Alternating Series Error Bound

3. Visual Connection

5. Concept Development

BC-only topic. Most alternating series do not have an easily computed exact sum. When their magnitudes decrease to zero, consecutive partial sums trap the sum and give a guaranteed truncation-error bound.

Notation for approximation and error

Let

The partial sum \(S_N\) is the approximation, while \(R_N\) is the signed remainder. The absolute error is \(|R_N|=|S-S_N|\).

Alternating Series Error Bound

If \(b_n>0\), \(b_{n+1}\le b_n\), and \(b_n\to0\), then

The error after \(N\) terms is no larger than the magnitude of the first omitted term. If decrease begins only eventually, use the bound for partial sums whose omitted tail lies entirely in that decreasing region.

Verify the hypotheses first

The bound is a consequence of the Alternating Series Test. Before using it, state that the terms alternate and that the positive magnitudes decrease to zero over the entire omitted tail. Alternating signs alone are not enough.

Why the first omitted term controls the error

For a positive-first alternating series with decreasing magnitudes, even partial sums increase toward \(S\), odd partial sums decrease toward \(S\), and

Because \(S_N\) and \(S_{N+1}\) lie on opposite sides of \(S\), the distance from \(S_N\) to \(S\) cannot exceed

A guaranteed interval for the sum

The symmetric error statement gives

A sharper one-sided interval comes from consecutive partial sums:

The sign of the error

The remainder \(R_N\) has the same sign as the first omitted term. For \(b_1-b_2+b_3-\cdots\),

• If \(N\) is even, the next term is positive, so \(R_N>0\) and \(S_N\) is an underestimate.
• If \(N\) is odd, the next term is negative, so \(R_N<0\) and \(S_N\) is an overestimate.

If the series begins with a negative term, reverse these parity conclusions and inspect the actual first omitted sign.

Finding how many terms are required

For an allowed error \(\varepsilon>0\), solve

1. Write the first omitted magnitude using the correct index.
2. Solve the inequality for \(N\).
3. Round in the direction that guarantees the inequality.
4. Check the smallest integer candidate in the original inequality.

“At most” versus “less than”

If the problem says error at most \(\varepsilon\), use \(b_{N+1}\le\varepsilon\). If it requires error less than \(\varepsilon\), use the strict inequality \(b_{N+1}<\varepsilon\). Equality can change the least acceptable integer.

Index versus number of terms

For a series starting at \(n=1\), \(S_N\) contains \(N\) terms and the first omitted index is \(N+1\). For a series starting at \(n=0\), the terms through index \(N\) total \(N+1\) terms, and the next index is still \(N+1\). Always count from the displayed starting index.

Constants and substituted values

Include every constant and substituted variable value in the first omitted magnitude. If \(c\sum(-1)^n b_n\) is approximated, then

For an alternating power series evaluated at \(x\), verify that the resulting numerical magnitudes decrease to zero before applying the bound.

A bound is not the exact error

The inequality says the actual error is somewhere between \(0\) and the next-term magnitude. Write “at most” or “bounded by,” not “the error equals \(b_{N+1}\).”

Decimal-place guarantees

To guarantee rounding to \(d\) decimal places, it is sufficient to make the error at most half of one unit in the \(d\)th decimal place:

Use the tolerance explicitly given by the problem when one is provided.

AP-style justification checklist

1. Identify the positive magnitude \(b_n\).
2. Verify that the evaluated series alternates and that \(b_n\) decreases to zero.
3. Name the Alternating Series Error Bound.
4. Display the magnitude of the first omitted term.
5. Compare it with the requested tolerance using the correct inequality.
6. State the resulting error bound, interval, or minimum number of terms in context.

6. Detailed Worked Example and Error Check

Example 1: Bound a known partial-sum error

For the alternating harmonic series, \(b_n=1/n\). After ten terms,

The actual error is at most \(1/11\); it is not asserted to equal \(1/11\).

Example 2: Trap the sum between consecutive partial sums

For \(1-1/2+1/3-\cdots\),

Because \(S_4\) is an even partial sum and \(S_5\) is the next odd partial sum,

Example 3: Determine overestimate or underestimate

In the positive-first alternating harmonic series, the first omitted term after \(S_8\) is \(+1/9\). Hence \(R_8>0\), so \(S_8<S\). After \(S_9\), the first omitted term is \(-1/10\), so \(S_9>S\).

Example 4: Error at most 0.001

To approximate the alternating harmonic sum with error at most \(0.001\), require

Thus \(N+1\ge1000\), and the least choice is \(N=999\) terms.

Example 5: A cubic denominator

For \(\sum_{n=1}^{\infty}(-1)^{n+1}/n^3\), require error at most \(10^{-4}\):

Because \(21^3=9261<10000<22^3=10648\), the least first omitted index is \(22\), so \(N=21\) terms suffice.

Example 6: A series starting at n = 0

Let \(S_N=\sum_{n=0}^{N}(-1)^n/(n+2)\). This sum contains \(N+1\) terms. The first omitted index is \(n=N+1\), whose magnitude is

Therefore \(|S-S_N|\le1/(N+3)\).

Example 7: The bound need not equal the error

For \(\sum_{n=0}^{\infty}(-1)^n/2^n\), the first four terms give \(S_3=5/8\), and the first omitted magnitude is \(1/16\). Thus

The geometric sum is \(S=2/3\), so the actual error is \(1/24\), which is smaller than the guaranteed bound.

Example 8: Magnitudes that become decreasing later

For \(\sum_{n=2}^{\infty}(-1)^n\ln n/n\), the magnitudes decrease for \(n>e\) and approach zero. For the partial sum through \(n=10\), the first omitted magnitude is

Example 9: A scaled alternating series

The Leibniz series gives \(\pi=4\sum_{n=0}^{\infty}(-1)^n/(2n+1)\). If \(m\) terms are retained, the first omitted index is \(n=m\), so

For error at most \(0.001\), require \(4/(2m+1)\le0.001\). The least integer is \(m=2000\) terms.

Example 10: When the bound cannot be used

For \(\sum(-1)^n(n+1)/(n+2)\), the magnitudes approach \(1\), not \(0\). The series diverges, and an expression such as \(|R_N|\le(n+2)/(n+3)\) is invalid because no infinite sum \(S\) exists.

Common errors

• Using the error bound without verifying the Alternating Series Test hypotheses.
• Using \(b_N\) instead of the first omitted magnitude \(b_{N+1}\).
• Calling the bound the exact error.
• Dropping a constant multiplier from the first omitted term.
• Confusing the final index with the number of included terms when the series starts at \(0\).
• Rounding \(N\) down and losing the guarantee.
• Using \(\le\) when the problem requires a strictly smaller error.
• Forgetting to substitute the specified \(x\)-value into a power-series term.
• Claiming every even partial sum is an underestimate without checking which sign comes first.
• Giving only a decimal approximation without the theorem-based inequality.
• Applying the alternating bound to a nonalternating or divergent series.