Why alternating terms help control error
An infinite series can be difficult to evaluate exactly. For some alternating series, however, you can establish both that the series converges and that a partial sum is close to its limit without finding the limit itself. The key is to check how the sizes of the terms behave: if they decrease toward zero, successive partial sums alternate around the sum, getting closer as they go.
That conclusion depends on specific conditions. Alternating signs alone do not guarantee convergence, and the fact that the terms approach zero alone is not enough to apply the alternating series test. The sequence of positive term magnitudes must also be nonincreasing.
The alternating series test
Write an alternating series in the form \(\sum_{n=1}^{\infty}(-1)^{n+1}b_n\), where each \(b_n\) is positive. Thus, the terms are \(b_1,-b_2,b_3,-b_4,\ldots\). The numbers \(b_n\) are called the term magnitudes.
In many AP problems, the magnitudes decrease from the first term onward. To check that condition, compare consecutive terms and show \(b_{n+1}\leq b_n\), or use another valid argument such as the fact that a function representing \(b_n\) is decreasing on the relevant range. To check the limit, evaluate \(\lim_{n\to\infty}b_n\) or bound \(b_n\) by a positive quantity that tends to zero.
Both checks matter. The test does not assert convergence merely because a series alternates. Nor does it say that every alternating series whose terms approach zero converges: the required decrease must also be established.
Why the test works
First assume the magnitudes are nonincreasing from the first term, so \(b_{n+1}\leq b_n\) for every \(n\), and \(b_n\to0\). Let the partial sums be \(S_1=b_1\), \(S_2=b_1-b_2\), and so on. Pairing consecutive terms shows how the odd- and even-numbered partial sums behave.
So the odd partial sums are nonincreasing. Similarly,
so the even partial sums are nondecreasing. Each even partial sum is no greater than each odd partial sum, since \(S_{2m+1}-S_{2m}=b_{2m+1}>0\) and the monotone behavior carries the comparison across indices. The odd partial sums are therefore bounded below by \(S_2\) and nonincreasing, and the even partial sums are bounded above by \(S_1\) and nondecreasing, so both approach limits. The gap between them is \(b_{2m+1}\), which tends to zero, so the limits are equal and the full sequence of partial sums converges.
If the magnitudes become nonincreasing only from some index \(K\) onward, apply this argument to the tail of the series beginning with term \(K\). Removing finitely many initial terms changes the sum by a fixed number but not whether the series converges.
The same reasoning gives the error bound. Apply the alternating-sum behavior to the terms left after \(S_N\). If the first omitted term is positive, the tail begins by moving upward, and its sum lies between zero and that first omitted term. If it is negative, the tail lies between that term and zero. Consequently:
The index is important: \(S_N\) includes terms 1 through \(N\), so the first omitted magnitude is \(b_{N+1}\), not \(b_N\). The bound is an inequality, not an assertion that the error equals the omitted term.
A practical method for an exam response
Express the terms as alternating signs times positive numbers \(b_n\).
Show \(b_{n+1}\leq b_n\) on the relevant range. If needed, state where the inequality begins to hold.
Show that \(\lim_{n\to\infty}b_n=0\). Only after both checks can you invoke the test.
For \(N\) included terms, use \(b_{N+1}\), the magnitude of the first omitted term.
The sum lies between \(S_N\) and \(S_N\) plus the first omitted signed term.
Worked examples
Worked Example: Bounding the error in an alternating harmonic partial sum
Consider \(\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}\). Verify the test conditions and use the first four terms to bound the sum.
Here \(b_n=\frac{1}{n}>0\). For every \(n\geq 1\), \(n+1>n\), so \(\frac{1}{n+1}<\frac{1}{n}\). Also, \(\lim_{n\to\infty}\frac{1}{n}=0\). The Alternating Series Test therefore guarantees convergence.
The first four terms give
The first omitted term is positive and has magnitude \(b_5=\frac15\). Thus the remainder is nonnegative and at most \(\frac15\):
In particular, \(|S-S_4|\leq \frac15\). This uses the fifth term because the partial sum includes four terms. The interval does not claim that the sum equals either endpoint.
Worked Example: Checking decrease before bounding a remainder
Consider \(\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n}{n^2+1}\). Verify the conditions, then bound the error after three terms.
