Remainder Estimates

11 min read

Textbook Reference

The remainder estimate comes directly from the integral test. The same rectangle picture that proved convergence now bounds the error.

Overview

Knowing a series converges is one thing. Knowing how close your partial sum is to the true total is another. The remainder estimate answers the second question.

Once the integral test has told you a positive decreasing series converges, you can approximate the sum $S$ by adding the first $n$ terms, getting the partial sum $s_n$. The leftover, everything you did not add, is called the remainder $R_n = S - s_n$. It is the error in your approximation. The question is: how big is that error?

The answer comes free from the same rectangles you drew for the integral test. The tail $a_{n+1} + a_{n+2} + \cdots$ is a staircase that hugs the curve $y = f(x)$ out past $x = n$, so the leftover is trapped between two integrals of the tail. That gives a hard numerical bound on the error without ever knowing the true sum. You can then turn the question around and ask the practical one: how many terms must I add to be sure the error is below some target accuracy? The bound answers that too. This is the moment the abstract idea of convergence becomes a usable numerical tool.

Prerequisite Check

• Apply the integral test, including evaluating $\displaystyle\int_n^{\infty} f(x)\,dx$ (previous lesson)
• Know that the sum of a series is the limit of its partial sums (the series lesson)
• Compute a definite integral such as $\displaystyle\int_n^{\infty}\frac{1}{x^3}\,dx = \frac{1}{2n^2}$
• Solve a simple inequality such as $\dfrac{1}{2n^2} < 0.0005$ for $n$

If these are solid, you are ready.

Quick Reference

The remainder. If $S = \displaystyle\sum_{n=1}^{\infty} a_n$ and $s_n$ is the $n$th partial sum, the remainder is \[ R_n = S - s_n = a_{n+1} + a_{n+2} + a_{n+3} + \cdots, \] the error when $s_n$ is used to approximate $S$.

Remainder Estimate for the Integral Test (formal gloss). Let $f$ be continuous, positive, and decreasing for $x \ge n$ with $f(k) = a_k$, and suppose $\sum a_k$ converges. Then \[ \int_{n+1}^{\infty} f(x)\,dx \;\le\; R_n \;\le\; \int_{n}^{\infty} f(x)\,dx. \] (In plain words: the leftover tail is at least the area starting one step later and at most the area starting at $n$.)

Bounds on the true sum. Adding $s_n$ to each side, \[ s_n + \int_{n+1}^{\infty} f(x)\,dx \;\le\; S \;\le\; s_n + \int_{n}^{\infty} f(x)\,dx. \] The midpoint of this interval estimates $S$ with error at most half the interval width.

Two questions this answers.

• How accurate is $s_n$? The error is at most $\displaystyle\int_n^{\infty} f(x)\,dx$.
• How many terms for accuracy $\varepsilon$? Solve $\displaystyle\int_n^{\infty} f(x)\,dx \le \varepsilon$ for $n$.

Key Concepts

1. Where the Bound Comes From

Reuse the integral-test picture, now for the tail of the series. The remainder is \[ R_n = a_{n+1} + a_{n+2} + a_{n+3} + \cdots, \] a staircase of rectangles of width $1$ and heights $a_{n+1}, a_{n+2}, \dots$, sitting out past $x = n$.

Upper bound. Place the rectangles so each lies under the curve $y = f(x)$ on $[n, \infty)$. Their total area is then at most the area under the curve: \[ R_n \le \int_n^{\infty} f(x)\,dx. \]

Lower bound. Place the rectangles so their tops lie above the curve on $[n+1, \infty)$. Their total area is then at least the area under the curve from $n+1$: \[ R_n \ge \int_{n+1}^{\infty} f(x)\,dx. \]

Putting the two together gives the Remainder Estimate. The error you make by stopping at $s_n$ is squeezed between two integrals you can compute.

2. Bounding the Error of a Partial Sum

Example 1 (estimate the error). Approximate $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^3}$ by its first ten terms, and bound the error.

