What the Remainder Bound Tells Us
An alternating series switches sign from one term to the next. When the magnitudes of its terms decrease toward zero, successive partial sums move back and forth around the sum. The distance from a partial sum to the actual sum is bounded by the magnitude of the next term—the first omitted term.
This gives more than a numerical estimate. It tells us which side of the sum a partial sum lies on, and it provides an interval guaranteed to contain the sum. To use the result correctly, however, we must check all of its conditions.
Definitions and Conditions
Consider an alternating series written with a positive first term:
where each \(a_n\) is nonnegative. The number \(a_n\) is the magnitude of the \(n\)th term; the sign is specified separately by \((-1)^{n-1}\). Define the \(N\)th partial sum and the remainder after \(N\) terms by
where \(S\) is the sum of the infinite series. Thus, \(R_N\) consists of all the terms starting with the \((N+1)\)st term. In particular, its first term has magnitude \(a_{N+1}\), not \(a_N\).
“Nonincreasing” allows two consecutive magnitudes to be equal. The limit-zero condition is essential: the magnitudes must eventually become arbitrarily small. Under these conditions the alternating series converges, and its remainder obeys the bound proved below.
Why the Remainder Is Bounded
First consider a tail that begins with a positive term, with magnitudes \(b_1,b_2,\ldots\) that are nonincreasing and tend to zero. Its finite partial sums alternate between sums with an even and an odd number of terms. Grouping adjacent terms shows why they stay within a short interval.
Every difference in these expressions is nonnegative because the magnitudes are nonincreasing. Therefore \(q_{2m}\geq 0\) and \(q_{2m+1}\leq b_1\). Also, \(q_{2m}\leq q_{2m+2}\): adding two terms adds \(b_{2m+1}-b_{2m+2}\geq 0\). The even partial sums increase and are bounded above by \(b_1\), so they converge. The odd partial sum differs from the preceding even one by \(b_{2m+1}\), which tends to zero. The odd partial sums therefore converge to the same limit. Call that limit \(q\). Since the even partial sums are nonnegative and the odd partial sums are at most \(b_1\), their common limit satisfies \(0\leq q\leq b_1\).
For a remainder after \(N\) terms, factor out the sign of its first omitted term. What remains is a tail beginning with a positive term, to which the argument above applies. This yields the theorem.
The sign formula follows because the first omitted term is \((-1)^N a_{N+1}\). For \(N=0\), the partial sum is \(S_0=0\), so the statement also covers the whole series. The theorem’s sign claim is sometimes phrased as “the sum lies on the side of the partial sum indicated by the next term.”
Using the Bound to Make an Estimate
After calculating \(S_N\), identify the next term before stating an error bound. If it is positive, then \(S\) is at or above \(S_N\); if it is negative, then \(S\) is at or below \(S_N\). In either case, the distance between \(S_N\) and \(S\) is at most the next term’s magnitude.
Worked Example: Bounding the Alternating Harmonic Series
Use four terms of \(1-\frac12+\frac13-\frac14+\cdots\) to bound its sum, which is \(\ln 2\).
The magnitudes \(a_n=\frac1n\) are positive, decrease as \(n\) increases, and tend to zero, so the theorem applies. The four-term partial sum is
The first omitted term is \(+\frac15\). The remainder is therefore nonnegative and no greater than \(\frac15\):
Thus, four terms give a guaranteed interval from \(\frac7{12}\) to \(\frac{47}{60}\). The check is that the next term has positive sign and magnitude \(\frac15\), exactly matching the direction and maximum width asserted by the theorem.
Worked Example: Bounding a Sum When the Next Term Is Negative
For \(\sum_{n=1}^{\infty}(-1)^{n-1}\frac1{n^2}\), find an interval containing the sum after three terms.
Here \(a_n=\frac1{n^2}\) is positive, decreasing, and tends to zero. Compute the partial sum:
The first omitted term is the fourth term, \(-\frac1{16}\). Thus the remainder is nonpositive, and its magnitude is at most \(\frac1{16}\):
The sum therefore lies between \(\frac{115}{144}\) and \(\frac{31}{36}\). The check: the next term is negative, so the sum must be no greater than \(S_3\), and the lower endpoint subtracts exactly the allowed maximum error.
Worked Example: Bounding a Taylor Approximation
Use the first two nonzero terms of the sine series to bound \(\sin(\frac12)\).
The series is \(\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\). At \(x=\frac12\), its first two terms give
For \(k\geq 0\), the magnitude of the term with \(x^{2k+1}\) is \(a_k=\frac{(1/2)^{2k+1}}{(2k+1)!}\). The ratio of the next magnitude to this one is
So the magnitudes decrease, and they tend to zero. The first omitted term is positive and equals \(\frac{(1/2)^5}{5!}=\frac1{3840}\). Hence
The check: the next term is positive, placing the true value above the two-term approximation; the possible increase is no more than that next term.
Common Mistake: Checking Only That Terms Decrease
A decreasing sequence of magnitudes is not enough by itself. For example, let \(a_n=1+\frac1n\). These magnitudes decrease, but their limit is \(1\), not zero. The alternating series \(a_1-a_2+a_3-a_4+\cdots\) does not converge: its terms do not approach zero. Indeed, successive partial sums differ by the next term, whose magnitude approaches \(1\), so those partial sums cannot approach one common limit.
A related indexing mistake is to use the last included term as the error bound. In the first example, after four terms the last included magnitude is \(\frac14\), but the theorem’s bound is \(\frac15\), the magnitude of the first omitted term. The correction is to write \(S_N\), identify the term with index \(N+1\), and use \(a_{N+1}\).
Why This Result Is Useful
The remainder theorem turns a finite calculation into a rigorous statement about an infinite sum. It is especially useful when the terms alternate and become smaller: a partial sum is not merely a plausible approximation but an endpoint of a guaranteed interval. The sign information can matter as much as the size bound, because it tells whether the approximation is an underestimate or an overestimate.
Before applying the result, verify that the series really alternates, that its magnitudes are nonincreasing from the point where the estimate begins, and that those magnitudes tend to zero. Then compute the finite partial sum and the first omitted term. The theorem gives a bound for that particular truncation; it does not say that the error equals the next term.
Check Your Understanding
Answer each question by identifying the relevant term magnitudes, partial sum, and remainder sign where appropriate.
- State the three conditions on \(a_n\) used in the alternating-series remainder theorem.
- After \(N\) terms, which magnitude gives the remainder bound: \(a_N\) or \(a_{N+1}\)? Explain why.
- If the first omitted term is negative, what does the theorem say about the relation between \(S\) and \(S_N\)?
- Why does the alternating series with magnitudes \(1+\frac1n\) fail to meet the theorem’s conditions?
- For \(1-\frac13+\frac15-\frac17+\cdots\), find the three-term partial sum and state the sign and maximum magnitude of its remainder.
