A Certified Approximation Workshop

From an Error Bound to a Certified Approximation

An alternating series can be useful even when its exact sum is not known in a convenient form. If its terms decrease in magnitude toward zero, the first omitted term bounds the error in a partial sum. That gives a practical way to decide how many terms to calculate.

A numerical answer is certified when the calculation comes with a valid bound that guarantees its stated accuracy. For decimal rounding, it is not enough to calculate several terms and observe that the answer appears stable. We need to show that the true sum lies in an interval that supports the requested rounding.

The Alternating-Series Remainder Bound

Consider a series whose terms alternate in sign and have nonnegative magnitudes \(a_n\). Assume that \(a_n\) is nonincreasing and that \(a_n\to 0\). Write its partial sum through index \(N\) as \(S_N\), and its sum as \(S\).

Theorem: If \(a_n\ge 0\), \(a_{n+1}\le a_n\), and \(a_n\to 0\), then the alternating series converges and \[ |S-S_N|\le a_{N+1}. \] The error \(S-S_N\) has the sign of the first omitted term. Thus, \(S\) lies between \(S_N\) and \(S_{N+1}\).

Here the indices start at \(0\), so \(S_N\) includes the terms with indices \(0,1,\ldots,N\), and the first omitted term has magnitude \(a_{N+1}\). If a series instead starts at \(1\), the same rule applies: identify the term immediately after the last included term. Checking this index carefully prevents a common off-by-one error.

The reason for the bound is that successive partial sums bracket the limit. For a series beginning with a positive term, the even-indexed partial sums decrease, the odd-indexed partial sums increase, and the gap between the paired sums is one term, whose magnitude tends to zero. Consequently, both subsequences approach the same limit, and that limit lies between any two consecutive partial sums. Their difference is the magnitude of the first omitted term, proving the stated bound. If the first term is negative, multiplying the series by \(-1\) gives the same conclusion, with the error sign reversed accordingly.

A Reliable Certification Procedure

For a requested tolerance \(\varepsilon>0\), it is sufficient to choose \(N\) so that \(a_{N+1}\le\varepsilon\). This guarantees \(|S-S_N|\le\varepsilon\). To claim a specific number of decimal places, do one additional check: find the interval between \(S_N\) and \(S_{N+1}\), then verify that the entire interval rounds to the same number.

Check the conditions.
Verify that the terms alternate, their magnitudes are nonincreasing, and those magnitudes tend to zero.

Choose the truncation.
Express the first omitted term in terms of the last included index, and make its magnitude no larger than the requested error tolerance.

Calculate the partial sum.
Keep exact fractions where practical, and only then convert to a decimal.

Certify the claim.
Use the sign and magnitude of the first omitted term to give an interval containing the sum. Check that the interval supports the tolerance or rounding claim.

For rounding to \(d\) decimal places, one unit in the last displayed place is \(10^{-d}\). A common sufficient tolerance is half of that, \(10^{-d}/2\), but the location of the partial sum also matters: an error bound alone does not rule out crossing a rounding midpoint. The interval check makes the certification explicit.

Worked Examples

Worked Example: Certifying Two Decimal Places in a Geometric Alternating Series

Approximate \[ S=\sum_{n=0}^{\infty}\frac{(-1)^n}{3^{n+1}} \] to two decimal places. Its term magnitudes \(a_n=1/3^{n+1}\) decrease to zero, so the alternating-series bound applies.

Rounding to two decimal places has a half-unit tolerance of \(0.005=1/200\). If we include terms through \(N\), the first omitted term is \(1/3^{N+2}\). We need \(1/3^{N+2}\le1/200\). For \(N=2\), the omitted term is \(1/81>1/200\); for \(N=3\), it is \(1/243<1/200\). Thus, \(N=3\) is the first index meeting this bound.

Calculate the partial sum, showing each step:

\begin{aligned} S_0&=\frac13,\\ S_1&=\frac13-\frac19=\frac29,\\ S_2&=\frac29+\frac1{27}=\frac7{27},\\ S_3&=\frac7{27}-\frac1{81}=\frac{21}{81}-\frac1{81} =\frac{20}{81}. \end{aligned}

The next term is positive, with magnitude \(1/3^5=1/243\). Therefore \(S\) lies between \(S_3\) and \(S_4\):

\frac{20}{81}=\frac{60}{243} \le S\le \frac{61}{243} =\frac{20}{81}+\frac1{243}.

In decimals, this interval is approximately \([0.246914,\,0.251029]\). Every number in it lies between \(0.245\) and \(0.255\), the rounding interval for \(0.25\) to two decimal places. Hence the certified approximation is \(S\approx0.25\) to two decimal places. The calculation also verifies the error bound: \(1/243\approx0.004115<0.005\).

Worked Example: Finding a Truncation for a Tolerance

Find the shortest truncation that certifies an error no greater than \(0.05\) for \[ S=\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^2}, \] then determine the nearest tenth. The magnitudes \(a_n=1/(n+1)^2\) decrease to zero.

