Tutorials › Calculus II › Why Alternating Remainders Have Predictable Signs

Home  /  Tutorials  /  Remainders and Error Estimates

Remainders and Error Estimates · Tutorial 858 of 998

Why Alternating Remainders Have Predictable Signs

Learn why alternating partial sums bracket the series sum and how the parity of the truncation index determines the error’s sign.

Advanced 9 min read

What You'll Learn

  • State the hypotheses for the alternating-series bracketing result.
  • Prove convergence by comparing the even and odd partial sums.
  • Determine which consecutive partial sum is the upper or lower bound.
  • Infer the sign of a truncation error from the first omitted term.
  • Apply the result to geometric, logarithmic, and sine series.
  • Identify why the last included term does not determine the error’s sign.

Why the Error Has a Direction

An approximation to a number may be too large or too small. For a general truncated series, the direction of the error can be difficult to identify. Alternating series have a useful structure: when the term magnitudes decrease toward zero, each new partial sum corrects the preceding one from the opposite side.

This gives more information than a bound on the error’s size. It tells us which side of the sum each partial sum lies on. In particular, the sum is between any two consecutive partial sums, and the first omitted term determines the direction of the truncation error.

Definition: An alternating series with decreasing magnitudes is a series of the form \[ \sum_{n=1}^{\infty}(-1)^{n-1}b_n =b_1-b_2+b_3-b_4+\cdots, \] where \(b_n\geq 0\), \(b_{n+1}\leq b_n\) for every \(n\), and \(\lim_{n\to\infty}b_n=0\). Its \(N\)th partial sum is \(S_N=\sum_{n=1}^{N}(-1)^{n-1}b_n\), and, when the series converges to \(S\), its truncation error is \(R_N=S-S_N\).

The condition that the magnitudes decrease means they never increase from one term to the next; they need not decrease strictly at every step. The limit condition rules out terms that remain a fixed positive size. Both conditions matter for the conclusion.

The displayed form begins with a positive term. If an alternating series begins with a negative term, multiplying every term by \(-1\) reduces it to this form; the conclusions about upper and lower bounds then reverse when translated back to the original series.

How the Partial Sums Move

Pairing terms reveals the pattern. From \(S_{2m}\) to \(S_{2m+2}\), the series adds a positive term and then a negative term. Their net contribution is nonnegative because the positive term is at least as large as the negative term in magnitude. Thus the even-indexed partial sums move upward. The odd-indexed partial sums move downward: the next two terms have a nonpositive combined contribution.

$$ S_{2m+2}-S_{2m}=b_{2m+1}-b_{2m+2}\geq 0, \qquad S_{2m+3}-S_{2m+1}=-b_{2m+2}+b_{2m+3}\leq 0. $$

Moreover, each even partial sum is no larger than the odd partial sum immediately after it, since their difference is the next nonnegative term. This creates nested bounds: the even partial sums rise, the odd partial sums fall, and the gap between a consecutive pair is a term whose magnitude tends to zero.

Proof of the Bracketing Result

Theorem: Suppose \(b_n\geq 0\), \(b_{n+1}\leq b_n\), and \(b_n\to 0\). Then \(\sum_{n=1}^{\infty}(-1)^{n-1}b_n\) converges to a sum \(S\). For every \(N\geq 1\), \(S\) lies between \(S_N\) and \(S_{N+1}\). In particular, \(R_N=S-S_N\) has the sign of the first omitted term, or is zero. Also, \(|R_N|\leq b_{N+1}\).

Here “lies between” includes the possibility of equality. The error’s sign can therefore be stated precisely: \(R_N\geq 0\) if \(N\) is even, and \(R_N\leq 0\) if \(N\) is odd. These are the signs of the next term, \((-1)^N b_{N+1}\).

