a good place to remember that the alternating test is a convergence test for numerical series, not an identity theorem for functions. A power series identity valid for $|x|<R$ may require separate justification at an endpoint. [quotetheorem:3833] This theorem explains why endpoint sums often match limiting function values when the endpoint series converges. For alternating series arising from Taylor expansions, it provides a bridge between convergence tests and function evaluation. ## Beyond and Connected Topics Alternating series sit between elementary convergence tests and the more delicate theory of conditional convergence. The immediate parent topic is [series](/page/Series), where partial sums, convergence, divergence, and standard comparison tests are developed without assuming any sign pattern. Alternating series add a special cancellation mechanism to that general framework. The natural internal continuation is Androma's [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes), where alternating series appear alongside sequences, limits, and ordinary convergence tests. The later [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology) perspective places series inside a broader analysis curriculum, where uniform convergence and function spaces begin to matter. The next mathematical direction is power series. Alternating signs frequently appear when a Taylor series is evaluated at a negative point or at a boundary point of its interval of convergence. Endpoint tests combine the alternating series test with radius-of-convergence methods and, at a deeper level, Abel-type theorems. Conditional convergence leads to rearrangement theory. The [Riemann rearrangement theorem](/theorems/3831) shows that the order of a conditionally convergent real series is part of its mathematical data. This theme later reappears in functional analysis, where unconditional convergence in infinite-dimensional spaces is stronger than ordinary convergence. Alternating estimates also appear in [numerical analysis](/page/Numerical%20Analysis). The next-term error bound is a simple certified error estimate, useful when evaluating slowly convergent expansions such as the alternating harmonic series. More advanced acceleration methods improve the speed, but the basic alternating bound remains a model example of rigorous approximation. ## References Androma, [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes). Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Tom M. Apostol, *Calculus, Volume I* (1967). Walter Rudin, *Principles of Mathematical Analysis* (1976). G. H. Hardy, *Divergent Series* (1949).