The first omitted magnitude after \(S_N\) is \(1/(N+2)^2\). At \(N=2\), it is \(1/16=0.0625>0.05\). At \(N=3\), it is \(1/25=0.04<0.05\). Thus, including terms through \(N=3\) is the shortest truncation that meets the requested tolerance.

Compute the partial sums exactly:

\begin{aligned} S_0&=1,\\ S_1&=1-\frac14=\frac34,\\ S_2&=\frac34+\frac19=\frac{27}{36}+\frac4{36}=\frac{31}{36},\\ S_3&=\frac{31}{36}-\frac1{16} =\frac{124}{144}-\frac9{144} =\frac{115}{144}. \end{aligned}

The next term is positive and equals \(1/25\). The true sum is therefore between the next two partial sums:

\frac{115}{144} \le S\le \frac{115}{144}+\frac1{25} =\frac{2875}{3600}+\frac{144}{3600} =\frac{3019}{3600}.

The endpoints are approximately \(0.798611\) and \(0.838611\). Both lie inside \([0.75,0.85)\), the rounding interval for \(0.8\) to the nearest tenth. Therefore \(S\) rounds to \(0.8\) to the nearest tenth. As a final check, the interval width is \(1/25=0.04\), exactly the first omitted term and within the \(0.05\) tolerance.

Worked Example: Certifying Three Decimal Places with Factorial Terms

Certify the sum to three decimal places for \[ S=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}, \] where \(0!=1\). From \(n=1\) onward the magnitudes \(1/n!\) decrease to zero; the first two magnitudes are equal, which is allowed by the nonincreasing condition.

For three decimal places, use the tolerance \(0.0005\). After including terms through \(N=5\), the first omitted magnitude is \(1/6!=1/720\approx0.001389\), which is too large. Through \(N=6\), the first omitted magnitude is \(1/7!=1/5040\approx0.0001984\), which is below \(0.0005\). Calculate through \(N=6\):

\begin{aligned} S_0&=1,\\ S_1&=1-1=0,\\ S_2&=0+\frac12=\frac12,\\ S_3&=\frac12-\frac16=\frac13,\\ S_4&=\frac13+\frac1{24}=\frac38,\\ S_5&=\frac38-\frac1{120}=\frac{45}{120}-\frac1{120} =\frac{44}{120}=\frac{11}{30},\\ S_6&=\frac{11}{30}+\frac1{720} =\frac{264}{720}+\frac1{720} =\frac{265}{720}=\frac{53}{144}. \end{aligned}

The first omitted term is negative. Hence the sum lies between \(S_7=S_6-1/5040\) and \(S_6\). With a common denominator,

S_7=\frac{53}{144}-\frac1{5040} =\frac{3710}{10080}-\frac2{10080} =\frac{3708}{10080}=\frac{103}{280}.

Thus \(0.367857\ldots\le S\le0.368056\ldots\). This interval lies inside \([0.3675,0.3685)\), the rounding interval for \(0.368\) to three decimal places. The certified answer is \(0.368\) to three decimal places. The interval width is \(1/5040\approx0.0001984\), confirming the error estimate.

A Common Pitfall: Rounding the Partial Sum Too Soon

It is tempting to round a partial sum to the requested number of places and assume the true sum rounds the same way. That conclusion needs an error check. In the factorial example, truncating at \(N=5\) gives \(S_5=11/30=0.366666\ldots\), which rounds to \(0.367\). But the remainder bound is \(1/720\approx0.001389\), larger than the \(0.0005\) half-unit tolerance for three decimal places. More decisively, the next term is positive, so

0.366666\ldots=S_5\le S\le S_6=0.368055\ldots.

This interval crosses the rounding midpoint \(0.3675\). Values below that midpoint round to \(0.367\), while values above it round to \(0.368\). Therefore \(S_5\) alone does not certify the three-decimal answer. Including one more term gives the narrower interval that does.

Another frequent error is to use the last included term rather than the first omitted one. If the truncation includes indices \(0\) through \(N\), the bound is \(a_{N+1}\), not \(a_N\). State the indices and write down the omitted term before selecting \(N\).

Why Certification Matters

A partial sum is a numerical estimate; a bound turns it into a controlled approximation. The distinction matters whenever a result must meet a stated accuracy, such as a tolerance or a number of decimal places. The alternating-series estimate is particularly useful because it requires only the first omitted term, not a calculation of the entire remaining tail.

Keep the claims precise. “The approximation is close” is not a certification. “The sum lies between these two consecutive partial sums, and every number in that interval rounds to \(0.368\)” is. The interval also reveals when a requested decimal cannot yet be justified, even if the displayed partial sum looks convincing.

Check Your Understanding

For each question, identify the first omitted term and use it, rather than visual stability of the partial sums, to justify your conclusion.

If terms through index \(N\) are included in an alternating series, what term bounds the absolute error?
For \(\sum_{n=0}^{\infty}(-1)^n/(n+1)^2\), what is the first omitted term after truncating at \(N=4\)?
Why does an error bound smaller than half a unit in the last decimal place not, by itself, always certify a particular rounded answer?
For the factorial series in the example, what is the sign of the error after truncating at \(N=6\)?
What additional interval check certifies that a sum rounds to a specified decimal?