When a Limit Comparison Does Not Decide the Integral
An improper integral can be difficult to classify directly, so it is natural to compare its integrand with a familiar model. The limit comparison test is decisive when the ratio tends to a finite, positive number. But a ratio tending to zero or infinity, or having no limit at all, does not automatically settle the question.
The useful next step is to ask what pointwise inequality the ratio does provide. That inequality may prove one direction of comparison, or it may suggest a better model that does resolve convergence.
What the Ratio Can Establish
Consider nonnegative functions \(f\) and \(g\) on an interval extending to an improper endpoint. For an infinite interval, suppose \(f(x)\ge 0\) and \(g(x)>0\) for all sufficiently large \(x\). For a finite endpoint, use the corresponding conditions sufficiently close to that endpoint. The same reasoning applies in either setting.
If \(f(x)/g(x)\to 0\), then \(f(x)\le g(x)\) sufficiently close to the endpoint. If \(f(x)/g(x)\to\infty\), then \(f(x)\ge g(x)\) sufficiently close to it. These inequalities give only one direction of comparison.
| Ratio behavior | What the comparison can establish | What remains undecided |
|---|---|---|
| \(f/g\to 0\) | If \(\int g\) converges, then \(\int f\) converges. | If \(\int g\) diverges, the smaller function \(f\) may converge or diverge. |
| \(f/g\to\infty\) | If \(\int g\) diverges, then \(\int f\) diverges. | If \(\int g\) converges, the larger function \(f\) may converge or diverge. |
| \(f/g\) has no limit | Useful upper or lower bounds on the ratio may still give a comparison. | The absence of a limit alone gives no convergence conclusion. |
In particular, these cases differ from the standard limit comparison result: a finite positive limit gives two-sided bounds by constant multiples of \(g\), so the two integrals have the same convergence behavior. A zero or infinite limit generally gives only a one-sided bound.
Why the One-Sided Conclusions Are Valid
Here is the reasoning for the case \(f/g\to 0\). Since \(f/g\) tends to zero, there is a point \(X\) beyond which \(0\le f(x)/g(x)\le 1\). Because \(g(x)>0\), this implies \(0\le f(x)\le g(x)\) there.
for every \(T>X\). If the improper integral of \(g\) converges, the right-hand side remains bounded as \(T\) increases. The increasing truncated integrals of the nonnegative function \(f\) are therefore bounded, so they have a finite limit. Thus \(\int_X^\infty f(x)\,dx\) converges. The part of the integral over any finite interval away from the improper endpoint does not affect this conclusion, provided it is proper and finite.
When \(f/g\to\infty\), the same argument gives \(f(x)\ge g(x)\) sufficiently far out. If \(\int g\) diverges, then the truncated integrals of \(f\) are at least as large as those of \(g\), and hence are unbounded. Therefore \(\int f\) diverges. At a finite singular endpoint, take truncated integrals approaching that endpoint and use the same inequalities.
Notice what the proof does not say. The inequality \(f\le g\) cannot force \(f\) to diverge when \(g\) diverges; a smaller nonnegative function may still have a finite integral. Likewise, \(f\ge g\) cannot force \(f\) to converge just because \(g\) does.
Worked Example: A Zero Ratio with a Divergent Comparator
Worked Example: Find a Better Model After a Zero Limit
Consider \(\int_e^\infty \frac{1}{x(\ln x)^2}\,dx\). First compare its integrand \(f(x)=1/[x(\ln x)^2]\) with \(g(x)=1/x\). The ratio is
But \(\int_e^\infty 1/x\,dx\) diverges, so this comparison alone is inconclusive. Instead, choose \(h(x)=1/[x(\ln x)^{3/2}]\). For \(x\ge e\), \(\ln x\ge 1\), and
The comparison integral converges. With \(u=\ln x\), so that \(du=dx/x\),
Thus \(0\le f\le h\) and the integral of \(h\) is finite, so the integral of \(f\) converges. As a check, an antiderivative of \(f\) is \(-1/\ln x\), since
Consequently, the original integral equals \(\lim_{R\to\infty}(1-1/\ln R)=1\), confirming convergence.
Worked Example: An Infinite Ratio with a Convergent Comparator
Worked Example: Use a Lower Bound to Prove Divergence
Compare \(f(x)=1/\sqrt{x}\) with \(g(x)=1/x^2\) on \([1,\infty)\). Their ratio is
The comparator \(g\) is integrable:
That fact does not decide whether the larger function \(f\) is integrable. Use the divergent model \(h(x)=1/x\) instead. For \(x\ge1\), \(\sqrt{x}\le x\), so \(1/\sqrt{x}\ge1/x\). Therefore,
As \(R\to\infty\), \(\ln R\to\infty\), so the integral of \(f\) diverges. Direct evaluation confirms this: \(\int_1^R x^{-1/2}\,dx=2(\sqrt{R}-1)\to\infty\). The lower bound by \(1/x\) correctly establishes divergence.
