Products and Quotients of Asymptotic Equivalents

Why Equivalent Terms Can Be Simplified

When studying a series with positive terms, an exact formula can sometimes be more complicated than the behavior that determines convergence. Asymptotic equivalence lets us replace a sequence by a simpler one when their ratio approaches $1$. This tutorial develops the rules that allow such replacements inside products, quotients, and fixed powers.

The key restriction is that the rules concern sequences that are positive for all sufficiently large indices. Positivity ensures that quotients and real powers are defined and that ratios can be compared without sign complications.

Definition: Let $u_n$ and $v_n$ be positive for all sufficiently large $n$. They are asymptotically equivalent, written $u_n\sim v_n$, if $$ \lim_{n\to\infty}\frac{u_n}{v_n}=1. $$

The notation $u_n\sim v_n$ does not mean the terms are equal. It says that their relative difference becomes negligible: the ratio $u_n/v_n$ approaches $1$. For instance, if $u_n\sim 3/n$, then $u_n$ can often be replaced by $3/n$ when determining the behavior of a positive series.

Rules for Products, Quotients, and Powers

Suppose $u_n\sim a_n$ and $v_n\sim b_n$, where all four sequences are positive for sufficiently large $n$. Their ratios satisfy $$ \frac{u_n}{a_n}\longrightarrow 1 \qquad\text{and}\qquad \frac{v_n}{b_n}\longrightarrow 1. $$ The ordinary limit laws now give rules for combining these equivalents.

Theorem: If $u_n\sim a_n$ and $v_n\sim b_n$, with all sequences positive eventually, then $$ u_nv_n\sim a_nb_n \qquad\text{and}\qquad \frac{u_n}{v_n}\sim\frac{a_n}{b_n}. $$ If $u_n\sim a_n$ and the sequences are positive eventually, then for every fixed real number $\alpha$, $$ u_n^\alpha\sim a_n^\alpha. $$

The quotient rule requires the denominator sequence to be nonzero eventually; the stated eventual positivity guarantees this. In the power rule, $\alpha$ must be fixed: it cannot change with $n$. Positive bases are required when $\alpha$ is an arbitrary real number.

Why the Rules Work

Each rule follows by writing the ratio that defines the claimed equivalence and expressing it in terms of ratios already known to approach $1$.

Product.
Divide the product by its proposed equivalent and factor the result into two known ratios: $$ \frac{u_nv_n}{a_nb_n} = \frac{u_n}{a_n}\frac{v_n}{b_n} \longrightarrow 1\cdot 1=1. $$ Therefore $u_nv_n\sim a_nb_n$.

Quotient.
Divide the quotient by its proposed equivalent: $$ \frac{u_n/v_n}{a_n/b_n} = \frac{u_n/a_n}{v_n/b_n} \longrightarrow \frac{1}{1}=1. $$ The denominator ratio is nonzero for all sufficiently large $n$, because it tends to $1$. Therefore $u_n/v_n\sim a_n/b_n$.

Fixed power.
For positive terms, $$ \frac{u_n^\alpha}{a_n^\alpha} = \left(\frac{u_n}{a_n}\right)^\alpha \longrightarrow 1^\alpha=1. $$ Here the last limit uses continuity of the function $x\mapsto x^\alpha$ at $x=1$, which holds for every fixed real $\alpha$ on positive inputs.

These calculations prove the rules directly from the definition of asymptotic equivalence. They also show why the hypotheses matter: each proof uses positive ratios, and the quotient proof needs a nonzero denominator.

The product rule can be applied repeatedly. For a fixed finite number of positive sequences, if $u_{j,n}\sim a_{j,n}$ for each $j$, then $$ \prod_{j=1}^{m}u_{j,n}\sim\prod_{j=1}^{m}a_{j,n}. $$ This follows by applying the two-factor product rule successively. The number $m$ must be fixed; this statement alone does not justify a product with a number of factors that grows with $n$.

Using Equivalents to Study Positive Series

For positive sequences $c_n$ and $d_n$, if $$ \lim_{n\to\infty}\frac{c_n}{d_n}=L \qquad\text{with}\qquad 0<L<\infty, $$ then the limit comparison test says that $\sum c_n$ and $\sum d_n$ either both converge or both diverge. In particular, $c_n\sim d_n$ gives $L=1$, so the two series have the same behavior. Any finite number of initial terms does not affect convergence.

Worked Example: Simplifying a Product of Rational Factors

Consider the positive terms $$ c_n= \frac{n+2}{n^2+1} \cdot \frac{3n^2+1}{n^3+4n}, \qquad n\ge 1. $$ We find an equivalent for each factor. For the first, $$ \frac{\dfrac{n+2}{n^2+1}}{\dfrac{1}{n}} = \frac{n(n+2)}{n^2+1} = \frac{n^2+2n}{n^2+1} \longrightarrow 1. $$ Thus $\dfrac{n+2}{n^2+1}\sim\dfrac{1}{n}$. For the second, $$ \frac{\dfrac{3n^2+1}{n^3+4n}}{\dfrac{3}{n}} = \frac{n(3n^2+1)}{3(n^3+4n)} = \frac{3n^3+n}{3n^3+12n} \longrightarrow 1. $$ Thus $\dfrac{3n^2+1}{n^3+4n}\sim\dfrac{3}{n}$. By the product rule, $$ c_n\sim \frac{1}{n}\cdot\frac{3}{n}=\frac{3}{n^2}. $$ The check is the ratio of the original term to the proposed equivalent: $$ \frac{c_n}{3/n^2} = \frac{n(n+2)}{n^2+1} \cdot \frac{3n^3+n}{3n^3+12n} \longrightarrow 1\cdot1=1. $$ Therefore the equivalence is verified. Since the terms are positive and $\sum 3/n^2$ converges, limit comparison shows that $\sum c_n$ converges.