Goal. Compute $s_{10}$, then bound the remainder $R_{10}$ with the upper integral.

Work. The tenth partial sum is \[ s_{10} = 1 + \frac{1}{2^3} + \frac{1}{3^3} + \cdots + \frac{1}{10^3} \approx 1.197532. \] With $f(x) = \dfrac{1}{x^3}$, the tail integral is \[ \int_n^{\infty}\frac{1}{x^3}\,dx = \lim_{t\to\infty}\left[-\frac{1}{2x^2}\right]_n^t = \frac{1}{2n^2}. \] So the remainder satisfies \[ R_{10} \le \int_{10}^{\infty}\frac{1}{x^3}\,dx = \frac{1}{2(10)^2} = \frac{1}{200} = 0.005. \]

Answer: $s_{10} \approx 1.197532$, with error at most $0.005$.

Recap. The error bound came entirely from one integral. The true sum is somewhere in $[s_{10}, s_{10} + 0.005]$, and you never needed to know it.

(Common error: reporting the partial sum as the exact total. The partial sum $s_{10}$ is an approximation; the remainder $R_{10}$ up to $0.005$ is the part you left off.)

3. How Many Terms for a Target Accuracy

Choosing the term count to hit a target accuracy is the practical direction, and the one used most often.

Example 2 (terms needed for a tolerance). How many terms of $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^3}$ are needed to ensure the partial sum is within $0.0005$ of the true sum?

Goal. Require the error bound to be below $0.0005$ and solve for $n$.

Work. The error satisfies $R_n \le \dfrac{1}{2n^2}$, so it is enough to make \[ \frac{1}{2n^2} < 0.0005. \] Solving, \[ n^2 > \frac{1}{2(0.0005)} = 1000 \quad\Longrightarrow\quad n > \sqrt{1000} \approx 31.6. \] So $n = 32$ terms guarantee accuracy within $0.0005$.

Answer: $32$ terms.

Recap. Set the error bound below the tolerance and solve. Because $n$ must be a whole number and $\sqrt{1000} \approx 31.6$, round up to $32$.

(Common error: rounding $31.6$ down to $31$. The inequality requires $n > 31.6$, so the smallest whole number that works is $32$. Always round up when finding the number of terms.)

4. A Sharper Estimate Using Both Bounds

Adding $s_n$ to the remainder inequality traps the true sum in a narrow interval, which beats using $s_n$ alone.

Example 3 (improved estimate). Use both bounds with $n = 10$ to estimate $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^3}$.

Goal. Sandwich the true sum between $s_{10} + \int_{11}^{\infty}$ and $s_{10} + \int_{10}^{\infty}$, then take the midpoint.

Work. Using $\displaystyle\int_n^{\infty}\frac{1}{x^3}\,dx = \frac{1}{2n^2}$ and $s_{10} \approx 1.197532$, \[ s_{10} + \frac{1}{2(11)^2} \;\le\; S \;\le\; s_{10} + \frac{1}{2(10)^2}, \] which is \[ 1.201664 \;\le\; S \;\le\; 1.202532. \] Estimating $S$ by the midpoint gives \[ S \approx 1.2021, \quad \text{with error at most half the interval width, under } 0.0005. \]

Answer: $S \approx 1.2021$ with error under $0.0005$.

Recap. The plain partial sum needed $32$ terms for this accuracy (Example 2); the two-sided estimate reaches the same accuracy with only $10$ terms. Sandwiching with both integrals reaches a given accuracy with fewer terms than the one-sided bound.

Inline Self-Check

Question. For $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$, the tail integral is $\displaystyle\int_n^{\infty}\frac{1}{x^2}\,dx = \frac{1}{n}$. What is the maximum error if you approximate the sum by $s_{100}$?

Common Errors Summary

Common Misconceptions

Leveled Practice

Attempt each problem before opening the answer.

