Why Compare Ratios?
Direct comparison can establish convergence or divergence when one integrand is bounded above or below by a familiar one. Sometimes, however, two integrands have similar endpoint behavior without an immediately useful inequality. Their ratio can reveal that similarity.
If the ratio of two nonnegative integrands approaches a fixed positive number near an improper endpoint, then neither integrand is becoming negligible relative to the other. The ratio supplies upper and lower bounds, and those bounds transfer convergence behavior from one integral to the other.
Improper Integrals and the Ratio Limit
An integral is improper at an endpoint when the interval is unbounded or the integrand is unbounded there. For a nonnegative function \(f\) on \((a,b]\) that may be unbounded at the finite endpoint \(a\), its improper integral is defined by
provided the limit exists and is finite. For a nonnegative function on \([a,\infty)\), the corresponding definition is
If the limit is finite, the improper integral converges; if the nonnegative truncated integrals grow without bound, it diverges. The same ideas apply at a finite right endpoint, using a limit from the left.
The Limit Comparison Theorem
The essential point is that a ratio close to \(L\) gives two inequalities. For example, once the ratio is between \(L/2\) and \(3L/2\), multiplying by the positive quantity \(g(x)\) gives
The constants need not be these particular choices. They are convenient because \(L\) is positive and finite, so both are positive and finite.
Why the Theorem Works
Here is the complete argument for an endpoint approached from the right, such as \(x\to a^+\). The proof for \(x\to b^-\) or \(x\to\infty\) uses the same inequalities, integrated over the corresponding truncated intervals.
Since \(f(x)/g(x)\to L\) and \(L>0\), the definition of the limit gives a point \(c>a\) such that, for \(a<x<c\), $$ \left|\frac{f(x)}{g(x)}-L\right|<\frac{L}{2}. $$ Thus \(L/2<f(x)/g(x)<3L/2\). Because \(g(x)>0\), multiplication preserves the inequalities: $$ \frac{L}{2}g(x)\le f(x)\le \frac{3L}{2}g(x). $$
For any \(t\) with \(a<t<c\), integrate the pointwise inequalities from \(t\) to \(c\): $$ \frac{L}{2}\int_t^c g(x)\,dx \le \int_t^c f(x)\,dx \le \frac{3L}{2}\int_t^c g(x)\,dx. $$ The integrals from \(c\) to any fixed interior point are finite by the hypotheses, so they cannot change whether the endpoint contribution is finite or unbounded.
If the improper integral of \(g\) converges, its truncated integrals near \(a\) remain bounded. The upper inequality then bounds the truncated integrals of \(f\), so the integral of nonnegative \(f\) converges. Conversely, rearranging the lower inequality gives $$ g(x)\le\frac{2}{L}f(x). $$ If the integral of \(f\) converges, this upper bound shows that the integral of \(g\) converges as well. Therefore, one converges exactly when the other does; equivalently, they diverge together.
Nonnegativity matters in this argument: it lets the inequalities transfer bounds on the truncated integrals directly. The theorem as stated is for nonnegative integrands, not for arbitrary functions whose positive and negative contributions may cancel.
Worked Examples
Worked Example: Comparing Two Integrands at a Finite Endpoint
Determine whether $$ \int_0^1 \frac{1+x}{\sqrt{x}}\,dx $$ converges by comparing its integrand with \(g(x)=1/\sqrt{x}\).
Both functions are positive for \(0<x\le1\). Calculate their ratio: $$ \frac{(1+x)/\sqrt{x}}{1/\sqrt{x}}=1+x, \qquad \lim_{x\to0^+}(1+x)=1. $$ The ratio limit is finite and positive, so the theorem says the two integrals have the same convergence behavior.
To check that behavior directly, evaluate the comparison integral using a truncated lower endpoint: $$ \int_t^1 \frac{1}{\sqrt{x}}\,dx = \left[2\sqrt{x}\right]_t^1 = 2-2\sqrt{t}. $$ Taking the limit gives \(\lim_{t\to0^+}(2-2\sqrt{t})=2\), which is finite. Therefore the original integral converges.
Its value can also be checked: $$ \int_t^1\frac{1+x}{\sqrt{x}}\,dx = \int_t^1(x^{-1/2}+x^{1/2})\,dx = \left[2\sqrt{x}+\frac{2}{3}x^{3/2}\right]_t^1. $$ As \(t\to0^+\), this tends to \(2+2/3=8/3\). Differentiating the antiderivative gives \(x^{-1/2}+x^{1/2}=(1+x)/\sqrt{x}\), confirming the computation.
