Mathwords logoMathwords

Remainder of a Series

Remainder of a Series

The difference between the nth partial sum and the sum of a series.

 

Example showing partial sums S₁–S₃ and remainders R₁–R₃ for Σ1/2ⁿ, with Sₙ=1/2+1/4+…+1/2ⁿ and Rₙ=1/2ⁿ⁺

 

 

See also

Alternating series remainder, integral test remainder, Taylor series remainder, convergent series, divergent series

Key Formula

Rn=S−Sn=∑k=1∞ak−∑k=1nak=∑k=n+1∞akR_n = S - S_n = \sum_{k=1}^{\infty} a_k - \sum_{k=1}^{n} a_k = \sum_{k=n+1}^{\infty} a_k
Where:
  • RnR_n = The remainder after n terms — the error when approximating S by S_n
  • SS = The exact sum of the infinite series
  • SnS_n = The nth partial sum, i.e., the sum of the first n terms
  • aka_k = The kth term of the series
  • nn = The number of terms included in the partial sum

Worked Example

Problem: Find the remainder R₃ for the geometric series ∑(1/2)^k from k = 1 to ∞ after summing the first 3 terms.
Step 1: Find the exact sum S of the infinite geometric series with first term a = 1/2 and common ratio r = 1/2.
S=a1−r=121−12=1212=1S = \frac{a}{1 - r} = \frac{\tfrac{1}{2}}{1 - \tfrac{1}{2}} = \frac{\tfrac{1}{2}}{\tfrac{1}{2}} = 1
Step 2: Compute the 3rd partial sum S₃ by adding the first 3 terms.
S3=12+14+18=4+2+18=78S_3 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{4 + 2 + 1}{8} = \frac{7}{8}
Step 3: Subtract the partial sum from the exact sum to get the remainder.
R3=S−S3=1−78=18R_3 = S - S_3 = 1 - \frac{7}{8} = \frac{1}{8}
Step 4: Verify: the remainder equals the sum of all terms from k = 4 onward. This is itself a geometric series starting at (1/2)⁴ = 1/16.
R3=∑k=4∞(12)k=1161−12=18  ✓R_3 = \sum_{k=4}^{\infty}\left(\frac{1}{2}\right)^k = \frac{\tfrac{1}{16}}{1 - \tfrac{1}{2}} = \frac{1}{8} \;\checkmark
Answer: The remainder after 3 terms is R₃ = 1/8, meaning the partial sum S₃ = 7/8 underestimates the true sum by 0.125.

Another Example

This example differs from the first because the exact sum is not computed directly. Instead, the Alternating Series Estimation Theorem provides an upper bound on |Rₙ| without needing to know S exactly — a common technique in calculus courses.

Problem: Use the alternating series estimation to bound the remainder when approximating ∑ (-1)^(k+1) / k! from k = 1 to ∞ using the first 4 terms.
Step 1: Write out the first several terms of the series.
∑k=1∞(−1)k+1k!=1−12+16−124+1120−⋯\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k!} = 1 - \frac{1}{2} + \frac{1}{6} - \frac{1}{24} + \frac{1}{120} - \cdots
Step 2: Compute the 4th partial sum S₄.
S4=1−12+16−124=24−12+4−124=1524=58S_4 = 1 - \frac{1}{2} + \frac{1}{6} - \frac{1}{24} = \frac{24 - 12 + 4 - 1}{24} = \frac{15}{24} = \frac{5}{8}
Step 3: By the Alternating Series Estimation Theorem, the absolute value of the remainder is at most the absolute value of the first omitted term, which is the k = 5 term.
∣R4∣≤∣(−1)65!∣=1120≈0.00833|R_4| \leq \left|\frac{(-1)^{6}}{5!}\right| = \frac{1}{120} \approx 0.00833
Step 4: State the conclusion: S₄ approximates the true sum with an error no larger than 1/120.
∣S−58∣≤1120\left|S - \frac{5}{8}\right| \leq \frac{1}{120}
Answer: After 4 terms, S₄ = 5/8 and the remainder satisfies |R₄| ≤ 1/120 ≈ 0.00833.

Frequently Asked Questions

What is the difference between the remainder of a series and the partial sum?
The nth partial sum Sₙ is the total you get by adding the first n terms. The remainder Rₙ is everything left over — all the terms from (n + 1) onward. Together they equal the exact sum: Sₙ + Rₙ = S. The partial sum is your approximation; the remainder is your error.
Can you find the remainder of a divergent series?
No. The remainder Rₙ = S − Sₙ requires the series to have a finite sum S. If a series diverges, neither S nor Rₙ exists. You can only discuss the remainder for convergent series.
How do you estimate the remainder when you don't know the exact sum?
Several techniques give upper bounds on |Rₙ| without knowing S. For alternating series, the Alternating Series Estimation Theorem bounds |Rₙ| by the first omitted term. For positive-term series, the Integral Test Remainder uses an improper integral to bound Rₙ. For functions represented by Taylor series, the Lagrange remainder formula provides bounds based on derivatives.

Remainder of a Series (Rₙ) vs. nth Partial Sum (Sₙ)

Remainder of a Series (Rₙ)nth Partial Sum (Sₙ)
DefinitionThe sum of all terms from index n + 1 to ∞The sum of the first n terms of the series
FormulaRₙ = S − SₙSₙ = a₁ + a₂ + ⋯ + aₙ
What it measuresThe error (how much is missing from the approximation)The approximation itself
Behavior as n → ∞Rₙ → 0 for a convergent seriesSₙ → S (the exact sum) for a convergent series
Requires convergence?Yes — Rₙ is only defined if S existsNo — Sₙ can be computed for any series

Why It Matters

The remainder appears throughout Calculus II and beyond whenever you approximate an infinite sum with finitely many terms. Bounding the remainder lets you determine how many terms you need to achieve a desired accuracy — for instance, computing e or π to a specific number of decimal places using series. Understanding the remainder is also essential for error analysis in Taylor polynomial approximations, which is a core topic on the AP Calculus BC exam.

Common Mistakes

Mistake: Confusing the remainder Rₙ with the nth term aₙ of the series.
Correction: The remainder Rₙ is the sum of all terms after the nth term (from aₙ₊₁ onward), not just a single term. For an alternating series, |Rₙ| ≤ |aₙ₊₁|, but Rₙ and aₙ₊₁ are not the same quantity.
Mistake: Trying to compute the remainder of a divergent series.
Correction: The formula Rₙ = S − Sₙ requires a finite sum S. If the series diverges, S does not exist, so the remainder is undefined. Always verify convergence before discussing the remainder.

Related Terms