Integral Test Remainder

Q: What is the difference between the integral test remainder and the remainder of a series?

The remainder of a series Rₙ = S − Sₙ is the exact error, which is usually unknown. The integral test remainder provides computable upper and lower bounds for Rₙ using improper integrals. It does not give you the exact remainder but traps it between two values you can calculate.

Q: When can you use the integral test remainder?

You can use it only when the series passes the integral test. This requires a function f(x) that is continuous, positive, and decreasing on [N, ∞) with f(k) = aₖ for each integer k ≥ N. If the series converges by a different test but does not meet these conditions, you cannot apply the integral test remainder bounds.

Q: How do you use the integral test remainder to improve an estimate of the sum?

After computing Sₙ and the bounds on Rₙ, you know the true sum S satisfies Sₙ + ∫(n+1 to ∞) f(x) dx ≤ S ≤ Sₙ + ∫(n to ∞) f(x) dx. A better estimate is the midpoint of these two bounds. This technique gives a tighter approximation than the partial sum alone.

Integral Test Remainder

For a series that converges by the integral test, this is a quantity that measures how accurately the nth partial sum estimates the overall sum.

Math formula showing integral test remainder bounds: the integral from n+1 to ∞ of f(x)dx ≤ Rₙ ≤ integral from n to ∞ of f(x)dx

Key Formula

\int_{n+1}^{\infty} f(x)\,dx \;\leq\; R_n \;\leq\; \int_{n}^{\infty} f(x)\,dx

Where:

$R_n$ = The remainder (error) after the nth partial sum, defined as R_n = S - S_n where S is the total sum and S_n is the nth partial sum.
$S$ = The exact sum of the infinite series.
$S_n$ = The nth partial sum, the sum of the first n terms of the series.
$f(x)$ = A continuous, positive, decreasing function on [n, ∞) such that f(k) = a_k for each positive integer k.
$n$ = The number of terms included in the partial sum.

Worked Example

Problem: Consider the series ∑(k=1 to ∞) 1/k². Use the integral test remainder to bound the error when approximating the sum with the first 10 terms.

Step 1: Identify the function f(x) = 1/x², which is continuous, positive, and decreasing on [1, ∞). The integral test applies.

f(x) = \frac{1}{x^2}

Step 2: Compute the upper bound of the remainder by evaluating the improper integral from n = 10 to ∞.

\int_{10}^{\infty} \frac{1}{x^2}\,dx = \left[-\frac{1}{x}\right]_{10}^{\infty} = 0 - \left(-\frac{1}{10}\right) = \frac{1}{10} = 0.1

Step 3: Compute the lower bound of the remainder by evaluating the improper integral from n+1 = 11 to ∞.

\int_{11}^{\infty} \frac{1}{x^2}\,dx = \left[-\frac{1}{x}\right]_{11}^{\infty} = 0 - \left(-\frac{1}{11}\right) = \frac{1}{11} \approx 0.0909

Step 4: Apply the integral test remainder bounds to get a two-sided inequality for R₁₀.

\frac{1}{11} \leq R_{10} \leq \frac{1}{10}

Step 5: Interpret the result: the error from using only 10 terms is between approximately 0.0909 and 0.1. The actual sum is π²/6 ≈ 1.6449, and S₁₀ ≈ 1.5498, giving R₁₀ ≈ 0.0951, which indeed falls within the bounds.

0.0909 \leq 0.0951 \leq 0.1 \;\checkmark

Answer: The remainder R₁₀ satisfies 1/11 ≤ R₁₀ ≤ 1/10, so the 10-term partial sum approximates the true sum with an error between about 0.091 and 0.1.

Another Example

This example reverses the problem: instead of bounding the error for a given n, you solve for the minimum n needed to achieve a desired accuracy.

Problem: For the series ∑(k=1 to ∞) 1/k³, determine how many terms n are needed so that the partial sum Sₙ approximates the true sum with an error less than 0.001.

