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Alternating and Absolute Convergence · Tutorial 660 of 998

Numerical Estimates from Alternating Tails

Use partial sums, alternating-tail bounds, and adjacent partial sums to produce numerical estimates with clearly stated error guarantees.

Intermediate 9 min read

What You'll Learn

  • Identify the first omitted term and use it to bound an alternating-series remainder.
  • Enclose a sum between consecutive partial sums and interpret the interval numerically.
  • Compute partial sums for the alternating harmonic series.
  • Estimate the sum of a rational-term alternating series and correct rounding errors.
  • Decide whether a stated number of decimal places is certified by an interval.
  • Use the midpoint of a certified interval to produce an estimate with a guaranteed error bound.

From a Convergent Series to a Numerical Estimate

Knowing that an infinite series converges does not by itself tell us its sum to a useful degree of accuracy. To estimate the sum, we calculate a finite partial sum and control what remains. For an alternating series with positive terms that decrease to zero, the first omitted term gives a direct bound on the remaining tail.

The key is to keep track of the index. If a partial sum ends at \(N\), the first omitted term is the term with index \(N+1\), not the term with index \(N\). This simple observation makes it possible to turn a calculation into a certified estimate.

Definition: For a series \(S=\sum_{n=1}^{\infty}(-1)^{n-1}b_n\), its \(N\)th partial sum is \(S_N=\sum_{n=1}^{N}(-1)^{n-1}b_n\). The remainder after \(N\) terms is \(R_N=S-S_N\), provided the series converges.

The Alternating-Tail Bound

Suppose \(b_n>0\), the sequence \(b_n\) is decreasing, and \(b_n\to 0\). The alternating-series test guarantees convergence. The remainder estimate says that after summing \(N\) terms, the error is no larger than the first omitted term:

$$ |R_N|\le b_{N+1}. $$

The remainder also has the sign of the first omitted term. Thus, when the next term is positive, \(S_N\) is a lower bound and \(S_{N+1}\) is an upper bound. When the next term is negative, the roles reverse. In either case, the sum lies between two consecutive partial sums.

Theorem: If \(b_n>0\), \(b_n\) is decreasing, and \(b_n\to0\), then the alternating series \(S=\sum_{n=1}^{\infty}(-1)^{n-1}b_n\) converges. Its remainder satisfies \(|S-S_N|\le b_{N+1}\), with the sign of \(S-S_N\) matching the sign of the \((N+1)\)st term.

Here is a justification of the bound. If the tail begins with a positive term, write it as \(R_N=b_{N+1}-b_{N+2}+b_{N+3}-\cdots\). Grouping consecutive terms gives nonnegative differences, so the tail is at least zero. Also, rewriting it as \(b_{N+1}-(b_{N+2}-b_{N+3})-(b_{N+4}-b_{N+5})-\cdots\) shows that it is at most \(b_{N+1}\). These inequalities hold first for finite alternating tails; passing to the limit is valid because the series converges. If the tail begins negative, apply the same argument to its negative. This proves both the sign and the bound.

In practical work, the adjacent partial sums often provide the most informative form of the result: they give an explicit interval containing the sum. If the terms decrease strictly, the sum lies strictly between those consecutive partial sums.

Worked Example: A Certified Estimate for the Alternating Harmonic Series

Worked Example: A Certified Estimate for the Alternating Harmonic Series

Consider \(S=1-\frac12+\frac13-\frac14+\cdots\). We calculate ten terms, then use the first omitted term to enclose the sum.

Group the ten terms into five positive pairs:

$$ \begin{aligned} S_{10} &=\left(1-\frac12\right) +\left(\frac13-\frac14\right) +\left(\frac15-\frac16\right) +\left(\frac17-\frac18\right) +\left(\frac19-\frac1{10}\right)\\ &=\frac12+\frac1{12}+\frac1{30}+\frac1{56}+\frac1{90}\\ &\approx 0.6456349206. \end{aligned} $$

Because \(N=10\) is even, the next term, \(1/11\), is positive. The remainder bound and its sign therefore give

$$ S_{10}<S<S_{10}+\frac1{11}. $$

The upper endpoint is

$$ S_{10}+\frac1{11} \approx 0.6456349206+0.0909090909 =0.7365440116\ \text{(rounded up)}. $$

Thus \(0.6456349206<S<0.7365440116\), with the lower endpoint rounded down and the upper endpoint rounded up at the tenth decimal place so the interval remains valid. The midpoint of this interval is \(S_{10}+\frac1{22}\approx0.6910894661\). Its distance from the sum is at most half the interval width, so this midpoint estimate has error at most \(\frac1{22}\). The interval is wide: it does not certify the sum to any particular decimal place. That limitation is part of the information the estimate provides.

As a check, adding the first omitted term gives \(S_{11}=S_{10}+\frac1{11}\), exactly the stated upper endpoint. The next term is negative, so the eleventh partial sum is above the infinite sum, as required.

