Deriving the Integral Test with Rectangle Bounds

From Rectangles to a Convergence Test

A decreasing function has a useful geometric property: on each interval between consecutive integers, its values lie between its values at the two endpoints. The area under its graph can therefore be bounded by rectangles whose heights come from the sequence of endpoint values.

Adding those rectangle bounds links the partial sums of a series to areas under the function. The resulting inequalities explain why, under the integral test's hypotheses, the series and the corresponding improper integral either both converge or both diverge.

Setup: Let \(f\) be positive, continuous, and nonincreasing on \([N,\infty)\), where \(N\) is a positive integer. Set \(a_n=f(n)\). The integral test compares \(\sum_{n=N}^{\infty}a_n\) with the improper integral \(\int_N^\infty f(x)\,dx\), defined as \(\lim_{m\to\infty}\int_N^m f(x)\,dx\), when this limit exists or is infinite.

The Rectangle Bounds on One Interval

Fix an integer \(k\ge N\). If \(x\) lies between \(k\) and \(k+1\), then \(k\le x\le k+1\). Since \(f\) is nonincreasing, its value at the left endpoint is at least its value at any point in the interval, and its value at the right endpoint is at most that value. Thus

f(k+1)\le f(x)\le f(k) \qquad (k\le x\le k+1).

The interval has width \(1\). Integrating the inequalities over it gives

f(k+1)\le \int_k^{k+1}f(x)\,dx\le f(k).

Geometrically, the rectangle of height \(f(k+1)\) lies below the graph, while the rectangle of height \(f(k)\) lies above it. The area under the curve is between their areas. Nonincreasing, rather than strictly decreasing, is enough: equality may occur, but the bounds still hold.

Adding the Bounds Across Many Intervals

To compare a finite portion of the series with an integral, apply the one-interval inequality for \(k=N,N+1,\ldots,m-1\), where \(m>N\), and add the resulting inequalities. The integral pieces combine to give the integral from \(N\) to \(m\). The function values on the two sides are shifted by one index:

\sum_{k=N+1}^{m} f(k) \le \int_N^m f(x)\,dx \le \sum_{k=N}^{m-1} f(k).

These are the finite rectangle bounds behind the test. The left-hand sum uses the right endpoints of the unit intervals; the right-hand sum uses the left endpoints. The different starting and ending indices are essential: the integral from \(N\) to \(m\) covers the intervals \([N,N+1]\) through \([m-1,m]\).

Finite-sum takeaway: For every integer \(m>N\), the integral over \([N,m]\) is no larger than the sum of left-endpoint heights and no smaller than the sum of right-endpoint heights.

Why the Bounds Prove Matching Convergence

We now show that the series and integral have the same convergence behavior. Because \(f(x)>0\), the integral \(\int_N^m f(x)\,dx\) increases as \(m\) increases. Because every \(f(n)>0\), the partial sums of the series increase as well. For either increasing quantity, convergence to a finite value follows if its values remain bounded above.

Suppose the integral converges.
From the lower rectangle bound, the partial sum after its first term satisfies \(\sum_{k=N+1}^{m}f(k)\le\int_N^m f(x)\,dx\). Therefore the full partial sum is at most \(f(N)+\int_N^\infty f(x)\,dx\), a finite bound.

Conclude that the series converges.
The series partial sums increase because their terms are positive, and Step 1 bounds them above. Hence they approach a finite limit.

Suppose instead that the series converges.
The upper rectangle bound gives \(\int_N^m f(x)\,dx\le\sum_{k=N}^{m-1}f(k)\). The right-hand side is bounded above by the full series sum, so the increasing integrals are bounded above.

Conclude that the integral converges.
The integrals \(\int_N^m f(x)\,dx\) increase with \(m\) and are bounded above. Their limit is finite, which is precisely convergence of the improper integral.

This argument also establishes the divergence conclusion. If the integral is infinite, the series cannot converge, since convergence of the series would imply convergence of the integral by Steps 3 and 4. If the series diverges, the integral cannot converge, since convergence of the integral would imply convergence of the series by Steps 1 and 2.

Theorem: If \(f\) is positive, continuous, and nonincreasing on \([N,\infty)\), with \(N\) an integer, then \(\sum_{n=N}^{\infty}f(n)\) converges if and only if \(\int_N^\infty f(x)\,dx\) converges. If one diverges, the other diverges.

The theorem concerns matching convergence behavior, not equality of the sum and the integral. The finite bounds use sums with slightly different endpoints, and they do not say that an infinite series equals an improper integral.

Worked Examples

Worked Example: A Convergent Series from \(f(x)=1/x^2\)

Consider \(\sum_{n=1}^{\infty}1/n^2\). Let \(f(x)=1/x^2\) for \(x\ge1\). It is positive and continuous, and \(f'(x)=-2/x^3<0\), so it is decreasing. Its antiderivative is \(-1/x\), since \(\frac{d}{dx}(-1/x)=1/x^2\). For an integer \(m>1\),

\int_1^m\frac{1}{x^2}\,dx =\left[-\frac{1}{x}\right]_1^m =1-\frac{1}{m}.

