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Integral and Direct Comparison · Tutorial 607 of 998

Integral-Test Bounds for Series Remainders

Learn how areas under a decreasing function bound the error remaining after a partial sum, and how to turn those bounds into a practical estimate.

Intermediate 9 min read

What You'll Learn

  • Define the remainder after the first N terms of a convergent positive series.
  • Derive both integral bounds for that remainder using rectangles.
  • Check the indexing carefully when deciding which integral is the upper or lower bound.
  • Estimate remainders for power, logarithmic, and arctangent series.
  • Use remainder bounds to place a guaranteed interval around a partial sum.
  • Identify a common off-by-one error in applying the integral bounds.

Why Bound a Series Remainder?

A convergent series has a finite sum, but its partial sums usually do not equal that sum exactly. The difference between the full sum and a partial sum is the remainder, or error, after that partial sum. The integral test gives more than a convergence conclusion: when its hypotheses hold, areas under a decreasing function place the remainder between two explicit integrals.

These estimates are useful when a series sum is difficult to compute exactly. They tell us how far the partial sum can be from the full sum, and therefore how much confidence to place in a numerical approximation.

The Remainder and Its Integral Bounds

Definition: Suppose \(\sum_{n=1}^{\infty} a_n\) converges, and let \(S_N=\sum_{n=1}^{N}a_n\) be its \(N\)th partial sum. The remainder after the first \(N\) terms is $$ R_N=\sum_{n=N+1}^{\infty}a_n. $$ Consequently, the full sum is \(S=S_N+R_N\).

For the integral-test bounds, assume there is a function \(f\) that is continuous, positive, and decreasing on \([N,\infty)\), and that \(a_n=f(n)\) for integers \(n\ge N\). Also assume the improper integral \(\int_N^\infty f(x)\,dx\) converges. The integral test then ensures that the corresponding series tail converges.

Theorem: Under these conditions, the remainder after the first \(N\) terms satisfies $$ \int_{N+1}^{\infty} f(x)\,dx \le R_N \le \int_N^{\infty} f(x)\,dx. $$

The two integrals differ by one unit in their lower limits. That difference reflects which rectangles are being compared with the terms in the tail. In particular, the integral starting at \(N\) gives the upper bound, while the one starting at \(N+1\) gives the lower bound.

Why the Bounds Hold

Because \(f\) is decreasing, its value at the right endpoint of an interval is no greater than its values earlier in the interval. Its value at the left endpoint is no less than its values later in the interval. Comparing each term \(f(k)\) with the area of a unit-width rectangle gives the required inequalities.

1
Compare a term with the interval to its left.
For \(k\ge N+1\) and \(x\in[k-1,k]\), decrease gives \(f(x)\ge f(k)\). Integrating over this interval, whose width is one, yields \(\int_{k-1}^{k}f(x)\,dx\ge f(k)\).
2
Add these comparisons.
For any \(M\ge N+1\), summing over \(k=N+1,\ldots,M\) gives \(\int_N^M f(x)\,dx\ge\sum_{k=N+1}^{M}f(k)\).
3
Pass to the infinite tail.
As \(M\) tends to infinity, the finite sum tends to \(R_N\), and the integral tends to \(\int_N^\infty f(x)\,dx\). Hence \(R_N\le\int_N^\infty f(x)\,dx\).
4
Compare a term with the interval to its right.
For \(x\in[k,k+1]\), decrease gives \(f(x)\le f(k)\), so \(\int_k^{k+1}f(x)\,dx\le f(k)\). Adding for \(k=N+1,\ldots,M\) and taking \(M\) to infinity gives \(\int_{N+1}^\infty f(x)\,dx\le R_N\).

Both limits are valid because the improper integral converges and the positive series tail converges. This argument also shows the geometric reason for the different endpoints: the rectangles to the left of each tail term give the upper estimate, and those to the right give the lower estimate.

Worked Example: A Remainder Bound for the Sum of Reciprocals Squared

Worked Example: Bounding the Tail of \(\sum 1/n^2\)

Estimate the remainder after the first five terms of \(\sum_{n=1}^{\infty}1/n^2\), then place the full sum in a guaranteed interval.

Here \(f(x)=1/x^2\), which is positive, continuous, and decreasing for \(x\ge5\). Its improper integral converges. Since \(N=5\), the remainder is \(R_5=\sum_{n=6}^{\infty}1/n^2\). The bounds give

$$ \int_6^\infty \frac{1}{x^2}\,dx \le R_5 \le \int_5^\infty \frac{1}{x^2}\,dx. $$

An antiderivative is \(-1/x\). Evaluating the lower integral gives

$$ \int_6^\infty \frac{1}{x^2}\,dx =\lim_{b\to\infty}\left[-\frac1x\right]_6^b =\lim_{b\to\infty}\left(-\frac1b+\frac16\right) =\frac16. $$

Similarly, the upper integral is

$$ \int_5^\infty \frac{1}{x^2}\,dx =\lim_{b\to\infty}\left(-\frac1b+\frac15\right) =\frac15. $$

Therefore \(1/6\le R_5\le1/5\), or approximately \(0.166667\le R_5\le0.200000\). The partial sum is

$$ S_5=1+\frac14+\frac19+\frac1{16}+\frac1{25} =\frac{3600+900+400+225+144}{3600} =\frac{5269}{3600} \approx1.463611. $$

Since the full sum is \(S_5+R_5\), adding the remainder bounds gives

$$ \frac{5269}{3600}+\frac16 \le S \le \frac{5269}{3600}+\frac15, \qquad \frac{5869}{3600}\le S\le\frac{5989}{3600}. $$

Thus \(1.630278\le S\le1.663611\), with endpoints rounded to six decimal places. As a check, the remainder bounds came from the integrals starting at 6 and 5, respectively, so the interval for \(S\) is the partial sum plus a nonnegative error between those values.