Level 1: Direct Application

Problem 1. For $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^4}$, the tail integral is $\displaystyle\int_n^{\infty}\frac{1}{x^4}\,dx = \frac{1}{3n^3}$. Bound the error in approximating the sum by $s_{10}$.

Problem 2. Using the same series $\displaystyle\sum 1/n^4$, how many terms ensure the error is below $0.00001$?

Problem 3. For $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$ (tail integral $\frac{1}{n}$), how many terms ensure the error is below $0.001$?

Level 2: Two-Sided Estimate

Problem 4. For $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^4}$ with $s_{10} \approx 1.082037$ and tail integral $\frac{1}{3n^3}$, give the two-sided bounds on the true sum using $n = 10$.

Problem 5. Explain why the two-sided estimate (using both integrals) reaches a given accuracy with fewer terms than the one-sided bound.

Level 3: Reasoning

Problem 6. A convergent $p$-series $\sum 1/n^p$ (with $p > 1$) has tail integral $\displaystyle\int_n^{\infty} x^{-p}\,dx = \dfrac{1}{(p-1)n^{p-1}}$. Find a formula for the number of terms $n$ needed so the error is below a tolerance $\varepsilon$.

Problem 7. Why does the remainder estimate require the series to converge first, and what tool establishes that?

Mastery Checklist

Mastery means doing all of the following without notes:

• Define the remainder $R_n = S - s_n$ as the error of a partial-sum approximation
• State the Remainder Estimate inequality with the correct integral limits
• Bound the error of a given partial sum with the upper integral
• Find the number of terms needed for a target accuracy, rounding up
• Use both bounds to sandwich the true sum and take the midpoint
• Explain why the estimate requires convergence and which test supplies it

Mental Model

Go back to the staircase and the curve from the integral test, but look only at the part past $x = n$, the terms you chose not to add. That leftover staircase is the error $R_n$. Slide it under the curve and its area is at most $\int_n^{\infty} f$; lift it above the curve starting one step later and its area is at least $\int_{n+1}^{\infty} f$. The leftover is pinned between those two areas.

This turns “the error is some unknown amount” into “the error is between these two numbers I can compute.” And because both numbers shrink as $n$ grows, you can dial $n$ up until the upper number drops below whatever accuracy you need. The midpoint trick is just noticing that if the true value lies in a known interval, the center of that interval is your best single guess, and you are off by at most half the interval’s width.

Connections

Built From

• The integral test: the same function $f$, the same conditions, and the same rectangle comparison are reused, now applied to the tail instead of the full series.
• Infinite series and partial sums: the remainder is the difference between the full sum and a partial sum, so the partial-sum definition is the starting point.

Leads To

• Taylor remainder and Taylor polynomials: the idea of bounding the error of a truncated sum reappears for Taylor series later in the chapter, where the remainder controls how well a polynomial approximates a function.

Use in Numerical Computation

The remainder estimate is where series convergence becomes computation. A computer cannot add infinitely many terms; it adds a finite number and stops, and the remainder estimate is exactly the guarantee that the stopping point is accurate enough. The “how many terms for accuracy $\varepsilon$” calculation is the design step behind any series-based numerical routine, from evaluating a special function to summing a physics model. The comparison between the one-sided and two-sided estimates is a first lesson in a central theme of numerical analysis: a smarter use of the same information can reach the same accuracy with far less work.

Audience Notes

For students who find math intimidating: this is two integrals and one inequality. Compute the tail integral, and you have both the error of a partial sum and the number of terms you need.

For students who want depth: the remainder estimate is the same comparison-of-areas argument as the integral test, applied to the tail. Recognizing that one picture does both jobs is the kind of structural insight that makes the chapter cohere.

For students aimed at a career in computing or engineering: the terms-needed-for-accuracy calculation is precisely how you budget a truncated series in code, and the two-sided midpoint estimate is a concrete example of squeezing more accuracy from the same number of terms.

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