Limit Laws for Products

Multiplying Two Convergent Sequences

The Limit of a Scalar Multiple concerns multiplying every term of a sequence by one fixed real number. A product of two sequences is different: both factors may vary with the index. If \(a_n\to L\) and \(b_n\to M\), we expect their termwise products \(a_nb_n\) to approach \(LM\). The challenge is to control the error when both factors change.

The key is that a convergent sequence is bounded, as established in the theorem Convergent Sequences Are Bounded. Thus one factor can be kept uniformly bounded while the other is close to its limit. We combine this fact with the Limit of a Sum and the Limit of a Scalar Multiple.

Theorem (Limit of a Product): Suppose \(a_n\to L\) and \(b_n\to M\). Then

a_nb_n\longrightarrow LM.

Proof. Since \(a_n\to L\), the theorem Convergent Sequences Are Bounded gives a real number \(K\geq0\) such that \(|a_n|\leq K\) for every \(n\). Let \(\varepsilon>0\). By the epsilon–N definition applied to \(b_n\to M\), there is an index after which

|b_n-M|<\frac{\varepsilon}{2(K+1)}.

Also, since \(a_n\to L\), there is an index after which

|a_n-L|<\frac{\varepsilon}{2(|M|+1)}.

By the result Combining Finitely Many Thresholds, we may take one index \(N\) large enough that both inequalities hold whenever \(n\geq N\). For such \(n\), write the product error as

$$ a_nb_n-LM=a_n(b_n-M)+M(a_n-L). $$

The triangle inequality and the absolute-value rule for products now give

\begin{aligned} |a_nb_n-LM| &\leq |a_n|\,|b_n-M|+|M|\,|a_n-L|\\ &\leq K|b_n-M|+|M|\,|a_n-L|\\ &<\frac{K\varepsilon}{2(K+1)} +\frac{|M|\varepsilon}{2(|M|+1)} <\varepsilon. \end{aligned}

The last inequality holds because \(K/(K+1)<1\) and \(|M|/(|M|+1)<1\); each of the two terms is therefore less than \(\varepsilon/2\). We have shown that for every \(\varepsilon>0\), there is an \(N\) such that \(|a_nb_n-LM|<\varepsilon\) whenever \(n\geq N\). Hence \(a_nb_n\to LM\). \(\square\)

The decomposition is the main technique in this proof. The term \(a_n(b_n-M)\) uses the boundedness of \(a_n\) to control the error in \(b_n\). The other term, \(M(a_n-L)\), is a fixed scalar multiple of an error that tends to zero. The added \(1\) in the denominators makes the choices valid even when \(K=0\) or \(M=0\).

Worked Examples with Product Limits

Worked Example: A Product with an Alternating Geometric Error

Define

a_n=2+\left(-\frac{1}{3}\right)^n, \qquad b_n=-1+\frac{2}{n+1}.

The geometric sequence \((-1/3)^n\) tends to zero because its ratio has absolute value less than \(1\). Also, \(2/(n+1)\to0\): given \(\varepsilon>0\), choose \(N\) with \(N+1>2/\varepsilon\); then for \(n\geq N\), \(0<2/(n+1)\leq2/(N+1)<\varepsilon\). By the Limit of a Sum and the Limit of a Difference, \(a_n\to2\) and \(b_n\to-1\). The Limit of a Product therefore gives

a_nb_n\longrightarrow 2(-1)=-2.

Expanding the terms gives another view of the result:

a_nb_n =\left(2+\left(-\frac{1}{3}\right)^n\right) \left(-1+\frac{2}{n+1}\right).

At \(n=0\), \(a_0=3\) and \(b_0=1\), so \(a_0b_0=3\). Substitution into the displayed product gives \((2+1)(-1+2)=3\). At \(n=1\), \(a_1=5/3\) and \(b_1=0\), so \(a_1b_1=0\); the product formula gives \((2-1/3)(-1+1)=0\). Early terms need not be close to \(-2\); the limit law concerns what happens for all sufficiently large indices.

Worked Example: A Bounded Oscillating Factor and a Vanishing Factor

Let

u_n=(-1)^n, \qquad v_n=\frac{1}{n+1}.

The sequence \(u_n\) does not converge: its even-indexed terms are \(1\), while its odd-indexed terms are \(-1\). It is nevertheless bounded, since \(|u_n|=1\) for every \(n\). The sequence \(v_n\) tends to zero. Indeed, given \(\varepsilon>0\), choose \(N\in\mathbb{N}_0\) such that \(N+1>1/\varepsilon\). Then for all \(n\geq N\),

|v_n|=\frac{1}{n+1}\leq\frac{1}{N+1}<\varepsilon.

Here the product is \(u_nv_n=(-1)^n/(n+1)\), and it tends to zero. This example is covered by the bounded-times-vanishing result proved below, even though the Limit of a Product does not apply: that theorem requires both sequences to converge.