Worked Example: A Quotient Whose Terms Do Not Approach Zero

Let $$ c_n= \frac{\dfrac{2n+1}{n^2+3}}{\dfrac{n+4}{n^3+1}}, \qquad n\ge 1. $$ For the numerator, $$ \frac{\dfrac{2n+1}{n^2+3}}{\dfrac{2}{n}} = \frac{n(2n+1)}{2(n^2+3)} = \frac{2n^2+n}{2n^2+6} \longrightarrow 1. $$ For the denominator, $$ \frac{\dfrac{n+4}{n^3+1}}{\dfrac{1}{n^2}} = \frac{n^2(n+4)}{n^3+1} = \frac{n^3+4n^2}{n^3+1} \longrightarrow 1. $$ The quotient rule therefore gives $$ c_n\sim \frac{2/n}{1/n^2}=2n. $$ To check, divide $c_n$ by $2n$. The result is the ratio of the two factor ratios just computed: $$ \frac{c_n}{2n} = \frac{\dfrac{(2n+1)/(n^2+3)}{2/n}} {\dfrac{(n+4)/(n^3+1)}{1/n^2}} \longrightarrow \frac{1}{1}=1. $$ Thus $c_n/(2n)\to1$, and since $2n\to\infty$, the terms $c_n$ do not approach zero. The series $\sum c_n$ diverges by the term test.

Worked Example: Applying a Fixed Fractional Power

For $n\ge1$, define the positive base $$ b_n=\frac{n+1}{n^2+2n} $$ and the series term $c_n=b_n^{3/2}$. First compare the base with $1/n$: $$ \frac{b_n}{1/n} = \frac{n(n+1)}{n^2+2n} = \frac{n^2+n}{n^2+2n} = \frac{n+1}{n+2} \longrightarrow 1. $$ Hence $b_n\sim 1/n$. The exponent $3/2$ is fixed, so the power rule gives $$ c_n=b_n^{3/2}\sim \left(\frac{1}{n}\right)^{3/2} =\frac{1}{n^{3/2}}. $$ The direct check is $$ \frac{c_n}{1/n^{3/2}} = \left(\frac{b_n}{1/n}\right)^{3/2} = \left(\frac{n+1}{n+2}\right)^{3/2} \longrightarrow 1^{3/2}=1. $$ Thus the proposed equivalent is correct. The comparison series $\sum 1/n^{3/2}$ converges because its exponent is greater than $1$. Limit comparison then shows that $\sum c_n$ converges.

A Common Pitfall: Letting the Power Vary

The power rule is for a fixed exponent. It is not valid to assume the same conclusion when the exponent depends on $n$. For example, $1+1/n\sim1$, but it does not follow that $$ \left(1+\frac{1}{n}\right)^n\sim 1^n=1. $$ The exponent here is $n$, not a fixed real number. In fact, the binomial theorem gives $$ \left(1+\frac{1}{n}\right)^n \ge 1+n\left(\frac{1}{n}\right)=2 $$ for every positive integer $n$, so these terms cannot approach $1$. The fixed-power theorem cannot be used to obtain the incorrect conclusion.

Another useful discipline is to preserve the conditions of the quotient rule. Before replacing numerator and denominator by equivalents, verify that the sequences involved are positive eventually and that the denominator is nonzero eventually. If either condition fails, the quotient argument above does not apply as stated.

A Reliable Simplification Process

When an expression for a positive series term contains several factors, use the rules in a deliberate order. Identify a simple equivalent for each factor, keep track of constant factors, and then combine the results. For a quotient, simplify the numerator and denominator separately before taking their quotient. For a power, verify that its exponent is fixed.

Check the domain conditions.
Confirm that the sequences are positive eventually and that any denominator is nonzero eventually.

Find factor equivalents.
For each factor, divide by the proposed simpler expression and verify that the ratio tends to $1$.

Combine using the rules.
Apply the product, quotient, or fixed-power rule only after its hypotheses have been checked.

Use the equivalent for the series question.
If the resulting comparison sequence is positive and the ratio tends to a finite positive number, apply limit comparison.

These rules matter because they turn complicated positive terms into simpler comparison terms without losing their series behavior. The equivalence itself must still be checked by a ratio limit; visual resemblance or matching leading pieces is not a substitute for that verification.

Check Your Understanding

For each question, identify the relevant condition or rule before calculating.

If $u_n\sim a_n$ and $v_n\sim b_n$, what ratio should be evaluated to verify $u_nv_n\sim a_nb_n$?
What positivity and denominator conditions are needed to use the quotient rule?
If $u_n\sim a_n$ and both sequences are positive eventually, does $u_n^{-2}\sim a_n^{-2}$? Explain which rule applies.
Suppose $c_n\sim 5/n^2$ and $c_n$ is positive eventually. What does limit comparison say about $\sum c_n$?
Why does the fixed-power rule not justify replacing $(1+1/n)^n$ by $1$?