Limit Laws for Scalar Multiplication

Multiplying a Sequence by a Fixed Scalar

The Limit of a Sum and the Limit of a Difference describe how limits behave under addition and subtraction. Another basic operation is multiplying every term of a sequence by the same real number. If \(a_n\) approaches \(L\), it is natural to expect \(c a_n\) to approach \(cL\), where \(c\) is fixed. This is the scalar multiplication law.

The word fixed matters: \(c\) is one real number, independent of \(n\). The scalar may be positive, negative, or zero. A negative scalar reverses the signs of the terms but does not obstruct convergence; zero makes every scaled term zero. We will prove the law from the epsilon–N definition, then examine how a nonzero scalar allows convergence to be recovered in the reverse direction.

Theorem (Limit of a Scalar Multiple): Suppose \(a_n\to L\), and let \(c\in\mathbb{R}\) be fixed. Then

c a_n\longrightarrow cL.

Proof. First consider \(c=0\). Then \(c a_n=0\) and \(cL=0\) for every \(n\), so the sequence \(c a_n\) is constant with value \(0\). The theorem on constant sequences shows that \(c a_n\to cL\).

Now suppose \(c\neq0\), and let \(\varepsilon>0\). Since \(a_n\to L\), the epsilon–N definition, applied with the positive tolerance \(\varepsilon/|c|\), gives an \(N\in\mathbb{N}_0\) such that whenever \(n\geq N\),

|a_n-L|<\frac{\varepsilon}{|c|}.

For every such \(n\), the absolute-value rule for products gives

|c a_n-cL| =|c(a_n-L)| =|c|\,|a_n-L| <|c|\frac{\varepsilon}{|c|} =\varepsilon.

Thus, for every \(\varepsilon>0\), there is an \(N\) such that \(|c a_n-cL|<\varepsilon\) for all \(n\geq N\). This is exactly the definition of \(c a_n\to cL\). The proof covers positive and negative \(c\), because the estimate uses \(|c|\). \(\square\)

Worked Examples with Scalar Multiplication

Worked Example: A Negative Scalar Changes the Sign of the Limit

Let

a_n=3+\left(-\frac{2}{5}\right)^n.

The geometric ratio has absolute value \(2/5<1\), so the theorem on geometric sequences with a contracting ratio gives \((-2/5)^n\to0\). By the Limit of a Sum theorem and the convergence of constant sequences, \(a_n\to3\). Apply the Limit of a Scalar Multiple with \(c=-4\):

-4a_n\longrightarrow -4\cdot3=-12.

The terms also confirm the form of the scaled sequence:

-4a_n =-4\left(3+\left(-\frac{2}{5}\right)^n\right) =-12-4\left(-\frac{2}{5}\right)^n.

For instance, at \(n=0\), \(a_0=4\) and \(-4a_0=-16\); the displayed formula gives \(-12-4=-16\). At \(n=1\), \(a_1=3-2/5=13/5\) and \(-4a_1=-52/5\); the formula gives \(-12+8/5=-52/5\). The scaled terms need not be close to their limit at early indices. The law says they eventually become arbitrarily close to \(-12\).

Worked Example: Scaling a Sequence with a Reciprocal Error

Define

b_n=2-\frac{1}{n+1}.

The reciprocal term tends to zero. Indeed, given \(\varepsilon>0\), choose \(N\in\mathbb{N}_0\) with \(N+1>1/\varepsilon\). For \(n\geq N\), \(n+1\geq N+1\), and hence

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

Thus \(1/(n+1)\to0\), and the Limit of a Difference gives \(b_n\to2\). Taking the fixed scalar \(c=3/2\), the scalar multiplication law yields

\frac{3}{2}b_n\longrightarrow \frac{3}{2}\cdot2=3.

Substitution verifies the terms:

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

At \(n=0\), the left side is \((3/2)(1)=3/2\), and the right side is \(3-3/2=3/2\). At \(n=2\), the left side is \((3/2)(5/3)=5/2\), and the right side is \(3-3/6=5/2\). The remaining term \(3/(2(n+1))\) tends to zero, in agreement with the limit law.

