Absolute, conditional, and alternating convergence

Two series can have exactly the same terms up to sign, and yet behave completely differently once you strip the signs away. The series $\sum (-1)^{n-1}/n^2$ survives having its signs removed — $\sum 1/n^2$ still converges. The series $\sum (-1)^{n-1}/n$ does not: remove the signs and you are left with the divergent harmonic series. Both of the original, signed series converge, but for different reasons, and calculus gives this difference a name: absolute versus conditional convergence. This node builds the vocabulary to tell them apart, and hands you the one dedicated tool for alternating series — the Leibniz test — along with the free error bound it carries. Everything here sits directly after the ratio, root and comparison tests in the sequences-and-series unit, and reuses them rather than replacing them.

The core insight

A series is absolutely convergent if the series of its absolute values converges (which forces the original to converge too), conditionally convergent if it converges only because of sign cancellation, and the Leibniz test settles convergence — with a free error bound — for any alternating series whose terms decrease to zero.

The core idea

Two kinds of convergence. Take any series $\sum a_n$ , whether or not its terms alternate in sign, and form the companion series of absolute values $\sum |a_n|$ . Three outcomes are possible: $\sum|a_n|$ converges (then $\sum a_n$ is called absolutely convergent); $\sum|a_n|$ diverges but $\sum a_n$ still converges (then $\sum a_n$ is conditionally convergent); or $\sum a_n$ itself fails to converge. The dividing line between the first two cases is a genuine theorem, not just a definition:

\sum |a_n| \text{ converges} \implies \sum a_n \text{ converges}.

where: $a_n$ is the general term (any sign), and $|a_n|$ its absolute value. The implication only runs one way — a convergent $\sum a_n$ says nothing about $\sum|a_n|$ , and that gap is exactly what conditional convergence occupies. Absolute convergence is the stronger, more robust property: it means the series would still converge even if every minus sign flipped to a plus, so there is no delicate cancellation holding it together.

Alternating series and the Leibniz test. A series whose terms strictly alternate sign,

\sum_{n=1}^{\infty} (-1)^{n-1} b_n = b_1 - b_2 + b_3 - b_4 + \cdots, \qquad b_n > 0,

is an alternating series. The Leibniz test gives a clean sufficient condition for convergence: if $\{b_n\}$ is (eventually) monotonically decreasing and $b_n \to 0$ , then $\sum (-1)^{n-1} b_n$ converges. Both hypotheses are needed — decreasing terms that do not shrink to zero, or terms that shrink to zero but wobble non-monotonically, can each break the guarantee. Geometrically, the partial sums $S_N$ zig-zag, first overshooting the limit $S$ , then undershooting it, by ever-smaller amounts, squeezing in from both sides.

Figure

Partial sums of an alternating series zig-zag above and below the limit and squeeze in as the number of terms grows

Classification workflow. Given any series, first test $\sum|a_n|$ using the ratio, root or comparison test. If it converges, the series is absolutely convergent — stop, that already gives convergence. If $\sum|a_n|$ diverges and the series alternates, apply Leibniz to the original series; if Leibniz succeeds, the series is conditionally convergent, and if nothing shows convergence, it diverges. A subtlety worth flagging: the ratio and root tests, applied in their usual form to $\sum a_n$ , actually operate on $|a_{n+1}/a_n|$ and $|a_n|^{1/n}$ , so an $L<1$ verdict from either test is secretly always a verdict about $\sum|a_n|$ — meaning ratio/root can only ever certify absolute convergence, never conditional convergence.

Derivation AThe Leibniz error bound for a truncated alternating series

The goal is to show that when an alternating series $\sum (-1)^{n-1} b_n$ satisfies the Leibniz hypotheses — $b_n$ eventually decreasing and $b_n \to 0$ — the error made by stopping after $N$ terms is never larger than the size of the very next, first-omitted term, $|S - S_N| \le b_{N+1}$ , a fact that turns the qualitative "it converges" into a quantitative, usable accuracy guarantee for estimating the true sum $S$ from a finite computation.

Step 1. Set up the partial sums. Let $S_N = \sum_{n=1}^{N} (-1)^{n-1} b_n$ and let $S$ be the (already established, by Leibniz) limit. Consider the even-indexed partial sums first: $S_{2k} = S_{2k-2} + b_{2k-1} - b_{2k}$ . Since $b_{2k-1} \ge b_{2k}$ (the sequence is decreasing), each step from $S_{2k-2}$ to $S_{2k}$ adds a non-negative amount, so the even partial sums $S_2, S_4, S_6, \dots$ form an increasing sequence.

Step 2. Track the odd-indexed partial sums the same way. $S_{2k+1} = S_{2k-1} - b_{2k} + b_{2k+1}$ , and since $b_{2k} \ge b_{2k+1}$ , each step subtracts a non-negative amount, so the odd partial sums $S_1, S_3, S_5, \dots$ form a decreasing sequence. Because $S_{2k+1} - S_{2k} = b_{2k+1} \ge 0$ , every odd partial sum sits above every even one; the odd sums decrease and the even sums increase, both trapped between $S_1$ and $S_2$ , and both squeeze toward the same limit $S$ — this is precisely the zig-zag picture above.

Step 3. Bound the error at an even cutoff $N = 2k$ . Since the even sums increase toward $S$ and $S$ is sandwiched below the next odd sum $S_{2k+1} = S_{2k} + b_{2k+1}$ , we get $S_{2k} \le S \le S_{2k+1} = S_{2k} + b_{2k+1}$ . Subtracting $S_{2k}$ throughout gives $0 \le S - S_{2k} \le b_{2k+1}$ , i.e. $|S - S_N| \le b_{N+1}$ for even $N$ .