Worked Example: Estimating a Rational-Term Alternating Series

Worked Example: Estimating a Rational-Term Alternating Series

Consider

$$ S=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{3n+1} =\frac14-\frac17+\frac1{10}-\frac1{13}+\cdots. $$

The positive term sizes \(b_n=1/(3n+1)\) decrease to zero, so the alternating-tail estimate applies. Add ten terms in order:

$$ \begin{aligned} S_1&=\frac14=0.2500000000,\\ S_2&=\frac14-\frac17=0.1071428571,\\ S_3&=S_2+\frac1{10}=0.2071428571,\\ S_4&=S_3-\frac1{13}=0.1302197802,\\ S_5&=S_4+\frac1{16}=0.1927197802,\\ S_6&=S_5-\frac1{19}=0.1400882013,\\ S_7&=S_6+\frac1{22}=0.1855427467,\\ S_8&=S_7-\frac1{25}=0.1455427467,\\ S_9&=S_8+\frac1{28}=0.1812570324,\\ S_{10}&=S_9-\frac1{31}=0.1489989679. \end{aligned} $$

These running totals are rounded to ten decimal places; each addition is made using the exact fractions, not the already-rounded display. Since the next term is \(1/34\), which is positive, the sum is between \(S_{10}\) and \(S_{11}\):

$$ S_{10}<S<S_{10}+\frac1{34}. $$

Using the exact \(S_{10}\) for the endpoint calculation gives

$$ 0.1489989679<S<0.1784107327, $$

where the lower endpoint is rounded down and the upper endpoint rounded up at the tenth decimal place. The midpoint is \(S_{10}+\frac1{68}\approx0.1637048503\), and the error of that estimate is at most \(\frac1{68}\approx0.0147058824\). The interval’s endpoints round differently to one decimal place, so the interval does not certify the sum to the nearest tenth, much less to the nearest hundredth. Rounding the midpoint to \(0.164\) is not, by itself, a certified rounding of the infinite sum.

For a check on the enclosure, the first omitted term adds \(1/34\) to \(S_{10}\), giving \(S_{11}\approx0.1784107326\). The following term is negative, so the infinite sum lies below \(S_{11}\), consistent with the stated bounds.

Decimal Accuracy Means Checking the Whole Interval

A decimal approximation and a certified decimal rounding are not the same thing. A partial sum may display many decimal places, but its error bound may be much larger than the last displayed digit. To certify a rounding, check that every value in the proven interval rounds to the same number of decimal places.

For example, to certify rounding to three decimal places, an interval must not cross a rounding boundary between thousandths. An error bound smaller than half of \(0.001\) is useful, but it is not always enough by itself: if the approximation is close to a rounding boundary, the interval could still cross it. Checking the two endpoints avoids that ambiguity.

Worked Example: Certifying Three Decimal Places

Consider the alternating geometric series \(S=\frac12-\frac14+\frac18-\frac1{16}+\cdots\). Its term sizes decrease to zero. After twelve terms, the finite geometric sum is

$$ S_{12}=\frac12-\frac14+\cdots-\frac1{2^{12}} =\frac13\left(1-\frac1{2^{12}}\right) =\frac{1365}{4096} =0.333251953125. $$

The thirteenth term is positive and equals \(1/2^{13}=1/8192\). Therefore

$$ 0.333251953125<S<0.333251953125+\frac1{8192} =0.3333740234375. $$

Both endpoints round to \(0.333\) to three decimal places. Consequently the sum is certified to round to \(0.333\) to three decimal places. As an additional check, the exact geometric sum is \(1/3\), and \(0.333251953125<1/3<0.3333740234375\), consistent with the enclosure.

A Common Indexing Mistake

A frequent error is to use the last included term as the error bound. In the alternating harmonic example, after ten terms, one might incorrectly write

$$ |S-S_{10}|\le\frac1{10}. $$

That is a valid but unnecessarily loose bound, because \(1/10\) is already included in \(S_{10}\). The first omitted term is \(1/11\), so the alternating-series remainder theorem gives the sharper bound

$$ |S-S_{10}|\le\frac1{11}. $$

The sign matters too. Since the next term after \(S_{10}\) is positive, the sum is above \(S_{10}\), not below it. Always identify the next term explicitly before writing either the error bound or the interval.

A Reliable Estimation Procedure

For a numerical estimate from an alternating series, use the following sequence of checks.

1
Check the term conditions.
Write the series as alternating signs times positive term sizes. Verify that the sizes decrease to zero.
2
Compute the partial sum.
Keep enough digits for the desired result, but perform the additions using exact fractions or unrounded values when feasible.
3
Find the first omitted term.
If the partial sum ends at \(N\), use \(b_{N+1}\) for the remainder bound.
4
Use its sign.
Determine whether the infinite sum lies above or below the partial sum, and write the corresponding interval.
5
Check decimal claims.
Round both interval endpoints to the requested precision. If they disagree, that precision is not certified by the interval.

Why These Estimates Matter

The alternating-series remainder bound is valuable because it turns a finite calculation into a guarantee about an infinite sum. It distinguishes a useful approximation from a numerical guess: the partial sum supplies the estimate, while the first omitted term supplies a mathematically justified error limit.

The bound can be conservative. It does not claim that the actual error equals the first omitted term; it says only that the error cannot exceed it. If a requested decimal precision is not certified, sum more terms and repeat the interval check. A justified conclusion may be that the current calculation is insufficient, rather than that a particular rounded value is correct.

Check Your Understanding

For each question, identify the partial sum, the first omitted term, and what the resulting bound does or does not certify.

  1. After \(N\) terms of an alternating series with decreasing positive term sizes \(b_n\), which term gives the standard remainder bound?
  2. For the alternating harmonic series, is the sum above or below \(S_9\)? Explain using the sign of the first omitted term.
  3. If an interval containing a sum has endpoints that round to different values to the nearest hundredth, what can you conclude about certified rounding?
  4. Why must the terms for a partial sum be added using unrounded values when a final decimal is reported?
  5. An estimate has a remainder bound of \(0.0004\). Why should you still check the interval endpoints before claiming a particular rounding to three decimal places?

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