Turning a Remainder Interval into an Estimate

If \(L\le R_N\le U\), then \(S_N+L\le S\le S_N+U\). The midpoint of this interval can be used as an estimate of \(S\); because the true sum lies in the interval, its distance from the midpoint is at most half the interval's width, \((U-L)/2\). This gives a guaranteed accuracy statement, not just a plausible numerical guess.

For the preceding example, the remainder interval has width \(1/5-1/6=1/30\). Its midpoint is \((1/6+1/5)/2=11/60\), so the midpoint estimate of the remainder has error at most \(1/60\). This conclusion follows directly from the fact that every number between the two endpoints is within half their separation of their midpoint.

Worked Example: A Logarithmic Tail

Worked Example: Bounding a Tail with a Logarithmic Factor

For the series \(\sum_{n=2}^{\infty}1/(n(\ln n)^2)\), bound the remainder after the first four terms.

Use \(f(x)=1/(x(\ln x)^2)\) for \(x\ge2\). It is positive and continuous there. Differentiating gives

$$ f'(x)=-\frac{\ln x+2}{x^2(\ln x)^3}<0 \qquad (x\ge2), $$

so \(f\) is decreasing. An antiderivative is \(-1/\ln x\), since its derivative is \(1/(x(\ln x)^2)\). The improper integral converges: its value from \(a>1\) to infinity is \(1/\ln a\), which is finite. The tail starts at \(n=5\), so

$$ \int_5^\infty\frac{dx}{x(\ln x)^2} \le R_4 \le \int_4^\infty\frac{dx}{x(\ln x)^2}. $$

Evaluating each integral gives

$$ \int_a^\infty\frac{dx}{x(\ln x)^2} =\lim_{b\to\infty}\left[-\frac1{\ln x}\right]_a^b =\lim_{b\to\infty}\left(-\frac1{\ln b}+\frac1{\ln a}\right) =\frac1{\ln a}. $$

Therefore

$$ \frac1{\ln5}\le R_4\le\frac1{\ln4}, \qquad 0.621335\le R_4\le0.721348, $$

where the decimal endpoints are rounded to six places. The lower endpoint uses \(a=5=N+1\); the upper uses \(a=4=N\), as required. The tail is positive and is confined to this interval, which checks both the direction and the interpretation of the estimate.

Worked Example: A Tail with an Arctangent Integral

Worked Example: Bounding a Tail of a Rational Series

Estimate the remainder after the first three terms of \(\sum_{n=1}^{\infty}1/(1+n^2)\).

Take \(f(x)=1/(1+x^2)\). It is positive and continuous for \(x\ge3\), and

$$ f'(x)=-\frac{2x}{(1+x^2)^2}<0 \qquad (x\ge3). $$

An antiderivative is \(\arctan x\), and \(\lim_{b\to\infty}\arctan b=\pi/2\). Thus the two integrals in the remainder bounds are

$$ \int_4^\infty\frac{dx}{1+x^2} =\frac{\pi}{2}-\arctan4 =\arctan\!\left(\frac14\right), $$

and

$$ \int_3^\infty\frac{dx}{1+x^2} =\frac{\pi}{2}-\arctan3 =\arctan\!\left(\frac13\right). $$

The identities follow because, for positive \(t\), \(\arctan t+\arctan(1/t)=\pi/2\). Therefore

$$ \arctan\!\left(\frac14\right) \le R_3\le \arctan\!\left(\frac13\right), \qquad 0.244979\le R_3\le0.321751. $$

The decimals are rounded to six places. For a check on the starting indices, \(R_3\) contains the terms with \(n\ge4\); the lower integral starts at 4 and the upper integral at 3. Both integral values are positive, and the lower value is less than the upper value, consistent with the proven bounds.

A Common Indexing Mistake

A frequent error is to use \(\int_N^\infty f(x)\,dx\) as the lower bound for the remainder after \(N\) terms. For example, for \(\sum 1/n^2\) after five terms, this would incorrectly claim

$$ R_5\ge\int_5^\infty\frac{dx}{x^2}=\frac15. $$

But the correct upper bound is \(R_5\le1/5\), not a lower bound. The lower bound is the integral starting at \(N+1=6\), which is \(1/6\). The error comes from comparing the wrong rectangles: the interval \([N,N+1]\) lies before the first omitted term \(f(N+1)\), and its area contributes to the upper estimate, not the lower one.

Takeaway: For the tail after the first \(N\) terms, write the remainder explicitly as \(\sum_{n=N+1}^\infty f(n)\). Then apply the bounds in this order: integral from \(N+1\) is the lower bound; integral from \(N\) is the upper bound.

Check Your Understanding

Use the tail definition and the integral bounds to answer each question.

  1. For a positive decreasing function with \(a_n=f(n)\), write the two integral bounds for the remainder after the first \(N\) terms.
  2. For \(\sum_{n=1}^{\infty}1/n^3\), what two integrals bound the remainder after the first four terms?
  3. Why does the integral starting at \(N\) provide an upper bound rather than a lower bound?
  4. If a remainder lies between \(0.12\) and \(0.18\), what is the midpoint estimate, and what maximum error is guaranteed for that estimate?
  5. State the continuity, positivity, monotonicity, and convergence conditions needed to apply the integral remainder bounds.

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