The rectangle bounds give \(\sum_{k=2}^{m}1/k^2\le 1-1/m\). Including the first term,

\sum_{k=1}^{m}\frac{1}{k^2} \le 1+1-\frac{1}{m} =2-\frac{1}{m} <2.

Thus the positive partial sums are bounded above by \(2\), so the series converges. As a check, the evaluated integral is consistent with its antiderivative, and the finite sum is bounded exactly as the right-endpoint rectangle inequality requires. No value for the sum is asserted by this argument.

Worked Example: A Divergent Series from \(f(x)=1/\sqrt{x}\)

Consider \(\sum_{n=1}^{\infty}1/\sqrt{n}\). Let \(f(x)=1/\sqrt{x}\) for \(x\ge1\). This function is positive and continuous, and \(f'(x)=-1/(2x^{3/2})<0\), so it is decreasing. The antiderivative \(2\sqrt{x}\) is correct because its derivative is \(1/\sqrt{x}\). Therefore

\int_1^m\frac{1}{\sqrt{x}}\,dx =\left[2\sqrt{x}\right]_1^m =2\sqrt{m}-2.

The upper rectangle bound says that this integral is at most the left-endpoint sum:

2\sqrt{m}-2 \le \sum_{k=1}^{m-1}\frac{1}{\sqrt{k}}.

As \(m\) increases, \(2\sqrt{m}-2\) increases without bound. The partial sums on the right must therefore also be unbounded, so the series diverges. This checks the divergence conclusion directly through the finite inequality: a series whose partial sums stayed bounded could not dominate these increasingly large integral values.

Worked Example: A Convergent Series from \(f(x)=1/(1+x^2)\)

Consider \(\sum_{n=1}^{\infty}1/(1+n^2)\). Define \(f(x)=1/(1+x^2)\) for \(x\ge1\). It is positive and continuous, and

f'(x)=\frac{-2x}{(1+x^2)^2}<0 \qquad (x\ge1),

so it is decreasing. An antiderivative is \(\arctan x\), since \(\frac{d}{dx}\arctan x=1/(1+x^2)\). Thus

\int_1^m\frac{1}{1+x^2}\,dx =\arctan(m)-\arctan(1) =\arctan(m)-\frac{\pi}{4}.

As \(m\to\infty\), this tends to \(\pi/2-\pi/4=\pi/4\), a finite value. The lower rectangle bound then gives

\sum_{k=2}^{m}\frac{1}{1+k^2} \le \int_1^m\frac{1}{1+x^2}\,dx <\frac{\pi}{4}.

Adding the first term, \(f(1)=1/2\), bounds every such partial sum by \(1/2+\pi/4\). The series converges. The antiderivative check and the finite endpoint inequality confirm each step; again, this establishes convergence, not equality between the sum and \(\pi/4\).

A Common Endpoint Mistake

A frequent error is to reverse the endpoint heights for a decreasing function. For example, on \([1,2]\), take \(f(x)=1/x\). Since \(f(2)=1/2\), an incorrect argument might claim that the area under the curve is at most the right-endpoint rectangle area \(1/2\): \(\int_1^2 1/x\,dx\le1/2\). But the integrand is at least \(1/2\) throughout the interval and strictly greater before \(x=2\). Hence its integral is greater than \(1/2\); indeed, it equals \(\ln 2\).

The right-endpoint rectangle is below the graph, not above it. The correct bounds are

\frac{1}{2} \le \int_1^2\frac{1}{x}\,dx =\ln 2 \le 1.

On each unit interval, the right endpoint gives the lower rectangle and the left endpoint gives the upper rectangle. Keeping that orientation straight prevents the corresponding finite-sum inequalities from being reversed.

What the Comparison Does—and Does Not—Say

The integral test is useful because it can replace a difficult question about infinitely many terms with a question about an improper integral. The proof works because monotonicity provides a reliable area bound on every unit interval, and positivity ensures that the partial sums and accumulated areas increase.

The starting index does not have to be \(1\). If the hypotheses hold from some integer \(N\) onward, the proof applies starting at \(N\). Adding or removing finitely many terms does not change whether a series converges: a finite initial sum changes partial sums by only a fixed amount. The same principle allows an integral comparison to begin at any suitable finite endpoint.

These bounds do not establish the integral test if the function is not positive and nonincreasing on the range being used. Nor do they compare the series sum and integral as if they were equal. Their conclusion is specifically about whether each has a finite limit. When applying the test, the hypotheses and the endpoint indices must match the inequalities being used.

Check Your Understanding

Use the rectangle bounds and their convergence argument to answer these questions.

For \(k\le x\le k+1\), which of \(f(k)\) and \(f(k+1)\) is the upper rectangle height when \(f\) is nonincreasing?
Write the finite bounds relating \(\int_N^m f(x)\,dx\) to sums of endpoint values.
Why does a finite improper integral imply that the positive series partial sums are bounded?
Why does convergence of the positive series imply that the integrals \(\int_N^m f(x)\,dx\) are bounded?
Do the rectangle bounds imply that the series sum equals the improper integral? Explain briefly.