Step 4. Bound the error at an odd cutoff $N = 2k-1$ by the mirror argument: the odd sums decrease toward $S$ , and $S$ is sandwiched above the next even sum $S_{2k} = S_{2k-1} - b_{2k}$ , giving $S_{2k-1} - b_{2k} = S_{2k} \le S \le S_{2k-1}$ , so $0 \le S_{2k-1} - S \le b_{2k}$ , i.e. $|S - S_N| \le b_{N+1}$ for odd $N$ too. Both parities give the same clean bound, so it holds for every $N$ : truncating after $N$ terms leaves an error no bigger than the magnitude of the first term you dropped. Verified: with $b_n$ decreasing and $b_n \to 0$ , both the even and odd partial-sum subsequences are monotone and bounded, hence convergent, and both limits coincide since $S_{2k+1}-S_{2k}=b_{2k+1}\to 0$ ; the sandwiching in Steps 3-4 is exact, giving $|S-S_N|\le b_{N+1}$ for every $N$ .

Worked example

Classify $\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n}$ — the alternating harmonic series — and use the first 4 terms to estimate its sum, quoting a guaranteed error bound.

Step 1. Test absolute convergence first. $\sum |a_n| = \sum 1/n$ is the harmonic series, which diverges (it is a $p$ -series with $p=1$ ). So the series is not absolutely convergent — Leibniz is now the only route to convergence.

Step 2. Apply Leibniz to $b_n = 1/n$ . Decreasing: $b_{n+1} = 1/(n+1) < 1/n = b_n$ for every $n \ge 1$ , so yes. Limit: $\lim_{n\to\infty} 1/n = 0$ , so yes. Both hypotheses hold, so $\sum (-1)^{n-1}/n$ converges.

Step 3. Classify. It converges by Leibniz, but $\sum|a_n|$ diverges, so the series is conditionally convergent — the standard textbook example of the phenomenon. (Its exact sum is $\ln 2 \approx 0.6931$ , obtainable from the Maclaurin series of $\ln(1+x)$ at $x=1$ , though that derivation belongs to a different node.)

Step 4. Compute $S_4$ and bound its error using Step 3 of the derivation above. $S_4 = 1 - \tfrac12 + \tfrac13 - \tfrac14 = \tfrac{12-6+4-3}{12} = \tfrac{7}{12} \approx 0.5833$ . The error bound is $|S - S_4| \le b_5 = 1/5 = 0.2$ . Checking: $|\ln 2 - 0.5833| \approx 0.110$ , comfortably inside the guaranteed bound of $0.2$ — the estimate $S_4 \approx 0.5833$ is trustworthy to within two decimal places of confidence, exactly as the bound promised.

Competitive stretch

How many terms $N$ of the alternating harmonic series are enough to guarantee an estimate of $\ln 2$ accurate to within $0.001$ ? Using $|S - S_N| \le b_{N+1} = 1/(N+1)$ , require $1/(N+1) \le 0.001$ , i.e. $N+1 \ge 1000$ , so $N \ge 999$ . Taking $N = 999$ terms guarantees accuracy within $0.001$ — a direct, term-counting consequence of the error bound, with no need to sum anything.

Common traps

Treating $\sum a_n$ convergent as automatically meaning $\sum|a_n|$ convergent. False — that gap between the two is precisely what conditional convergence is. Only the reverse implication (absolute $\Rightarrow$ convergent) is a theorem.
Applying Leibniz while skipping the monotonicity check. $b_n \to 0$ alone does not suffice; $\{b_n\}$ must also be decreasing. A sequence that trends to zero while oscillating in size can fail Leibniz even though it alternates.
Using the error bound outside its licence. $|S - S_N| \le b_{N+1}$ only applies once the Leibniz hypotheses have actually been verified for that series — it cannot be borrowed for a series that has not been shown to satisfy them.
Skipping straight to Leibniz without testing $\sum|a_n|$ first. That order is what lets you distinguish absolute from conditional convergence, which is usually the actual question being asked.

Ask the tutor: Want to walk through why the ratio test can only ever detect absolute convergence, or try classifying a mixed-sign series of your own choosing?

Why does absolute convergence imply convergence but not the other way around?

Absolute convergence is the stronger statement: if $\sum|a_n|$ converges, the positive and negative parts of $\sum a_n$ are each individually well-behaved, so no amount of sign-flipping can break convergence — that is the content of the theorem. Conditional convergence is different: the series converges only because positive and negative terms cancel in a delicate, order-dependent way. Strip the signs (equivalently, take absolute values) and that cancellation is gone, so $\sum|a_n|$ is free to diverge. The one-directional arrow reflects that absolute convergence is a robustness property that conditional convergence simply does not have.

Do I need the series to actually alternate sign to use the Leibniz test?

Yes, in its standard statement the Leibniz test is specifically about series of the form $\sum (-1)^{n-1} b_n$ with $b_n > 0$ — genuinely alternating series. If a series has mixed signs but not in a strict alternating pattern, Leibniz simply does not apply, and you fall back on testing $\sum|a_n|$ directly (ratio, root, comparison) to check for absolute convergence; there is no general shortcut for irregular sign patterns at this level.

If Leibniz shows a series converges, does that also tell me the value of the sum?

No — Leibniz only certifies that a limit exists; it does not compute it. Finding the exact sum, when possible at all, requires a separate technique, such as recognizing the series as a known Maclaurin expansion evaluated at a specific point (the alternating harmonic series equals $\ln 2$ this way). What Leibniz does hand you for free is the error bound $|S-S_N|\le b_{N+1}$ , which lets you estimate the unknown sum numerically to any desired accuracy just by adding enough terms, even without knowing a closed form.

Absolute, conditional, and alternating convergence

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads