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Proof of the Product Limit Law

Learn how to control the error in a product and use that estimate to prove that products of convergent sequences converge to the product of their limits.

Intermediate 9 min read

What You'll Learn

  • Decompose the error between a termwise product and its proposed limit
  • Prove a quantitative estimate using boundedness of a convergent sequence
  • Choose epsilon–N thresholds that control both parts of the product error
  • Identify where the nonzero or zero values of the limits matter
  • Apply the product limit law to oscillating and vanishing sequences

Controlling the Error in a Product

The Limit of a Product was stated in “Limit Laws for Products”: if \(a_n\to L\) and \(b_n\to M\), then \(a_nb_n\to LM\). The central issue in proving this law is that the error in a product is not just the error in one factor. Both factors vary with \(n\), and their deviations from their limits interact.

A useful starting point is to subtract the proposed limit \(LM\) from the product \(a_nb_n\) and rearrange. The resulting identity separates the error into two parts. One part contains \(b_n-M\), multiplied by \(a_n\); the other contains \(a_n-L\), multiplied by the fixed number \(M\). Convergence controls the differences, while boundedness controls the varying multiplier \(a_n\).

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

For example, expanding the right-hand side gives \(a_nb_n-a_nM+Ma_n-ML=a_nb_n-LM\), as required. We will turn this decomposition into an estimate and then use it to prove the product limit law directly from the epsilon–N definition.

A Quantitative Product Error Estimate

The proof needs a bound on the sequence multiplying \(b_n-M\). Since \(a_n\to L\), the theorem Convergent Sequences Are Bounded says that \((a_n)\) is bounded. This is a result from earlier in the course; we use it here without re-proving it.

Lemma (Product Error Estimate): Suppose \((a_n)\) is bounded, with \(|a_n|\leq K\) for every \(n\), where \(K\geq0\). Let \(b_n\) be a real sequence, and let \(L,M\in\mathbb{R}\). Then, for every \(n\in\mathbb{N}_0\),
$$ |a_nb_n-LM|\leq K|b_n-M|+|M||a_n-L|. $$

Proof. Use the identity above, the triangle inequality, and the assumed bound on \(a_n\):

$$ \begin{aligned} |a_nb_n-LM| &=|a_n(b_n-M)+M(a_n-L)|\\ &\leq |a_n||b_n-M|+|M||a_n-L|\\ &\leq K|b_n-M|+|M||a_n-L|. \end{aligned} $$

The first inequality is the Triangle Inequality, established in “Proofs Involving Absolute Values”; the second uses \(|a_n|\leq K\). This proves the estimate. \(\square\)

The estimate is useful because it reduces one product error to a sum of two errors whose behavior is already understood. It also identifies why boundedness matters: even if \(b_n-M\) is small, multiplying it by an arbitrarily large \(a_n\) might not produce a small quantity. In the product limit law, convergence of \(a_n\) supplies the needed bound.

Proof of the Product Limit Law

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 sequence \((a_n)\) is bounded by Convergent Sequences Are Bounded. Choose \(K\geq0\) such that \(|a_n|\leq K\) for every \(n\). Set

$$ C=\max\{K,|M|,1\}. $$

Then \(C>0\), \(K\leq C\), and \(|M|\leq C\). Let \(\varepsilon>0\). By \(b_n\to M\), there is an index \(N_1\) such that

$$ |b_n-M|<\frac{\varepsilon}{2C} $$

whenever \(n\geq N_1\). By \(a_n\to L\), there is an index \(N_2\) such that

$$ |a_n-L|<\frac{\varepsilon}{2C} $$

whenever \(n\geq N_2\). Take \(N=\max\{N_1,N_2\}\). For every \(n\geq N\), the Product Error Estimate gives

$$ \begin{aligned} |a_nb_n-LM| &\leq K|b_n-M|+|M||a_n-L|\\ &\leq C|b_n-M|+C|a_n-L|\\ &<C\frac{\varepsilon}{2C}+C\frac{\varepsilon}{2C}\\ &=\varepsilon. \end{aligned} $$

Thus, for every \(\varepsilon>0\), there is an \(N\) such that \(|a_nb_n-LM|<\varepsilon\) whenever \(n\geq N\). By the epsilon–N definition of convergence, \(a_nb_n\to LM\). \(\square\)

The proof does not require either limit to be nonzero. If \(M=0\), the term \(|M||a_n-L|\) in the estimate is simply zero, and the other term is still controlled because \((a_n)\) is bounded. If \(L=0\), no separate argument is needed: the same proof applies. The nonzero-limit condition needed for reciprocals and quotients is not needed for products.

