From an Error Tolerance to a Number of Terms
When an alternating series converges, its partial sums can approximate its infinite sum. In applications, it is useful to know how many terms are enough to meet a specified error tolerance. The alternating-series remainder bound turns this into a concrete task: make the first omitted term no larger than the tolerance, then solve the resulting inequality for the number of terms retained.
The key detail is that the term omitted after \(N\) terms is indexed \(N+1\), not \(N\). Once the inequality is solved, the index must also be rounded in the direction that preserves the bound. These details determine whether a proposed truncation is actually certified.
The Bound and the Truncation Index
Consider an alternating series whose terms have magnitudes \(b_n\), where \(b_n\) is nonincreasing and tends to zero. For example, its signs might begin \(+,-,+,-,\ldots\). If \(S\) is the sum and \(S_N\) is the partial sum of its first \(N\) terms, the alternating-series remainder bound says that the remainder \(R_N=S-S_N\) satisfies
The bound applies when the term magnitudes decrease (at least from the relevant point onward) and tend to zero, as required by the alternating-series result. In particular, it applies to the examples below. The number \(N\) is the number of terms retained; \(b_{N+1}\) is the magnitude of the first term left out.
To find this index, solve \(b_{N+1}\le\varepsilon\) for \(N\), taking account of the fact that \(N\) must be a whole number. A real-valued threshold is not itself a valid number of terms unless it is an integer. Usually the answer is the least integer satisfying the inequality, not simply the nearest integer.
This is a sufficient error guarantee, not a claim that the actual error for every smaller index is too large. The actual remainder can be smaller than the bound. If the bound fails at \(N-1\), we know only that this method has not certified \(N-1\); we have not proved that its actual error exceeds the tolerance.
Why the First Omitted Term Controls the Error
The reason this method works is the alternating structure. After the first omitted term, the remaining terms have alternating signs and nonincreasing magnitudes. Their successive additions offset one another. The remainder has the sign of the first omitted term, and its magnitude cannot exceed that first term's magnitude. This is the alternating-series remainder result, also called the alternating-series error bound.
In the usual notation, if the series begins with a positive term, then the remainder after \(N\) terms has the sign of the \((N+1)\)st term. Its size is bounded by \(b_{N+1}\). Thus making that first omitted magnitude at most \(\varepsilon\) is sufficient to guarantee the requested error. The bound depends on the alternating-series hypotheses; it should not be applied to an arbitrary series just because some terms change sign.
Write down the magnitude \(b_n\) of the \(n\)th term and note that \(N\) retained terms leave term \(N+1\) first.
Require \(b_{N+1}\le\varepsilon\), using the first-omitted-term bound.
Rearrange carefully, preserving the inequality direction. Convert the result into the least allowable integer \(N\).
Substitute the chosen \(N\) into \(b_{N+1}\), then check that the preceding index fails the bound when it is a valid index.
Worked Examples
Worked Example: The Alternating Harmonic Series
How many terms of \(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n}\) are sufficient to guarantee an error of at most \(0.01\)?
Here \(b_n=\frac{1}{n}\), and these magnitudes decrease to zero. The alternating-series remainder bound applies. After \(N\) terms, the first omitted magnitude is
Require this to be at most \(0.01=\frac{1}{100}\):
Since both denominators are positive, multiplying by \(100(N+1)\) gives \(100\le N+1\), so \(N\ge99\). The least integer candidate is \(N=99\). Its first omitted term is \(b_{100}=\frac{1}{100}=0.01\), so the bound guarantees \(|R_{99}|\le0.01\). For \(N=98\), the first omitted term is \(b_{99}=\frac{1}{99}\), and \(\frac{1}{99}>0.01\), since \(100>99\). Thus 99 terms is the smallest number certified by this bound.
Check the indexing: 99 terms are retained, so term 100 is the first omitted one.
Worked Example: An Alternating \(p\)-Series
Find the smallest number of terms certified to approximate \(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n^2}\) with error at most \(0.001\).
