Why an Alternating-Series Error Bound Needs Care
An alternating pattern can make a series look as though its terms cancel neatly. When the term magnitudes decrease to zero, that cancellation gives a useful bound on the error made by stopping after finitely many terms. But the bound is not a consequence of alternating signs alone. Its hypotheses matter, and its index must match the terms actually included.
Three checks prevent many false estimates: confirm that the magnitudes are nonincreasing, confirm that they tend to zero, and identify the first term left out of the partial sum. The examples below show what can go wrong when any of these checks is skipped.
The Remainder Bound and Its Conditions
Consider a series whose terms alternate in sign and have magnitudes \(b_n\):
For the alternating-series remainder bound, the magnitudes must be nonincreasing and tend to zero. If \(S\) is the sum and \(S_N=\sum_{n=1}^{N}(-1)^{n-1}b_n\) is the partial sum through the \(N\)th term, the remainder is \(R_N=S-S_N\).
The index \(N+1\) is essential: when the partial sum ends at \(N\), the first omitted term has magnitude \(b_{N+1}\). The bound is an upper bound, not generally an exact value for the error.
Why the Bound Holds
To see why the hypotheses matter, consider the tail after \(S_N\), and first suppose its first term is positive. Changing the overall sign of the tail, if needed, does not change its magnitude. Write its finite partial tails as \(T_m=b_{N+1}-b_{N+2}+\cdots+(-1)^{m-1}b_{N+m}\).
For an even number of terms, \(T_{2k}=(b_{N+1}-b_{N+2})+\cdots+(b_{N+2k-1}-b_{N+2k})\). Each pair is nonnegative because the magnitudes are nonincreasing. Thus \(T_{2k}\geq 0\).
For even \(m\), \(T_m=b_{N+1}-(b_{N+2}-b_{N+3})-\cdots-(b_{N+m-2}-b_{N+m-1})-b_{N+m}\leq b_{N+1}\). For odd \(m\), \(T_m=b_{N+1}-(b_{N+2}-b_{N+3})-\cdots-(b_{N+m-1}-b_{N+m})\leq b_{N+1}\). All subtracted quantities are nonnegative.
The even partial tails increase and are bounded above by \(b_{N+1}\); the odd partial tails decrease and are bounded below by zero. Their difference is a single term, which tends to zero because \(b_n\to 0\). They therefore approach the same limit, the tail \(R_N\). Hence \(0\leq R_N\leq b_{N+1}\) when the first omitted term is positive. If it is negative, the tail has the opposite sign and the same magnitude bound.
This argument uses both monotonicity and the zero limit. If either is missing, the pairing and limiting conclusions may fail. A sign pattern by itself does not justify the theorem.
When the Magnitudes Increase
If magnitudes increase somewhere, the alternating-series theorem does not apply. That alone does not prove the series diverges, but it does mean the theorem cannot certify the claimed remainder. In fact, a convergent alternating series can still have a tail larger than its first omitted term when the magnitudes are not nonincreasing.
Worked Example: An Increasing Magnitude Makes the Claimed Bound Fail
Define \(b_1=1\), \(b_2=\frac{1}{10}\), \(b_3=1\), and for \(n\geq 4\) let \(b_n=\frac{1}{100}\left(\frac12\right)^{n-4}\). Consider \(\sum_{n=1}^{\infty}(-1)^{n-1}b_n\). The magnitudes increase from \(b_2=\frac{1}{10}\) to \(b_3=1\), so they are not nonincreasing.
After one term, \(S_1=b_1=1\). The first omitted term is \(-b_2=-\frac{1}{10}\), so a mistaken application of the remainder bound would claim \(|R_1|\leq\frac{1}{10}\). Compute the actual tail instead. The terms starting at \(n=4\) form a geometric series beginning with \(-\frac{1}{100}\) and having ratio \(-\frac12\), so their sum is \(\frac{-1/100}{1-(-1/2)}=-\frac{1}{150}\). Therefore \(R_1=-\frac{1}{10}+1-\frac{1}{150}=\frac{67}{75}\).
Thus \(|R_1|=\frac{67}{75}>\frac{1}{10}=b_2\), and the proposed estimate fails. This series does converge: after its first three terms, it has a convergent geometric tail. The failure is specifically the estimate, not the existence of a sum. Direct subtraction also checks the result: the sum is \(S_1+R_1=1+\frac{67}{75}=\frac{142}{75}\), and subtracting \(S_1\) gives \(\frac{67}{75}\) as claimed.