Step 1: Set up the upper bound of the remainder and require it to be less than 0.001.

R_n \leq \int_{n}^{\infty} \frac{1}{x^3}\,dx < 0.001

Step 2: Evaluate the improper integral.

\int_{n}^{\infty} \frac{1}{x^3}\,dx = \left[-\frac{1}{2x^2}\right]_{n}^{\infty} = \frac{1}{2n^2}

Step 3: Solve the inequality 1/(2n²) < 0.001 for n.

\frac{1}{2n^2} < 0.001 \implies 2n^2 > 1000 \implies n^2 > 500 \implies n > \sqrt{500} \approx 22.36

Step 4: Since n must be a positive integer, round up to n = 23.

n = 23

Answer: You need at least 23 terms to guarantee the error is less than 0.001.

Frequently Asked Questions

What is the difference between the integral test remainder and the remainder of a series?

The remainder of a series Rₙ = S − Sₙ is the exact error, which is usually unknown. The integral test remainder provides computable upper and lower bounds for Rₙ using improper integrals. It does not give you the exact remainder but traps it between two values you can calculate.

When can you use the integral test remainder?

You can use it only when the series passes the integral test. This requires a function f(x) that is continuous, positive, and decreasing on [N, ∞) with f(k) = aₖ for each integer k ≥ N. If the series converges by a different test but does not meet these conditions, you cannot apply the integral test remainder bounds.

How do you use the integral test remainder to improve an estimate of the sum?

After computing Sₙ and the bounds on Rₙ, you know the true sum S satisfies Sₙ + ∫(n+1 to ∞) f(x) dx ≤ S ≤ Sₙ + ∫(n to ∞) f(x) dx. A better estimate is the midpoint of these two bounds. This technique gives a tighter approximation than the partial sum alone.

Integral Test Remainder vs. Alternating Series Remainder

	Integral Test Remainder	Alternating Series Remainder
Applies to	Series with positive, decreasing terms where f(x) is continuous	Alternating series satisfying the conditions of the alternating series test
Bound type	Two-sided: lower and upper bounds via improper integrals	One-sided: \|Rₙ\| ≤ aₙ₊₁ (the next term)
Formula	∫(n+1 to ∞) f(x) dx ≤ Rₙ ≤ ∫(n to ∞) f(x) dx	\|Rₙ\| ≤ \|aₙ₊₁\|
Computation required	Evaluate an improper integral	Evaluate the (n+1)th term of the series

Why It Matters

The integral test remainder appears frequently in Calculus II (AP Calculus BC) when you study series convergence and need to estimate sums to a specified accuracy. It is also essential in numerical analysis and scientific computing, where you must know how many terms of a series to compute to guarantee a result within a given tolerance. Understanding this concept bridges the gap between knowing a series converges and being able to use it for practical calculations.

Common Mistakes

Mistake: Using ∫(n to ∞) for both the upper and lower bounds, or using ∫(n+1 to ∞) for both.

Correction: The lower bound starts at n+1 and the upper bound starts at n. The two integrals have different lower limits: ∫(n+1 to ∞) f(x) dx ≤ Rₙ ≤ ∫(n to ∞) f(x) dx. Mixing these up invalidates the estimate.

Mistake: Applying the integral test remainder to a series whose terms are not eventually positive and decreasing.

Correction: The integral test (and its remainder bounds) require f(x) to be continuous, positive, and decreasing on [n, ∞). If the terms alternate in sign or are not decreasing, use a different remainder estimate such as the alternating series remainder.

Related Terms

Integral Test — The convergence test that enables this remainder bound
Remainder of a Series — The general concept of error Rₙ = S − Sₙ
nth Partial Sum — The finite sum Sₙ whose error is being estimated
Improper Integral — Used to compute the upper and lower bounds
Convergent Series — The series must converge for the remainder to exist
Convergence Tests — Family of tests including the integral test
Series — The infinite sum being approximated
Sum — The exact value S that Sₙ + Rₙ equals