Using an Alternating Sum as an Approximation
An infinite series can be useful even when only finitely many of its terms are calculated. If the series converges, its first \(N\) terms give a partial sum that approximates the full sum. The practical question is: how large must \(N\) be to ensure that the approximation is accurate enough?
For an alternating series whose term magnitudes decrease to zero, the remainder has a particularly simple estimate. Its size is controlled by the first term left out. This makes it possible to choose a truncation index by solving an inequality rather than by guessing and checking the actual infinite tail.
The First-Omitted-Term Estimate
Suppose \(b_n\) is positive, decreases as \(n\) increases, and approaches zero. The Alternating Series Test establishes convergence for the alternating series under these conditions. The remainder estimate gives more information: it controls the error of every partial sum.
The index matters: after \(N\) terms, the first omitted term has index \(N+1\), not \(N\). If the series begins with a negative term instead, its remainder has the sign of that first omitted term, but the absolute-error estimate is unchanged.
Why the Estimate Works
It is helpful to look at the tail beginning with its first omitted term. For simplicity, suppose that term is positive; if it is negative, multiply the whole tail by \(-1\). The tail has the form
Group successive terms, first starting with the positive term. Each difference in the resulting pairs is nonnegative, so the tail is nonnegative. Alternatively, isolate the first term and group the remaining tail in pairs:
Each subtracted difference is nonnegative, so \(R_N\leq b_{N+1}\). Together these arguments show \(0\leq R_N\leq b_{N+1}\), which gives the absolute-error bound. With strictly decreasing terms, the paired differences are positive, so the remainder is strictly between zero and the first omitted term. This proves both the sign statement and the strict version of the estimate.
Turning an Accuracy Requirement into an Index
To guarantee a requested absolute error less than \(\varepsilon\), where \(\varepsilon>0\), apply the remainder estimate to the first omitted term. A simple sufficient condition is \(b_{N+1}<\varepsilon\): then \(|R_N|\leq b_{N+1}<\varepsilon\). When the terms decrease strictly, the stronger estimate \(|R_N|<b_{N+1}\) means that \(b_{N+1}\leq\varepsilon\) is also sufficient.
Check that the series alternates, its term magnitudes are positive and decreasing (at least from the index relevant to the tail), and those magnitudes approach zero.
Use \(|R_N|\leq b_{N+1}\), or the strict bound when the magnitudes decrease strictly.
Find an integer \(N\) that makes the first omitted term small enough. Remember that the omitted-term index is \(N+1\).
A value obtained from the first-omitted-term estimate is a sufficient count certified by that estimate. It need not be the fewest terms that achieve the actual error.
Worked Examples
Worked Example: Alternating Harmonic Terms and a Strict Accuracy Goal
Consider \(S=\sum_{n=1}^{\infty}(-1)^{n+1}/n\). The magnitudes \(b_n=1/n\) are positive, strictly decreasing, and approach zero. We want to guarantee that the error is less than \(0.01\).
After \(N\) terms, the first omitted magnitude is \(b_{N+1}=1/(N+1)\). The simple condition \(b_{N+1}<0.01\) gives
Thus \(N=100\) is the smallest count for which this particular strict first-omitted-term condition holds: its first omitted term is \(1/101\), which is less than \(0.01\). But the series decreases strictly, so the strict remainder bound also applies. With \(N=99\), the first omitted term is \(b_{100}=1/100=0.01\), and therefore
So 99 terms do guarantee the requested strict accuracy. This calculation establishes that 99 suffices; it does not establish that 99 is the fewest possible number of terms for the actual error.
Check: For \(N=99\), the tail begins \(1/100-1/101+1/102-\cdots\), so the sign is positive, as the remainder sign rule predicts, and its magnitude is strictly less than \(1/100\).
Worked Example: A Series with Squared Denominators
Consider \(S=\sum_{n=1}^{\infty}(-1)^{n+1}/n^2\), and require an error strictly less than \(0.0001\). Here \(b_n=1/n^2\) is positive, strictly decreasing, and tends to zero.