1
Show the even partial sums increase.
For \(m\geq 0\), the change from \(S_{2m}\) to \(S_{2m+2}\) is \(b_{2m+1}-b_{2m+2}\), which is nonnegative by the decreasing-magnitude condition. (Set \(S_0=0\) for this convergence argument.)
2
Show the odd partial sums decrease.
The change from \(S_{2m+1}\) to \(S_{2m+3}\) is \(-b_{2m+2}+b_{2m+3}\), which is nonpositive. Also, \(S_{2m}\leq S_{2m+1}\), since \(S_{2m+1}-S_{2m}=b_{2m+1}\geq 0\). Thus the even partial sums are bounded above by \(S_1\), and the odd partial sums are bounded below by \(S_0=0\) only when that bound is applicable; more generally, they are bounded below by \(S_2\), since the odd subsequence decreases and \(S_3\geq S_2\), with the same ordering continuing for later pairs.
3
Take the two subsequence limits.
By monotone convergence for real sequences, the even and odd partial sums each have a finite limit. Their difference at matching indices is \(S_{2m+1}-S_{2m}=b_{2m+1}\to 0\), so their limits are equal. The full sequence of partial sums therefore converges to their common value \(S\).
4
Place the sum between consecutive partial sums.
The even partial sums increase toward \(S\), while the odd partial sums decrease toward \(S\). If \(N\) is even, \(S_N\leq S\leq S_{N+1}\). If \(N\) is odd, \(S_{N+1}\leq S\leq S_N\). In either case, \(S\) lies between the consecutive partial sums.

The ordering in the convergence argument can be seen directly for the odd subsequence: every odd partial sum is at least the following even partial sum, and every even partial sum is at least \(S_0=0\) only if the terms make that true. To avoid relying on that unnecessary lower bound, note instead that the odd partial sums are bounded below by \(S_2\): \(S_1\geq S_2\), and the odd subsequence decreases to values no lower than the even partial sums, which increase from \(S_0=0\). More simply, boundedness follows from \(S_{2m+1}=S_{2m}+b_{2m+1}\), with the even sums bounded and \(0\leq b_{2m+1}\leq b_1\). Hence both monotone subsequences have finite limits.

Finally, the distance between the consecutive partial sums is exactly \(b_{N+1}\). Since \(S\) lies within that interval, its distance from \(S_N\) cannot exceed the interval’s length. This proves \(|R_N|\leq b_{N+1}\) as a consequence of the bracketing, rather than as a separate estimate.

Worked Examples

Worked Example: A Geometric Alternating Series

Consider \(1-\frac12+\frac14-\frac18+\cdots\). Its magnitudes are \(b_n=(\frac12)^{n-1}\), which decrease to zero, so the theorem applies. The third partial sum is

$$S_3=1-\frac12+\frac14=\frac{4}{4}-\frac{2}{4}+\frac{1}{4}=\frac34.$$

The fourth is

$$S_4=\frac34-\frac18=\frac68-\frac18=\frac58.$$

The geometric-series formula gives

$$S_N=\frac{1-(-\frac12)^N}{1+\frac12}=\frac23\left(1-(-\frac12)^N\right),\qquad \lim_{N\to\infty}S_N=\frac23.$$

Thus \(\frac58\leq\frac23\leq\frac34\): the sum lies between \(S_4\) and \(S_3\). For \(N=3\), the error is

$$R_3=\frac23-\frac34=\frac{8}{12}-\frac{9}{12}=-\frac{1}{12}.$$

The first omitted term is \(-\frac18\), so the error is negative, as predicted. Its magnitude also satisfies \(\frac{1}{12}\leq\frac18\). The calculation checks both the location and the direction of the error.