The magnitudes \(b_n=\frac{1}{n^2}\) decrease to zero, so the remainder bound applies. The condition is
All quantities being compared are positive. Taking reciprocals reverses the comparison, giving \((N+1)^2\ge1000\). Because \(N+1\) is positive, this is equivalent to \(N+1\ge\sqrt{1000}\). Since \(\sqrt{1000}\approx31.622\), the least integer value of \(N+1\) that works is 32. Therefore \(N=31\).
Check the selected index: the first omitted term is \(b_{32}=\frac{1}{32^2}=\frac{1}{1024}\approx0.0009765625\), which is less than \(0.001\). For one fewer retained term, \(N=30\), the first omitted term is \(b_{31}=\frac{1}{31^2}=\frac{1}{961}\approx0.0010405827\), which is greater than \(0.001\). Thus 31 terms is the smallest number certified by the bound. The displayed decimals are comparisons only; the exact fractions already establish them because \(1024>1000\), whereas \(961<1000\).
Worked Example: A Linear Denominator
Determine the least number of terms that the first-omitted-term bound certifies for \(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{3n+2}\) with tolerance \(0.005\).
The term magnitudes \(b_n=\frac{1}{3n+2}\) decrease as \(n\) increases and tend to zero. With \(N\) terms retained, require
The denominator is positive, so cross-multiplication gives \(200\le3N+5\). Subtracting 5 gives \(195\le3N\), and dividing by 3 gives \(65\le N\). Thus the least candidate is \(N=65\).
At \(N=65\), the first omitted term is \(b_{66}=\frac{1}{3(66)+2}=\frac{1}{200}=0.005\), meeting the tolerance exactly. At \(N=64\), the first omitted term is \(b_{65}=\frac{1}{3(65)+2}=\frac{1}{197}\). Since \(197<200\), \(\frac{1}{197}>\frac{1}{200}=0.005\), so the bound does not certify 64 terms. Therefore 65 is the smallest index certified by this bound.
Rounding and Boundary Cases
When solving the inequality, rounding must respect its direction. For instance, if algebra gives \(N\ge 12.4\), the least integer that satisfies it is \(N=13\), not 12. If the result is \(N\ge12\), then \(N=12\) already works. In examples involving square roots or other non-integer thresholds, compare nearby integer candidates directly in the original inequality. This avoids mistakes from approximate decimal values.
The examples used a tolerance meaning “error at most \(\varepsilon\),” so the condition is \(|R_N|\le\varepsilon\), and equality in the bound is acceptable. If a problem instead requires error strictly less than \(\varepsilon\), use \(b_{N+1}<\varepsilon\). This distinction can change the answer when the first omitted term equals the tolerance exactly. In the linear-denominator example, 65 terms meet an at-most tolerance of \(0.005\); they do not establish a strict error bound below \(0.005\) using this estimate.
Common Pitfall: Using the Last Retained Term
A frequent indexing error is to test \(b_N\le\varepsilon\) after retaining \(N\) terms. But \(b_N\) is the magnitude of the last term included. The remainder starts with \(b_{N+1}\). For example, for the alternating harmonic series with tolerance \(0.01\), the incorrect inequality \(\frac{1}{N}\le0.01\) suggests \(N\ge100\). This tests the last retained term, not the first omitted term. The correct condition is \(\frac{1}{N+1}\le0.01\), which gives \(N\ge99\). Verify the meaning of the index before solving.
Another important limitation is that the least certified index need not be the true least index whose actual error is within tolerance. The bound gives a reliable guarantee, but it may overestimate the error. If its inequality fails at an index, that only means this particular guarantee is insufficient there; it does not determine the actual remainder.
Check Your Understanding
For each question, use the first-omitted-term bound and state what your result certifies.
- For the alternating harmonic series, how many terms are certified for tolerance \(0.02\)? Check the selected index and the preceding index.
- For \(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n^2}\), set up the inequality for tolerance \(0.0004\). What is the least certified index?
- Why is the first omitted magnitude \(b_{N+1}\), rather than \(b_N\), after \(N\) terms have been retained?
- If a bound equals the tolerance exactly, does it meet an “at most” requirement? What if the requirement is “strictly less than”?
- If the first-omitted-term bound fails for \(N-1\), what can and cannot be concluded about the actual error at that index?