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Alternating and Absolute Convergence · Tutorial 658 of 998

The Sign of the Alternating-Series Error

Use the parity of the last included term to determine the direction of an alternating-series error and bracket the sum between consecutive partial sums.

Intermediate 9 min read

What You'll Learn

  • Define the remainder as the series sum minus a chosen partial sum.
  • Determine whether an alternating partial sum overestimates or underestimates the sum.
  • Prove that the remainder has the sign of the first omitted term.
  • Locate the series sum between consecutive partial sums.
  • Avoid confusing the remainder with the approximation error defined in the opposite order.

Introduction

An alternating series changes sign from one term to the next. When its terms decrease in magnitude toward zero, successive partial sums move from one side of the series sum to the other. This means that the parity of the last term included tells us whether a partial sum is an overestimate or an underestimate.

Knowing the direction of the error is useful even when we do not know the exact sum. It gives a one-sided statement about the approximation and, by comparing consecutive partial sums, a precise interval containing the sum.

Partial Sums and the Remainder

Consider an alternating series of the form

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

where \(b_n\geq 0\), the sequence \(b_n\) is nonincreasing, and \(\lim_{n\to\infty}b_n=0\). Under these conditions, the Alternating-Series Test guarantees that the series converges. Write its sum as \(S\), and define the \(N\)th partial sum by

$$ s_N=\sum_{n=1}^{N}(-1)^{n-1}b_n. $$

We will use the remainder convention \(R_N=S-s_N\). Thus, a positive remainder means the sum is above the partial sum; a negative remainder means it is below. The first omitted term is \((-1)^N b_{N+1}\): it is positive when \(N\) is even and negative when \(N\) is odd.

Definition: The remainder after \(N\) terms is \(R_N=S-s_N\). For an alternating series that satisfies the conditions above, \(R_N\) has the same weak sign as the first omitted term. Consequently, an odd partial sum \(s_N\) is an overestimate, while an even partial sum \(s_N\) is an underestimate.

Here, “odd” and “even” refer to the number \(N\) of terms included, not to the sign of the partial sum. Because \(R_N=S-s_N\), the sign of \(R_N\) directly tells us which quantity is larger.

Why the Remainder Has the Sign of the First Omitted Term

The result follows by grouping consecutive terms of the tail. Since the series converges, its tail \(R_N\) is a well-defined limit. Multiply it by \((-1)^N\), which changes the sign if \(N\) is odd and leaves it unchanged if \(N\) is even. The result starts with a positive term:

$$ (-1)^N R_N=b_{N+1}-b_{N+2}+b_{N+3}-b_{N+4}+\cdots. $$

Each pair \(b_{N+1}-b_{N+2}\), \(b_{N+3}-b_{N+4}\), and so on is nonnegative, because the magnitudes do not increase. The even-length partial sums of this transformed tail are therefore nonnegative. The odd-length partial sums are also nonnegative: each is an even-length partial sum followed by one nonnegative term. Taking their limit gives \((-1)^N R_N\geq 0\), so \(R_N\) has the sign of the first omitted term, or is zero.

The same grouping also shows that the transformed tail cannot exceed its first term. Indeed, subtracting the transformed tail from \(b_{N+1}\) gives another alternating tail beginning with a positive term:

$$ b_{N+1}-(-1)^N R_N = b_{N+2}-b_{N+3}+b_{N+4}-b_{N+5}+\cdots \geq 0. $$

Combining the two inequalities gives the familiar remainder bound together with its direction:

$$ 0\leq (-1)^N R_N\leq b_{N+1}. $$

This derivation explains the sign without requiring us to know the exact value of \(S\). It depends on the alternating signs, nonincreasing magnitudes, and convergence to zero; without the required conditions, the conclusion need not hold.

The Sum Lies Between Consecutive Partial Sums

The next partial sum differs from \(s_N\) by exactly the first omitted term:

$$ s_{N+1}=s_N+(-1)^N b_{N+1}. $$

The remainder \(R_N=S-s_N\) has the same direction as this step, and its magnitude is no greater than \(b_{N+1}\). Therefore, starting at \(s_N\), the sum \(S\) lies no farther along that step than \(s_{N+1}\). In interval notation, this says

$$ \min(s_N,s_{N+1})\leq S\leq \max(s_N,s_{N+1}). $$

In particular, if \(N\) is even, the omitted term is positive and \(s_N\leq S\leq s_{N+1}\). If \(N\) is odd, the omitted term is negative and \(s_{N+1}\leq S\leq s_N\). The two consecutive partial sums give a bracket regardless of which is the larger one.

Theorem: For a convergent alternating series with nonnegative, nonincreasing term magnitudes tending to zero, the remainder \(S-s_N\) points in the direction of the first omitted term. Hence \(S\) lies between \(s_N\) and \(s_{N+1}\). Odd partial sums overestimate the sum; even partial sums underestimate it.

Worked Examples

Worked Example: Bracketing the Alternating Harmonic Series

Consider \(\sum_{n=1}^{\infty}(-1)^{n-1}/n\). Its magnitudes \(b_n=1/n\) decrease to zero, so the sign and bracketing results apply. Find the bracket using four and five terms.