The first omitted magnitude is \(b_{N+1}=1/(N+1)^2\). Requiring this to be strictly less than the target gives
For \(N=100\), the first omitted term is \(1/101^2=1/10201\), which is less than \(1/10000=0.0001\). The standard estimate therefore guarantees the requested accuracy. For \(N=99\), the first omitted term is \(1/100^2=0.0001\). Because the terms decrease strictly, the strict estimate gives
Thus 99 terms also suffice for the actual strict accuracy requirement, even though \(N=100\) is the first count for which the first omitted term itself is strictly below the target. Do not confuse those two statements.
Check: At \(N=100\), the tail begins with the negative term \(-1/101^2\), so \(R_{100}<0\). Its magnitude is bounded by the positive quantity \(1/101^2\), consistent with the remainder estimate.
Worked Example: A Sharper Check When the First Omitted Term Is Too Large
Consider \(S=\sum_{n=1}^{\infty}(-1)^{n+1}/(2n-1)\), and ask whether 49 terms guarantee error less than \(0.01\). Its magnitudes decrease strictly to zero. After 49 terms the tail begins
The first-omitted-term estimate gives \(|R_{49}|<1/99\), but \(1/99>0.01\). That estimate alone does not certify the requested accuracy. A sharper bound comes from taking the first three terms of this tail:
The remaining tail after these three terms starts with a negative term. By the sign property of the alternating-series remainder, that remaining tail is negative; hence \(R_{49}<q_3\). Also, \(R_{49}>0\), since the original tail starts positive. To verify the numerical target without relying on rounded decimals, compare exact fractions:
It follows that
Therefore 49 terms do guarantee the strict accuracy requirement, even though the first-omitted-term bound alone does not show it. This does not prove that 49 is the fewest count for the actual error; it proves only that 49 suffices by this sharper estimate.
Check: The tail starts positive, so its remainder is positive. The three-term tail sum is an upper bound because the tail after those three terms starts negative; both signs agree with the alternating-series remainder property.
Common Pitfall: Treating a Sufficient Bound as an Exact Error
The first-omitted-term estimate is a guarantee, not usually an exact calculation of the remainder. If it gives \(|R_N|\leq b_{N+1}\), then \(b_{N+1}\) is an upper bound on the error. It does not say that the error equals \(b_{N+1}\), nor does it show that the error is larger than any chosen tolerance when the bound exceeds that tolerance.
For example, in the third example, \(1/99>0.01\) does not imply that the 49-term error exceeds \(0.01\). It only means that the first-omitted-term estimate is too coarse to decide the question. The three-term tail calculation gives a stronger upper bound and establishes that the error is less than \(0.01\).
Likewise, when solving \(b_{N+1}<\varepsilon\), describe the result as the smallest \(N\) certified by that displayed condition, if it is the smallest such integer. Do not call it the fewest terms for the actual accuracy unless the actual error has also been analyzed. Strict decrease can itself sharpen the conclusion: if \(b_{N+1}=\varepsilon\), then \(|R_N|<\varepsilon\) under the strict remainder estimate.
Why This Estimate Matters
Alternating-series error control turns an infinite calculation into a finite one with a stated guarantee. It is useful whenever the desired result is an approximation rather than a symbolic description of the entire sum. The method also explains why alternating sums can converge rapidly in practice: successive terms partly cancel, leaving a tail no larger than its first term.
For an exam solution, make the logic visible. State the alternating-series conditions, identify \(b_{N+1}\), solve the inequality carefully, and state whether the count is sufficient by the standard bound, by its strict refinement, or by a sharper tail estimate. That wording distinguishes a proved accuracy guarantee from an unsupported claim about the minimum number of terms.
Check Your Understanding
For each question, attend to which estimate actually proves the claimed accuracy.
- For an alternating series with decreasing magnitudes \(b_n\), what term controls the error after \(N\) terms?
- If \(b_n\) decreases strictly and \(b_{N+1}=\varepsilon\), what strict error conclusion follows?
- For \(b_n=1/n\), what is the smallest \(N\) for which \(1/(N+1)<0.02\)? What does this count certify?
- Why does a first-omitted-term bound larger than a tolerance fail to prove that the actual error exceeds the tolerance?
- What additional fact about the terms makes \(|R_N|<b_{N+1}\) rather than only \(|R_N|\leq b_{N+1}\)?