Worked Example: Comparing Integrands on an Infinite Interval
Determine whether $$ \int_1^\infty \frac{2x+1}{x^2+1}\,dx $$ converges by comparing its integrand with \(g(x)=1/x\).
For \(x\ge1\), both functions are positive, and $$ \frac{(2x+1)/(x^2+1)}{1/x} = \frac{2x^2+x}{x^2+1} = \frac{2+1/x}{1+1/x^2}. $$ As \(x\to\infty\), \(1/x\to0\) and \(1/x^2\to0\), so this ratio tends to \(2\). The limit comparison theorem applies.
The comparison integral diverges: $$ \int_1^R\frac{1}{x}\,dx = [\ln x]_1^R = \ln R, $$ and \(\ln R\to\infty\) as \(R\to\infty\). The theorem therefore gives divergence of the original integral.
For a direct check, an antiderivative of the original integrand is \(\ln(x^2+1)+\arctan x\), since its derivative is $$ \frac{2x}{x^2+1}+\frac{1}{1+x^2} = \frac{2x+1}{x^2+1}. $$ Thus $$ \int_1^R\frac{2x+1}{x^2+1}\,dx = \ln(R^2+1)-\ln 2+\arctan R-\frac{\pi}{4}. $$ As \(R\to\infty\), the logarithmic term tends to infinity and the arctangent term remains bounded. The integral diverges, as the theorem predicted.
Worked Example: Using the Theorem to Establish Convergence
Consider $$ \int_1^\infty \frac{x+1}{x^3}\,dx $$ and compare its integrand with \(g(x)=1/x^2\).
For \(x\ge1\), both functions are positive. Their ratio is $$ \frac{(x+1)/x^3}{1/x^2} = \frac{x+1}{x} = 1+\frac{1}{x}, $$ which tends to \(1\) as \(x\to\infty\). The theorem gives the same convergence behavior for both integrals.
The comparison integral converges: $$ \int_1^R\frac{1}{x^2}\,dx = \left[-\frac{1}{x}\right]_1^R = 1-\frac{1}{R}, $$ and its limit is \(1\). Hence the integral of \((x+1)/x^3\) converges too.
In this case, direct evaluation confirms the result: $$ \int_1^R\frac{x+1}{x^3}\,dx = \int_1^R(x^{-2}+x^{-3})\,dx = \left[-\frac{1}{x}-\frac{1}{2x^2}\right]_1^R = \frac{3}{2}-\frac{1}{R}-\frac{1}{2R^2}. $$ The limit is \(3/2\). Differentiating \(-1/x-1/(2x^2)\) gives \(1/x^2+1/x^3=(x+1)/x^3\), checking the antiderivative.
A Common Mistake: Treating Any Finite Limit as Enough
The required ratio limit is not merely finite: it must be strictly positive. A limit of zero does not meet the theorem’s hypotheses. For example, on \([1,\infty)\), let \(f(x)=1/x^2\) and \(g(x)=1/x\). Then
It would be incorrect to conclude that the integrals have the same behavior merely because this limit is finite. In fact, $$ \int_1^\infty\frac{1}{x^2}\,dx=1 \qquad\text{but}\qquad \int_1^\infty\frac{1}{x}\,dx=\infty. $$ The first converges and the second diverges. The positive lower bound used in the theorem’s proof is unavailable when the ratio tends to zero.
How to Apply the Theorem Reliably
The practical task is to select a comparison function \(g\) whose improper integral is already understood, then compute \(f/g\). Before drawing a conclusion, check each condition:
- The two functions are being compared at the same improper endpoint.
- They are nonnegative near that endpoint, and the comparison function \(g\) is positive there.
- The ratio limit exists and lies strictly between zero and infinity.
- The convergence or divergence of the comparison integral is known.
Once those checks hold, no exact evaluation of the more complicated integral is needed. The ratio limit supplies two-sided constant bounds, and comparison transfers convergence behavior in both directions. This is the feature that distinguishes limit comparison from a one-sided estimate: neither integrand can be substantially smaller than the other near the endpoint.
Check Your Understanding
For each question, use the theorem’s hypotheses carefully and state what conclusion follows.
- What conditions on \(L\) and \(g\) are needed when \(f(x)/g(x)\to L\) at an improper endpoint?
- Explain why a finite positive ratio limit produces both an upper and a lower bound for \(f\) in terms of \(g\).
- If the integral of \(g\) converges and \(f/g\to L\) with \(0<L<\infty\), what does the theorem imply about the integral of \(f\)?
- Why does a ratio limit of zero not justify using this theorem?
- If the comparison integral diverges and the ratio limit is finite and positive, what can be concluded about the other integral?