The first terms verify the formula directly. At \(n=0\), \(u_0v_0=1\cdot1=1\); at \(n=1\), \(u_1v_1=(-1)(1/2)=-1/2\); and at \(n=2\), \(u_2v_2=1\cdot(1/3)=1/3\). Their absolute values are \(1/(n+1)\), which tend to zero.

Worked Example: Multiplying Two Sequences Approaching Finite Values

Consider

c_n=3+\frac{1}{n+1}, \qquad d_n=2-\frac{1}{n+1}.

Since \(1/(n+1)\to0\), the Limit of a Sum and the Limit of a Difference give \(c_n\to3\) and \(d_n\to2\). Thus \(c_nd_n\to6\) by the Limit of a Product. Expanding the product confirms the error terms:

c_nd_n =\left(3+\frac{1}{n+1}\right) \left(2-\frac{1}{n+1}\right) =6-\frac{1}{n+1}-\frac{1}{(n+1)^2}.

At \(n=0\), \(c_0=4\) and \(d_0=1\), so the product is \(4\); the expanded expression gives \(6-1-1=4\). At \(n=1\), \(c_1=7/2\) and \(d_1=3/2\), so the product is \(21/4\); the expanded expression gives \(6-1/2-1/4=21/4\). Both error terms tend to zero, in agreement with the product limit.

A Bounded Sequence Times a Sequence Tending to Zero

The preceding oscillating example illustrates a useful result that does not require both factors to converge. Boundedness alone is enough for one factor, provided the other tends to zero. This gives a flexible way to handle products whose factors have different kinds of behavior.

Theorem (Bounded Sequence Times a Null Sequence): Suppose \((u_n)\) is bounded and \(v_n\to0\). Then

u_nv_n\longrightarrow0.

Proof. Because \((u_n)\) is bounded, there is a number \(C\geq0\) such that \(|u_n|\leq C\) for every \(n\). If \(C=0\), then \(|u_n|=0\) and hence \(u_n=0\) for every \(n\). Therefore \(u_nv_n=0\) for every \(n\), so the product sequence tends to zero.

Now suppose \(C>0\), and let \(\varepsilon>0\). Since \(v_n\to0\), there is an \(N\in\mathbb{N}_0\) such that \(|v_n|<\varepsilon/C\) whenever \(n\geq N\). For every such \(n\),

|u_nv_n-0| =|u_n|\,|v_n| \leq C|v_n| <C\frac{\varepsilon}{C} =\varepsilon.

This is the epsilon–N definition of \(u_nv_n\to0\), proving the theorem. \(\square\)

The theorem applies whenever one factor stays bounded, even if that factor does not settle near a single value. In particular, it explains why multiplying the alternating sequence \((-1)^n\) by \(1/(n+1)\) produces a sequence tending to zero: the oscillation remains, but its size is reduced by a factor that becomes arbitrarily small.

Why the Hypotheses Matter

In the Limit of a Product, both sequences must converge. A convergent product does not, in general, show that either factor converges. For example, let \(x_n=(-1)^n\) and \(y_n=(-1)^n\). Neither sequence converges, since each takes the values \(1\) and \(-1\) infinitely often. But their product is

$$ x_ny_n=(-1)^n(-1)^n=1 $$

for every \(n\), so the product converges to \(1\). Thus the product law is a forward implication based on convergence of both factors; it is not a general method for recovering convergence of the factors from convergence of their product.

A second pitfall is to estimate the product error as though both factors were fixed scalars. In \(a_nb_n-LM\), the factor \(a_n\) is not fixed. The proof handles this by first using boundedness of the convergent sequence \((a_n)\). Without a bound, the product of a small error and a changing factor need not be small. For instance, \(n\cdot(1/n)=1\): the second factor tends to zero, but the first is unbounded, so the bounded-times-null theorem does not apply.

When using these results, identify precisely which factor is known to be bounded and which factor tends to zero, or check that both factors converge before applying the Limit of a Product. These distinctions prevent a valid estimate from being used outside its hypotheses.

Check Your Understanding

Use the product limit law and the bounded-times-null theorem to answer the following questions.

If \(a_n\to-2\) and \(b_n\to5\), what is the limit of \(a_nb_n\)?
In the proof of the Limit of a Product, why is it useful to know that one convergent factor is bounded?
If \((u_n)\) is bounded and \(v_n\to0\), what can be concluded about \(u_nv_n\), even if \((u_n)\) does not converge?
Does convergence of \(a_nb_n\) imply that \(a_n\) and \(b_n\) both converge? Give an example or a reason.
Why can the bounded-times-null theorem not be applied to \(u_n=n\) and \(v_n=1/n\)?

What You'll Learn