Set \(b_n=\frac{n}{n^2+1}\), which is positive. To check decrease, compare consecutive terms. Since the denominators are positive, cross-multiplication preserves the inequality:
The right-hand side minus the left-hand side is
for \(n\geq 1\). Therefore \(b_{n+1}<b_n\). Also, \(0<b_n=\frac{n}{n^2+1}<\frac1n\), and \(\frac1n\to0\), so \(b_n\to0\). The alternating series test applies.
The first three terms sum to
The next term is negative, with magnitude \(b_4=\frac{4}{17}\). Thus the sum lies between \(S_3-\frac{4}{17}\) and \(S_3\):
The remainder is nonpositive and has magnitude at most \(\frac4{17}\), so \(|S-S_3|\leq\frac4{17}\). The interval’s lower endpoint is positive because \(14/85>0\), consistent with the bound and the calculated partial sum.
Worked Example: An exact partial sum and a one-sided estimate
Define \(b_1=1\) and \(b_n=\frac{1}{n^2+1}\) for \(n\geq2\), and consider \(\sum_{n=1}^{\infty}(-1)^{n+1}b_n\). Verify the conditions and bound the sum using the first four terms.
The first magnitude is \(b_1=1\), and \(b_2=\frac15<1\). For \(n\geq2\), \(n+1>n\) and both are positive, so \((n+1)^2+1>n^2+1\). Taking reciprocals gives \(b_{n+1}<b_n\). Also, for \(n\geq2\), \(0<b_n=\frac{1}{n^2+1}<\frac{1}{n^2}\), so \(b_n\to0\). The finite initial term does not affect the limit or convergence. The test applies.
Writing out the first four terms gives
The next term is positive, with magnitude \(b_5=\frac{1}{5^2+1}=\frac1{26}\). The remainder is therefore between zero and \(\frac1{26}\):
Thus \(|S-S_4|\leq\frac1{26}\). As a check on the fraction conversion, \(\frac{143}{170}+\frac1{26}\) uses common denominator \(2210\), since \(170\cdot13=2210\) and \(26\cdot85=2210\). The resulting upper endpoint is greater than \(143/170\), as required by the positive next term.
Common mistakes and how to correct them
A frequent off-by-one error is to use the last included term as the error bound. For the alternating harmonic series, after four terms the last included magnitude is \(b_4=\frac14\), but the first omitted magnitude is \(b_5=\frac15\). The correct statement is \(|S-S_4|\leq\frac15\), not \(|S-S_4|\leq\frac14\). Although the latter happens to be a weaker upper bound here, it does not state the alternating series remainder bound correctly; identify the first omitted term before substituting.
Another mistake is to check only that \(b_n\to0\). For instance, in the alternating series with \(b_n=1\), the magnitudes do not tend to zero, so the test cannot establish convergence. More generally, a zero limit without the required decrease does not satisfy the hypotheses of this test. A complete justification states both checks: nonincreasing magnitudes and limit zero.
Finally, the error estimate is not the value of the remainder. If the first omitted term is positive, the sum is at least \(S_N\) and at most \(S_N+b_{N+1}\); it does not follow that \(S=S_N+b_{N+1}\). The omitted term supplies a bound, not an exact correction.
Why the remainder bound is useful
The bound lets you decide how many terms are enough before carrying out a long calculation. If a problem asks for an approximation with error no greater than a positive tolerance \(\varepsilon\), choose \(N\) so that \(b_{N+1}\leq\varepsilon\), after verifying the alternating series test conditions. This guarantees \(|S-S_N|\leq\varepsilon\).
For example, if the magnitudes are \(b_n=\frac1n\) and the requested error is at most \(0.01\), require \(\frac1{N+1}\leq0.01\). This is equivalent to \(N+1\geq100\), so \(N\geq99\) terms suffice. The choice \(N=99\) gives a first omitted magnitude of \(\frac1{100}=0.01\). In an exam response, make the index clear and connect the selected \(N\) to the first term left out.
Check Your Understanding
For each question, identify the relevant term magnitudes and explain how the alternating series test or remainder bound applies.
- For \(\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n+2}\), state how you would verify that the magnitudes decrease and approach zero.
- If a partial sum contains six terms, which term magnitude supplies the alternating-series error bound?
- For the series in Question 1, determine the sign of the remainder after four terms.
- What two conditions on \(b_n\) must be checked before invoking the alternating series test?
- If \(b_{N+1}=0.004\), what error guarantee does the remainder bound provide for \(S_N\)?