Worked Example: Bracketing a Logarithm

For \(0<x\leq 1\), the logarithm series is \(x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots\). At \(x=\frac12\), its term magnitudes are \(b_n=(\frac12)^n/n\). They decrease because

$$\frac{b_{n+1}}{b_n}=\frac{(\frac12)^{n+1}}{n+1}\cdot\frac{n}{(\frac12)^n}=\frac{n}{2(n+1)}<1.$$

They also tend to zero. The series therefore brackets its sum between consecutive partial sums. Through two terms,

$$S_2=\frac12-\frac{(\frac12)^2}{2}=\frac12-\frac18=\frac38.$$

The next term is \((\frac12)^3/3=1/24\), so

$$S_3=\frac38+\frac{1}{24}=\frac{9}{24}+\frac{1}{24}=\frac{5}{12}.$$

Since this series sums to \(\ln(1+\frac12)=\ln(\frac32)\), the result is

$$\frac38\leq\ln\left(\frac32\right)\leq\frac{5}{12}.$$

After \(S_2\), the omitted terms begin positively, so \(S-S_2\geq0\). After \(S_3\), they begin negatively, so \(S-S_3\leq0\). The endpoints are exactly the two computed partial sums, confirming the one-sided locations.

Worked Example: The Direction of the Sine Approximation Error

The Maclaurin series for \(\sin(1)\) is \(1-\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!}+\cdots\). Its magnitudes decrease to zero: each next magnitude divides the preceding one by the product of the next two positive integers. The first two partial sums are

$$S_2=1-\frac{1}{6}=\frac56,$$

and

$$S_3=\frac56+\frac{1}{120}=\frac{100}{120}+\frac{1}{120}=\frac{101}{120}.$$

The theorem places the sum between them:

$$\frac56\leq\sin(1)\leq\frac{101}{120}.$$

For a further check on the upper endpoint, the next partial sum is

$$S_4=\frac{101}{120}-\frac{1}{5040}=\frac{4242}{5040}-\frac{1}{5040}=\frac{4241}{5040}.$$

Thus \(S_4\leq\sin(1)\leq S_3\). In particular, \(R_3=\sin(1)-S_3\leq0\), and the first omitted term is negative. The partial sums and the sign of the error agree with the theorem.

A Common Sign Mistake

The error’s sign is determined by the first omitted term, not by the last term that was included. For example, \(S_3\) in the sine calculation ends with the positive term \(1/120\). A tempting but incorrect argument is, “The last included term is positive, so \(S_3\) is below the sum and the error is positive.” That confuses the included term with the tail.

The correction is to inspect the next term. Here it is \(-1/5040\), so the remainder after \(S_3\) is nonpositive. Indeed, the bracketing result gives \(S_4\leq\sin(1)\leq S_3\), and therefore \(\sin(1)-S_3\leq0\). The next term, not the last included term, points toward the tail.

Why the Bracket Is Useful

A one-sided conclusion matters when an approximation will be used in an inequality, a numerical estimate, or a later argument. Knowing only that the error is small may not settle whether an approximate value is above or below the exact value. Bracketing supplies that direction without requiring the exact value of the sum.

The parity rule is a practical way to remember the result. With a series that starts positive, an even-numbered partial sum is a lower bound, and an odd-numbered partial sum is an upper bound. Equivalently, if \(N\) terms are retained, the error points in the direction of term \(N+1\). The magnitude bound follows because the exact sum lies in an interval of length \(b_{N+1}\).

These conclusions require nonnegative, nonincreasing magnitudes tending to zero. Alternation alone is not enough: if the magnitudes do not decrease, the paired differences need not have the signs used in the proof. The assumptions give the partial sums their orderly motion and make the predictable error direction possible.

Check Your Understanding

Use the hypotheses and the order of the partial sums to answer each question.

  1. For a series beginning with a positive term, which of \(S_6\) and \(S_7\) is the upper bound for the sum?
  2. If \(N\) is odd, what is the sign of \(R_N=S-S_N\), and what determines it?
  3. Why do the even and odd partial sums approach the same limit?
  4. What additional conclusion about \(|R_N|\) follows from the distance between \(S_N\) and \(S_{N+1}\)?
  5. In the sine example, why is it incorrect to use the sign of \(1/120\) to determine the sign of \(R_3\)?

Previous

Next

Back to Tutorials