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Alternating Series Remainder

Alternating Series Remainder

A quantity that measures how accurately the nth partial sum of an alternating series estimates the sum of the series. If an alternating series is not convergent then the remainder is not a finite number.

Consider the following alternating series (where ak > 0 for all k) and/or its equivalents.

\[\sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)}^{k + 1}}{a_k}} = {a_1} - {a_2} + {a_3} - {a_4} + \cdots \]

If the series converges to S, then the nth partial sum Sn and the corresponding remainder Rn can be defined as follows.

\[{S_n} + {R_n} = S\] \[{S_n} = \sum\limits_{k = 1}^n {{{\left( { - 1} \right)}^{k + 1}}{a_k}} \] \[{R_n} = \sum\limits_{k = n + 1}^\infty {{{\left( { - 1} \right)}^{k + 1}}{a_k}} \]

This gives us the following

\[{R_n} = S - \sum\limits_{k = 1}^n {{{\left( { - 1} \right)}^{k + 1}}{a_k}} \]

If the series converges to S by the alternating series test, then the remainder Rn can be estimated as follows for all n ≥ N:

\[\left| {{R_n}} \right| \le {a_{n + 1}}\]

Note that the alternating series test requires that the numbers a1, a2, a3, ... must eventually be nonincreasing. The number N is the point at which the values of an become non-increasing.

an ≥ an +1 for all n ≥ N, where N ≥ 1.

 

See also

Remainder of a series, convergence tests, divergent series

Key Formula

∣Rn∣=∣S−Sn∣≤an+1\left| R_n \right| = \left| S - S_n \right| \le a_{n+1}
Where:
  • RnR_n = The remainder (error) after summing the first n terms
  • SS = The exact sum the infinite alternating series converges to
  • SnS_n = The nth partial sum, i.e., the sum of the first n terms
  • an+1a_{n+1} = The absolute value of the first omitted term (the (n+1)th term), which serves as the error bound

Worked Example

Problem: Estimate the sum of the alternating series ∑k=1∞(−1)k+1k3\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^3} using the first 4 terms, and find an upper bound for the error.
Step 1: Verify the series satisfies the alternating series test. Here ak=1k3a_k = \frac{1}{k^3}. The terms are positive, decreasing, and approach 0 as k→∞k \to \infty, so the test applies.
ak=1k3,lim⁡k→∞ak=0,ak+1<aka_k = \frac{1}{k^3}, \quad \lim_{k \to \infty} a_k = 0, \quad a_{k+1} < a_k
Step 2: Compute the 4th partial sum S4S_4 by adding the first four terms.
S4=11−18+127−164=1−0.125+0.03704−0.01563≈0.89641S_4 = \frac{1}{1} - \frac{1}{8} + \frac{1}{27} - \frac{1}{64} = 1 - 0.125 + 0.03704 - 0.01563 \approx 0.89641
Step 3: Apply the Alternating Series Remainder bound. The first omitted term is a5a_5.
∣R4∣≤a5=153=1125=0.008\left| R_4 \right| \le a_5 = \frac{1}{5^3} = \frac{1}{125} = 0.008
Step 4: Interpret the result. The true sum SS lies within 0.008 of our partial sum.
0.89641−0.008≤S≤0.89641+0.0080.89641 - 0.008 \le S \le 0.89641 + 0.008
Answer: The partial sum S4≈0.89641S_4 \approx 0.89641 approximates the series sum with an error of at most 1125=0.008\frac{1}{125} = 0.008.

Another Example

This example works backward—finding the minimum number of terms needed for a desired accuracy—rather than computing the error for a given partial sum.

Problem: How many terms of the series ∑k=1∞(−1)k+1k\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} are needed to approximate the sum with an error less than 0.01?
Step 1: Identify ak=1ka_k = \frac{1}{k}. The series is the alternating harmonic series, which converges to ln⁡2\ln 2 by the alternating series test.
∑k=1∞(−1)k+1k=ln⁡2\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = \ln 2
Step 2: Set up the remainder inequality. We need ∣Rn∣≤an+1<0.01|R_n| \le a_{n+1} < 0.01.
1n+1<0.01\frac{1}{n+1} < 0.01
Step 3: Solve for nn.
n+1>100⇒n>99⇒n≥100n + 1 > 100 \quad \Rightarrow \quad n > 99 \quad \Rightarrow \quad n \ge 100
Step 4: Verify: using 100 terms gives a101=1101≈0.0099<0.01a_{101} = \frac{1}{101} \approx 0.0099 < 0.01. This confirms the bound is satisfied.
∣R100∣≤1101≈0.0099\left| R_{100} \right| \le \frac{1}{101} \approx 0.0099
Answer: You need at least 100 terms to guarantee the error is less than 0.01.

Frequently Asked Questions

What is the difference between the Alternating Series Remainder and the Lagrange Remainder?
The Alternating Series Remainder applies specifically to convergent alternating series and bounds the error by the first omitted term an+1a_{n+1}. The Lagrange Remainder applies to Taylor polynomial approximations of functions and involves a derivative evaluated at some unknown point. They address different types of approximation problems, though a Taylor series can sometimes be alternating, in which case you might use either bound.
When can you use the Alternating Series Remainder?
You can use it only when the series passes the Alternating Series Test. That means the series must alternate in sign, the absolute values of the terms aka_k must eventually be non-increasing, and lim⁡k→∞ak=0\lim_{k \to \infty} a_k = 0. If any of these conditions fail, this remainder estimate does not apply.
Does the Alternating Series Remainder give the exact error?
No, it gives an upper bound on the error, not the exact error. The actual error ∣Rn∣|R_n| is usually smaller than an+1a_{n+1}. The bound guarantees that the error is no worse than an+1a_{n+1}, which is useful because you rarely know the exact sum SS.

Alternating Series Remainder vs. Lagrange Remainder (Taylor)

Alternating Series RemainderLagrange Remainder (Taylor)
Applies toConvergent alternating series satisfying the Alternating Series TestTaylor polynomial approximations of differentiable functions
Error bound formula∣Rn∣≤an+1|R_n| \le a_{n+1} (first omitted term)∣Rn(x)∣≤M(n+1)!∣x−c∣n+1|R_n(x)| \le \frac{M}{(n+1)!}|x - c|^{n+1} where MM bounds the (n+1)(n+1)th derivative
Ease of useVery simple—just evaluate the next termOften harder—requires finding a bound on a higher-order derivative
Typical course contextSeries and sequences unit in Calculus II / BCTaylor and Maclaurin series unit in Calculus II / BC

Why It Matters

The Alternating Series Remainder appears frequently on the AP Calculus BC exam, where you are asked to bound the error of a partial sum or determine the number of terms needed for a given accuracy. In scientific computing and engineering, it provides a quick error check when evaluating series like ln⁡2\ln 2, e−xe^{-x}, or arctan⁡x\arctan x via their alternating series representations. Understanding this bound also builds intuition for why alternating series are often "well-behaved" compared to series of all positive terms.

Common Mistakes

Mistake: Using ana_n instead of an+1a_{n+1} as the error bound.
Correction: The bound is the first omitted term, an+1a_{n+1}, not the last included term ana_n. If you sum nn terms, the error is bounded by term number n+1n+1.
Mistake: Applying the remainder bound without verifying the Alternating Series Test conditions.
Correction: The bound ∣Rn∣≤an+1|R_n| \le a_{n+1} is only valid when the terms aka_k are eventually non-increasing and approach zero. Always check these conditions first. If the terms are not eventually decreasing, this estimate can be invalid.

Related Terms