Compute the fourth partial sum term by term:

$$ s_4 =1-\frac12+\frac13-\frac14 =\frac{6}{6}-\frac{3}{6}+\frac{2}{6}-\frac{1}{6} =\frac{4}{6} =\frac23. $$

Since \(4\) is even, the first omitted term is positive, and \(s_4\) underestimates \(S\). The next partial sum is

$$ s_5=s_4+\frac15 =\frac23+\frac15 =\frac{10}{15}+\frac{3}{15} =\frac{13}{15}. $$

Thus the sum satisfies \(\frac23\leq S\leq\frac{13}{15}\). Checking the bracket, \(s_5-s_4=\frac{13}{15}-\frac{10}{15}=\frac15\), exactly the positive fifth term, so the two endpoints are consecutive partial sums as required.

Worked Example: Checking the Direction with a Geometric Series

Consider \(\sum_{n=1}^{\infty}(-1)^{n-1}(1/2)^n\). Its terms are \(\frac12,-\frac14,\frac18,-\frac1{16},\ldots\). The magnitudes decrease to zero. The geometric-series formula gives the sum

$$ S=\frac{\frac12}{1-(-\frac12)}=\frac{\frac12}{\frac32}=\frac13. $$

Using two terms gives

$$ s_2=\frac12-\frac14=\frac24=\frac14. $$

Since \(N=2\) is even, \(s_2\) should underestimate \(S\). Indeed, \(\frac14<\frac13\), and the remainder is \(R_2=S-s_2=\frac13-\frac14=\frac1{12}>0\), matching the positive first omitted term \(\frac18\). The next partial sum is

$$ s_3=s_2+\frac18=\frac14+\frac18=\frac38. $$

Since \(N=3\) is odd, \(s_3\) should overestimate \(S\). Indeed, \(\frac13<\frac38\), and \(R_3=\frac13-\frac38=\frac8{24}-\frac9{24}=-\frac1{24}<0\), matching the negative first omitted term \(-\frac1{16}\). The sum lies between the two consecutive values: \(\frac14\leq\frac13\leq\frac38\).

Worked Example: An Odd Partial Sum for an Alternating Reciprocal-Square Series

Consider \(\sum_{n=1}^{\infty}(-1)^{n-1}/n^2\), whose magnitudes \(b_n=1/n^2\) decrease to zero. We can locate the sum using the third and fourth partial sums without evaluating the infinite series.

$$ s_3=1-\frac14+\frac19 =\frac{36}{36}-\frac{9}{36}+\frac{4}{36} =\frac{31}{36}. $$

The third partial sum is odd, so it overestimates \(S\). The next term is \(-1/16\), giving

$$ s_4=s_3-\frac1{16} =\frac{124}{144}-\frac{9}{144} =\frac{115}{144}. $$

Therefore the sum is bracketed by these values:

$$ \frac{115}{144}\leq S\leq\frac{31}{36}. $$

To check the ordering, \(\frac{31}{36}=\frac{124}{144}\), and \(115/144<124/144\). Their difference is \(\frac{124}{144}-\frac{115}{144}=\frac9{144}=\frac1{16}\), the magnitude of the fourth term. The odd partial sum is the upper endpoint, as predicted.

A Common Sign Mistake

A frequent error is to use the sign of the last included term instead of the sign of the first omitted term, or to forget which order is used to define the remainder. For example, in the geometric-series example, the two-term sum is \(s_2=\frac14\). A mistaken argument might say, “The last included term is negative, so \(s_2\) is an overestimate.” But the computed sum is \(S=\frac13\), and \(\frac14<\frac13\): it is an underestimate.

The correction is to use the stated remainder convention and inspect the next term. Here \(R_2=S-s_2\), and the first omitted term is \(+\frac18\). Thus \(R_2>0\), so \(S>s_2\). If instead an error is defined as \(s_N-S\), its sign is the opposite. Keeping these definitions separate prevents a correct parity rule from being applied backward.

When the Direction Is Useful

A two-sided error bound tells us how far an approximation could be from the sum; the sign tells us which side of the sum the approximation lies on. Together, they provide a certified interval even when the exact sum is unknown. If \(N\) is odd, the partial sum is the upper endpoint and the next partial sum is the lower endpoint. If \(N\) is even, their roles reverse.

The result applies when the alternating-series conditions hold for the tail being considered. If magnitudes decrease only after some initial terms, apply the conclusion only once the omitted terms begin in that decreasing part. A sign pattern by itself is not enough: the required control on the magnitudes is what justifies the grouping argument.

Check Your Understanding

Answer each question using the convention \(R_N=S-s_N\).

  1. For a qualifying alternating series, does an odd partial sum overestimate or underestimate the sum?
  2. If \(N\) is even, what is the sign of the first omitted term, and what does that imply about \(S-s_N\)?
  3. Why does the sum lie between \(s_N\) and \(s_{N+1}\)?
  4. For the alternating series with terms \(1,-1/3,1/5,-1/7,\ldots\), is \(s_3\) above or below the sum? Identify the first omitted term.
  5. What conditions on the term magnitudes are needed for the sign conclusion?

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