Worked Examples Using the Product Limit Law

Worked Example: A Product with an Oscillating Factor

Define

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

Since \(\left|(-1)^n/(n+1)\right|=1/(n+1)\to0\), the Limit of a Sum and the Limit of a Scalar Multiple give \(a_n\to2\). Also \(2/(n+1)\to0\), so \(b_n\to-3\). The Limit of a Product therefore gives

$$ a_nb_n\longrightarrow 2(-3)=-6. $$

The first terms show that the alternating sign is included in the calculation. At \(n=0\), \(a_0=2+1=3\) and \(b_0=-3+2=-1\), so \(a_0b_0=-3\). Substitution into the product expression gives \((2+1)(-3+2)=3(-1)=-3\). At \(n=1\), \(a_1=2-1/2=3/2\) and \(b_1=-3+1=-2\), so \(a_1b_1=-3\); direct substitution gives \((2-1/2)(-3+1)=(3/2)(-2)=-3\). The initial products need not equal the limiting value \(-6\).

Worked Example: A Null Sequence Times a Bounded Oscillating Sequence

Let

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

We have \(u_n\to0\), and \(v_n\) is bounded because \(|v_n|=1\) for every \(n\). The theorem Bounded Sequence Times a Null Sequence, established in “Limit Laws for Products,” says that \(u_nv_n\to0\). This does not follow from the product limit law, since \(v_n\) does not converge (so the hypothesis fails). Here the bounded-sequence result is the appropriate tool.

The terms are \(u_0v_0=1\), \(u_1v_1=-1/2\), and \(u_2v_2=1/3\), because \(u_0=1,v_0=1\); \(u_1=1/2,v_1=-1\); and \(u_2=1/3,v_2=1\). Their signs alternate, but their absolute values are \(1/(n+1)\), which tend to zero. This example illustrates a related technique: one factor need not converge if it is bounded and the other factor tends to zero.

Worked Example: A Product Whose Limit Is Zero

Consider

$$ x_n=\frac{n+2}{n+1}, \qquad y_n=\frac{3}{n+1}. $$

Since \(x_n=1+1/(n+1)\), we have \(x_n\to1\). Also \(y_n\to0\). The product limit law gives \(x_ny_n\to1\cdot0=0\). Direct multiplication gives the same sequence in the form

$$ x_ny_n=\frac{3(n+2)}{(n+1)^2}. $$

At \(n=0\), the original factors are \(x_0=2\) and \(y_0=3\), whose product is \(6\); substitution into the simplified expression gives \(3(2)/1^2=6\). At \(n=1\), the factors are \(x_1=3/2\) and \(y_1=3/2\), whose product is \(9/4\); the simplified expression gives \(3(3)/2^2=9/4\). Although the products at these indices are not close to zero, the limit law concerns all sufficiently large indices.

Worked Example: Why Boundedness Cannot Be Omitted from the Estimate

Let \(r_n=n+1\) and \(s_n=1/(n+1)\). Then \(s_n\to0\), but \(r_n\) is unbounded. The products satisfy

$$ r_ns_n=(n+1)\frac{1}{n+1}=1 $$

for every \(n\). For instance, at \(n=0\) the factors are \(1\) and \(1\), with product \(1\); at \(n=2\) they are \(3\) and \(1/3\), again with product \(1\). Thus the product does not tend to zero. This does not contradict the product limit law: \(r_n\) does not converge to a real number, so its hypotheses are not satisfied. It does show why smallness of one factor alone is not enough when the other factor can grow without bound.

How to Organize a Product-Limit Proof

When proving a limit for a product from the epsilon–N definition, the main task is to find a decomposition that separates the changing factors from their limits. The identity

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

does this by using one varying factor and one fixed limiting factor in each term. The estimate then has a clear structure: convergence makes each difference small, and boundedness keeps the varying multiplier under control.

1
Subtract the proposed limit.
Write \(a_nb_n-LM\) and rearrange it into terms involving \(b_n-M\) and \(a_n-L\).
2
Find a bound for the varying factor.
Use the fact that a convergent sequence is bounded to choose \(K\) with \(|a_n|\leq K\).
3
Control each error term.
Choose thresholds for both convergent sequences so the two contributions together are less than \(\varepsilon\).

A common mistake is to write “both factors are close to their limits, so their product is close to the product of the limits” without estimating the error. The statement is true, but the proof must account for the size of the multiplier. Another mistake is to assume the limit \(LM\) is nonzero; the proof works equally well when one or both limits are zero. The indispensable facts are convergence of both sequences and the resulting boundedness of at least one factor.

Check Your Understanding

Use the product error estimate and the product limit law to answer these questions.

  1. What identity decomposes \(a_nb_n-LM\) into terms involving \(b_n-M\) and \(a_n-L\)?
  2. Where is boundedness used in the proof of the product limit law?
  3. If \(a_n\to-2\) and \(b_n\to0\), what is the limit of \(a_nb_n\)?
  4. Why does \(u_n\to0\) alone not guarantee that \(u_nv_n\to0\) when \(v_n\) is unbounded?
  5. Can the product limit law be applied when one factor is \((-1)^n\)? Explain what additional information would be needed to analyze a product involving this sequence.

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