Alternating Series Remainder Estimates

From Convergence to a Quantitative Error Estimate

The Alternating Series Test shows that a positive-first alternating series converges when its nonnegative magnitudes decrease to zero. Its proof also gives an even–odd bracketing of the sum: even partial sums lie at or below the sum, and odd partial sums lie at or above it. These facts let us say more than that a partial sum is close to the answer. They tell us how large the error can be and, with the parity known, on which side of the answer the partial sum lies.

Throughout, let \(b_n\geq0\), \(b_{n+1}\leq b_n\), and \(b_n\to0\). Consider the positive-first series \(\sum_{n=1}^{\infty}(-1)^{n-1}b_n\), with partial sums \(S_N=\sum_{n=1}^{N}(-1)^{n-1}b_n\) and sum \(S\). For \(N\geq1\), define the remainder after \(N\) terms by

The Remainder Bound for a Convergent Alternating Series, established earlier in this course, gives \(|R_N|\leq b_{N+1}\). The estimate is useful on its own, but the even–odd bracketing theorem gives additional information: the remainder is between zero and the first omitted term, allowing for equality at either endpoint. We will also obtain a sharper enclosure that uses the next omitted term.

The First Omitted Term Locates the Remainder

The first term not included in \(S_N\) is \((-1)^N b_{N+1}\). If \(N\) is even, this term is nonnegative and \(S_N\leq S\). If \(N\) is odd, it is nonpositive and \(S\leq S_N\). In either case, the partial sum and the full sum are ordered in the direction indicated by the first omitted term.

Proof. Suppose first that \(N\) is even. The even–odd partial-sum bracketing theorem gives \(S_N\leq S\leq S_{N+1}\). Subtracting \(S_N\) throughout yields $$ 0\leq R_N\leq S_{N+1}-S_N=b_{N+1}, $$ because \(N+1\) is odd and the \((N+1)\)st term is \(+b_{N+1}\).

Now suppose that \(N\) is odd. The bracketing theorem gives \(S_{N+1}\leq S\leq S_N\). Subtracting \(S_N\) gives $$ S_{N+1}-S_N\leq R_N\leq0. $$ Here \(S_{N+1}-S_N=-b_{N+1}\), so \(-b_{N+1}\leq R_N\leq0\). These two cases prove the stated bounds. If the remainder is nonzero, the relevant interval forces it to have the sign of the first omitted term. The intervals include zero, so they do not rule out \(R_N=0\). \(\square\)

The absolute-value estimate follows from either case: \(|R_N|\leq b_{N+1}\). More importantly for locating the sum, the theorem says \(S_N\leq S\) when \(N\) is even and \(S_N\geq S\) when \(N\) is odd. These are weak inequalities, not strict ones. A tail can cancel completely, so a nonzero first omitted term does not guarantee a nonzero remainder.

A Sharper Enclosure Using Two Omitted Terms

The first omitted term gives a one-sided bound, while the next partial sum on the opposite side of \(S\) gives a second bound. Together they make a narrower interval. The result below refines the usual remainder estimate without requiring a formula for the infinite sum.

Proof. If \(N\) is even, then \(N+1\) and \(N+2\) are odd and even, respectively. The bracketing theorem gives \(S_{N+2}\leq S\leq S_{N+1}\). Subtract \(S_N\). Since $$ S_{N+2}-S_N=b_{N+1}-b_{N+2} \quad\text{and}\quad S_{N+1}-S_N=b_{N+1}, $$ we obtain \(b_{N+1}-b_{N+2}\leq R_N\leq b_{N+1}\).

If \(N\) is odd, the bracketing theorem gives \(S_{N+1}\leq S\leq S_{N+2}\). Subtracting \(S_N\) and using $$ S_{N+1}-S_N=-b_{N+1} \quad\text{and}\quad S_{N+2}-S_N=-b_{N+1}+b_{N+2} $$ gives \(-b_{N+1}\leq R_N\leq-b_{N+1}+b_{N+2}\). This proves both cases. Since the magnitudes are nonincreasing, \(b_{N+1}-b_{N+2}\geq0\) and \(-b_{N+1}+b_{N+2}\leq0\), consistent with the signed bound. \(\square\)

The enclosure uses the next partial sum beyond \(S_{N+1}\) to tighten the bound on the side away from \(S_N\). Its interval has width \(b_{N+2}\): subtracting the two endpoints in either case gives \(b_{N+2}\). Thus the two-term enclosure can be substantially narrower than the one-term interval when \(b_{N+2}\) is small.

Worked Examples

Equality and the Meaning of the Sign

The word “sign” in a remainder estimate requires care. The theorem places \(R_N\) in an interval with zero as one endpoint. It follows that a nonzero remainder has the sign of the first omitted term, but the remainder itself may be zero. This possibility occurs even when the omitted term is nonzero; strict inequalities would therefore be incorrect under the hypotheses used here.

If the alternating series begins with a negative term, it is the negative of the positive-first series with the same magnitudes. Negating the partial sums, sum, and remainder reverses all inequalities. The absolute error bound is unchanged, while the direction of the signed estimate reverses. It is often simplest to apply the positive-first result and then negate the resulting interval.

Choosing How Many Terms to Use

For a requested error tolerance \(\varepsilon>0\), choose \(N\) so that \(b_{N+1}<\varepsilon\). The Remainder Bound then guarantees \(|S-S_N|\leq b_{N+1}<\varepsilon\). If only a non-strict guarantee \(|S-S_N|\leq\varepsilon\) is needed, it suffices to have \(b_{N+1}\leq\varepsilon\). A small first omitted term is a convenient stopping rule because it certifies the error without requiring the exact sum.

When a one-sided estimate is useful, parity supplies it at no additional cost: for even \(N\), \(S_N\leq S\), and for odd \(N\), \(S_N\geq S\). When a tighter numerical interval is desired, use the first two omitted magnitudes as in the two-term enclosure. These are bounds rather than formulas for the exact remainder, and equality can occur at either endpoint. The estimates say how reliably a finite computation represents the sum; they do not determine the sum itself.