When the Ratio Has No Limit
A ratio without a limit is not automatically useless. It may still be bounded in a way that gives the comparison needed. If \(f\) and \(g\) are nonnegative and \(g>0\) near the endpoint, an eventual upper bound \(f/g\le M\), where \(M\) is a fixed positive constant, implies \(f\le Mg\). Thus convergence of \(\int g\) implies convergence of \(\int f\). An eventual lower bound \(f/g\ge m>0\) implies \(f\ge mg\), so divergence of \(\int g\) implies divergence of \(\int f\).
If both bounds hold, \(m\le f/g\le M\), the two integrals have the same convergence behavior. This can happen even when the ratio oscillates and has no limit.
Worked Example: An Oscillating Ratio with Useful Bounds
Consider \(f(x)=(2+\sin x)/x^2\) on \([1,\infty)\), and take \(g(x)=1/x^2\). Then
This ratio has no limit: at \(x_n=\pi/2+2\pi n\) its value is \(3\), while at \(y_n=3\pi/2+2\pi n\) its value is \(1\); both sequences tend to infinity. Nevertheless, \(-1\le\sin x\le1\) gives
The upper comparison is integrable because \(\int_1^\infty 3/x^2\,dx=3\). Hence \(\int_1^\infty f(x)\,dx\) converges. The ratio's failure to have a limit does not prevent a conclusive comparison: its uniform upper bound is enough.
Signed Functions Require Extra Care
The direct comparison conclusions above require nonnegative integrands. If \(f\) changes sign, an inequality involving \(f\) does not by itself control cancellation in its integral, and a signed ratio may not be suitable for the nonnegative comparison test. To establish absolute convergence, compare \(|f|\) with a nonnegative integrable function. Absolute convergence means that \(\int |f(x)|\,dx\) converges; it guarantees convergence of the signed improper integral as well.
Worked Example: Compare the Absolute Value
Consider the signed integrand \(f(x)=\sin x/x^2\) on \([1,\infty)\). The ratio \(f(x)/(1/x^2)=\sin x\) has no limit, so use an absolute-value bound instead. Since \(|\sin x|\le1\),
The integral of the upper bound is
Therefore, \(\int_1^\infty |\sin x|/x^2\,dx\le1\), so the signed integral converges absolutely. This argument establishes absolute convergence; it does not infer convergence from the signed ratio alone.
A Reliable Strategy
When a limit comparison is inconclusive, keep the question narrow: what does the ratio imply as an inequality, and is that the direction needed? Then choose a model whose known behavior and inequality point the right way.
Confirm that the functions being compared are nonnegative and that the denominator is positive near the improper endpoint. For signed integrands, decide whether the goal is absolute convergence and compare absolute values.
A ratio tending to zero gives an eventual upper bound; a ratio tending to infinity gives an eventual lower bound. A ratio with no limit may still have a useful uniform upper or positive lower bound.
For convergence, seek an integrable upper bound. For divergence of a nonnegative integrand, seek a divergent lower bound.
If the original comparator has the wrong convergence behavior, choose another familiar model, such as a power or a power with a logarithmic factor, and verify the pointwise inequality.
Common Mistake: Treating a Zero Ratio as Proof of Convergence
Consider \(f(x)=1/(x\ln x)\) and \(g(x)=1/x\) on \([e,\infty)\). The ratio \(f/g=1/\ln x\) tends to zero, and \(\int_e^\infty g(x)\,dx\) diverges. It is incorrect to conclude that \(\int f\) converges merely because \(f/g\to0\).
In fact, direct integration shows the opposite:
The correction is to use the ratio only for the conclusion it supports: \(f\le g\) eventually, which does not settle convergence when \(g\) diverges. A different comparison or a direct evaluation is needed.
Why These Cases Matter
Recognizing an inconclusive comparison prevents an invalid conclusion and helps focus the next step. The goal is not to force every ratio into a limit-comparison test. It is to obtain a valid upper bound for convergence or a valid lower bound for divergence, while respecting nonnegativity and the endpoint conditions.
Check Your Understanding
For each question, identify what the comparison establishes and what additional information may be needed.
- If \(f,g\ge0\), \(f/g\to0\), and \(\int g\) converges, what can you conclude about \(\int f\)?
- If \(f,g\ge0\), \(f/g\to0\), and \(\int g\) diverges, does the ratio decide whether \(\int f\) converges?
- Suppose \(f/g\) has no limit but \(0\le f/g\le 4\) eventually. Which convergence conclusion follows from convergence of \(\int g\)?
- What pointwise inequality is useful for proving that a nonnegative improper integral diverges?
- Why can comparing a signed function directly with a nonnegative function be insufficient, and what should be compared to prove absolute convergence?