When the Magnitudes Do Not Approach Zero
A necessary condition for any series to converge is that its terms approach zero. For an alternating series, this requires \(b_n\to 0\). Decreasing magnitudes are not enough: a positive decreasing sequence can approach a positive number instead.
Worked Example: Decreasing Magnitudes with a Nonzero Limit
Take \(b_n=1+\frac{1}{n}\). These magnitudes decrease, but \(\lim_{n\to\infty}b_n=1\), not zero. The terms are \(2,-\frac{3}{2},\frac{4}{3},-\frac{5}{4},\ldots\), so their magnitudes do not tend to zero.
The partial sums begin \(S_1=2\), \(S_2=2-\frac32=\frac12\), and \(S_3=\frac12+\frac43=\frac{11}{6}\). More decisively, \(S_n-S_{n-1}=(-1)^{n-1}(1+\frac1n)\), whose magnitude tends to \(1\), not zero. A convergent sequence of partial sums must have successive differences tending to zero. Therefore the series does not converge, and there is no finite sum \(S\) or remainder \(R_N=S-S_N\) to estimate.
This check catches the problem before any error bound is attempted: if the terms fail to approach zero, the series diverges. Alternation cannot repair that failure.
Last Included Term or First Omitted Term?
Suppose a partial sum includes exactly \(N\) terms, with indices \(1\) through \(N\). The last included magnitude is \(b_N\); the first omitted magnitude is \(b_{N+1}\). The theorem bounds the error by the first omitted magnitude. A later magnitude, such as \(b_{N+2}\), is smaller under the theorem's hypotheses, but generally is not a valid replacement.
Worked Example: A One-Index Shift Gives a False Bound
Consider the geometric alternating series \(\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac12\right)^{n-1}=1-\frac12+\frac14-\frac18+\cdots\). Its magnitudes decrease to zero. After two terms, \(S_2=1-\frac12=\frac12\). The first omitted term is \(+\frac14\), so the correct bound is \(|R_2|\leq\frac14\).
The exact sum is \(S=\frac{1}{1-(-1/2)}=\frac23\). Hence \(R_2=S-S_2=\frac23-\frac12=\frac16\). The next term after the first omitted one has magnitude \(\frac18\), but \(\frac16>\frac18\). Thus using \(b_4=\frac18\) as the bound after two terms is false. The correct estimate checks: \(\frac16\leq\frac14\).
The calculation distinguishes all three positions: the last included term has magnitude \(b_2=\frac12\), the first omitted term has magnitude \(b_3=\frac14\), and the term after that has magnitude \(b_4=\frac18\). The remainder bound uses \(b_3\), not \(b_4\).
A Practical Check Before Using an Estimate
Before reporting an alternating-series error bound, write down the partial sum and the terms just after it. Then verify the assumptions against the full magnitude sequence, not merely the few values that appear in the calculation.
- Alternation: Do the signs actually alternate in the series being considered?
- Monotonicity: Is \(b_{n+1}\leq b_n\) for every index from which the theorem is being used?
- Limit: Does \(b_n\to 0\)?
- Index: If the sum includes terms through \(N\), is the proposed bound \(b_{N+1}\), the first omitted magnitude?
- Meaning: Is the estimate a bound on \(|R_N|\), rather than a claim that the remainder equals that term?
A failed check means the alternating-series remainder bound has not been justified. It does not always mean the series diverges or that no other estimate is possible. It means a different argument is needed before claiming an error bound.
Check Your Understanding
For each question, identify which hypothesis or indexing issue determines whether the estimate is justified.
- A series alternates and \(b_n\) decreases, but \(b_n\to 2\). Can its alternating-series remainder bound be applied? Why?
- A partial sum includes terms through \(n=6\). Which magnitude appears in the standard remainder bound, provided its hypotheses hold?
- Why does a single increase in the magnitudes prevent a direct application of the alternating-series remainder theorem?
- For the series in the geometric worked example, what is the actual remainder after one term, and what is the first omitted magnitude?
- If a proposed error bound is smaller than the exact remainder, what does that show about the proposed estimate?