Limit Comparison on Infinite Intervals

Why Compare Integrands at Infinity?

An improper integral on an infinite interval asks whether the accumulated area under an integrand remains finite as the interval grows without bound. Sometimes the integrand is complicated, but its behavior for large \(x\) resembles a simple power such as \(1/x^2\), \(1/x\), or \(x\). Comparing the two functions by their ratio turns that resemblance into a precise test.

The key is to compare the tail, not necessarily the whole interval. A finite initial portion does not affect convergence if the integrand is integrable there. What matters is whether the area from some sufficiently large point onward stays finite.

In a comparison, assume the functions are continuous and nonnegative for all sufficiently large \(x\). If needed, choose a larger starting point \(A\) so that these conditions hold on \([A,\infty)\). Provided the functions are integrable on every finite interval before \(A\), convergence or divergence is determined by the tail beginning at \(A\).

The Power-Function Models

For \(p\in\mathbb{R}\), the standard power tail is \(g(x)=1/x^p\), for \(x\geq 1\). Its integral has a sharp threshold:

For \(p\ne 1\), integrating to \(R\) gives \((R^{1-p}-1)/(1-p)\). This has a finite limit as \(R\to\infty\) exactly when \(1-p<0\), or \(p>1\). For \(p=1\), the integral is \(\ln R\), which grows without bound. Thus \(1/x^2\) is a convergent model, while \(1/x\) and \(1/\sqrt{x}\) are divergent models.

A model should reflect the leading growth or decay of the integrand. For example, if the dominant terms suggest that an integrand behaves like a constant times \(1/x^3\), use \(1/x^3\) as the comparator. The ratio tests whether this choice really captures the tail closely enough.

Limit Comparison on an Infinite Interval

The ratio need not approach \(1\). Any finite, strictly positive limit shows that each function is eventually bounded above and below by positive constant multiples of the other.

Why the Test Works

Here is a direct justification. Since \(f(x)/g(x)\to L\) with \(L>0\), the definition of a limit, applied with tolerance \(L/2\), gives a number \(B\geq A\) such that for \(x\geq B\),

Consequently,

Because \(g(x)>0\), multiplication by \(g(x)\) preserves the inequalities:

Integrating these inequalities from \(B\) to \(R\) compares the accumulated tails. If \(\int_B^\infty g(x)\,dx\) converges, the upper bound shows that the integrals of \(f\) over \([B,R]\) remain bounded as \(R\to\infty\); because \(f\geq0\), they increase to a finite limit, so the integral of \(f\) converges. If \(\int_B^\infty g(x)\,dx\) diverges, the lower bound shows that the integrals of \(f\) also grow without bound. This proves both directions of the theorem.

Worked Examples

Zero and Infinite Ratio Limits

A finite positive limit gives a two-sided comparison. A zero or infinite limit gives only a one-sided comparison, so the comparator’s classification matters. The following consequences are useful when choosing a power model:

For the first consequence, the zero limit implies \(f(x)\leq g(x)\) eventually, so convergence follows by direct comparison. For the second, the infinite limit implies \(f(x)\geq g(x)\) eventually, so divergence follows by direct comparison. In the other pairings—zero ratio with a divergent model, or infinite ratio with a convergent model—the ratio alone does not settle the question.

A Common Mistake: A Finite Ratio Is Not a Convergence Test by Itself

A common error is to find a finite ratio limit and then declare the integral convergent. For example, in the second worked example the ratio to \(1/x\) tends to \(2\). The incorrect conclusion would be: “The ratio limit is finite, so the integral converges.” The theorem does not say this. It says the two integrals have the same classification. Since \(\int_1^\infty 1/x\,dx\) diverges, the integral being tested diverges as well.

Another source of error is choosing a model with the wrong power. The useful question is not merely whether the ratio has a limit; ask whether that limit and the model’s known integral together answer the question. If a zero or infinite limit pairs with the wrong comparator classification, try a different power.

Why This Method Matters

Limit comparison replaces a difficult integral calculation with two manageable tasks: identify a simple tail model and evaluate a ratio limit. It is particularly effective when direct integration is unavailable or unnecessarily complicated. The power threshold makes the final decision quick once the model is chosen: decay faster than \(1/x\), in the precise power sense \(1/x^p\) with \(p>1\), gives a convergent model; \(1/x\) or slower power decay gives a divergent one.

The ratio must be formed using the actual functions, and its limit must be interpreted alongside the comparator’s integral. Nonnegativity is important because the comparisons concern accumulated area; cancellation from changing signs would require additional reasoning. Always verify the tail conditions, compute the limit carefully, and state the comparator’s convergence or divergence explicitly.