Worked Example: Recovering a Limit from a Scaled Sequence

Suppose \(a_n=-3+1/(n+1)\), and let \(x_n=-2a_n\). As in the preceding example, \(1/(n+1)\to0\), so the Limit of a Difference gives \(a_n\to-3\). The scalar multiplication law then predicts \(x_n\to6\). The terms satisfy

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

For \(n=0\), \(a_0=-2\), so \(x_0=-2(-2)=4\); the formula gives \(6-2=4\). For \(n=1\), \(a_1=-5/2\), so \(x_1=5\); the formula gives \(6-2/2=5\). Since \(2/(n+1)\to0\), this expression also confirms \(x_n\to6\).

The reverse reasoning is useful too: if \(x_n=-2a_n\) is known to converge to \(6\), then \(a_n\to-3\). The theorem below justifies this conclusion in general. Its key requirement is that the scalar be nonzero.

When Can Convergence Be Recovered?

Multiplication by a nonzero scalar can be undone by multiplying by its reciprocal. This gives a converse to the scalar multiplication law: if \(c\neq0\), then convergence of \(c a_n\) implies convergence of \(a_n\). This conclusion fails for \(c=0\), because the zero multiple is the constant zero sequence regardless of the behavior of the original sequence.

Theorem (Convergence Under a Nonzero Scaling): Let \(c\in\mathbb{R}\) with \(c\neq0\), and let \((a_n)\) be a real sequence. The sequence \((a_n)\) converges if and only if \((c a_n)\) converges. Moreover, if \(c a_n\to K\), then \(a_n\to K/c\).

Proof. Suppose first that \(a_n\) converges to some \(L\). The Limit of a Scalar Multiple gives \(c a_n\to cL\), so \((c a_n)\) converges.

Conversely, suppose \(c a_n\to K\). Since \(c\neq0\), the reciprocal \(1/c\) is a real number. For every \(n\),

a_n=\frac{1}{c}(c a_n).

Apply the Limit of a Scalar Multiple to the convergent sequence \(c a_n\), using the fixed scalar \(1/c\). It follows that

a_n=\frac{1}{c}(c a_n) \longrightarrow \frac{1}{c}K =\frac{K}{c}.

This proves both directions and identifies the limit in the reverse direction. \(\square\)

The nonzero hypothesis is essential. For example, \(a_n=n\) does not converge, but \(0a_n=0\) for every \(n\), and the scaled sequence converges to \(0\). Thus a zero scalar can erase all information about the original sequence’s behavior.

Worked Example: Why the Zero Scalar Has a Different Role

Take \(a_n=n\). This is an arithmetic sequence with nonzero common difference, so the earlier convergence criterion for arithmetic sequences shows that it does not converge. Yet for \(c=0\), every scaled term is

c a_n=0\cdot n=0.

The sequence \((c a_n)\) is therefore constant and converges to \(0\). The scalar multiplication law remains valid: since \(a_n\) does not have a finite limit, its hypothesis is not met here. The example instead illustrates why the converse theorem requires \(c\neq0\). A convergent zero multiple says nothing about whether the original sequence converges.

Scope and Common Pitfalls

The scalar multiplication law concerns a single fixed number \(c\) applied at every index. It does not by itself establish a result for a varying factor \(c_n\), as in \(c_n a_n\). In that expression, both factors may change with \(n\), so a separate argument and additional hypotheses are needed. Limit laws for products address that kind of expression.

A second common mistake is to assume that a negative scalar changes convergence into divergence. The proof shows why this is incorrect: distance from the proposed limit is measured by

|c a_n-cL|=|c|\,|a_n-L|.

The sign of \(c\) does not affect this distance. If \(c<0\), the terms are reflected across zero, and the limit is reflected in the same way; the absolute error is multiplied by the positive number \(|c|\).

Finally, a limit should be scaled along with the terms. If \(a_n\to L\), the conclusion is \(c a_n\to cL\), not \(L\) unless \(cL=L\). For example, if \(a_n\to4\) and \(c=-3\), the scaled limit is \(-12\). When using the converse, divide the known limit by \(c\), retaining its sign: if \(c a_n\to K\) and \(c<0\), then the limit of \(a_n\) is still \(K/c\), which may be positive or negative depending on \(K\).

Check Your Understanding

Use the scalar multiplication law and the results proved here to answer the following questions.

If \(a_n\to-5\), what is the limit of \(7a_n\)?
Why does the proof for \(c\neq0\) use the tolerance \(\varepsilon/|c|\) rather than \(\varepsilon/c\)?
If \(-3a_n\to12\), what is the limit of \(a_n\), and which theorem justifies the conclusion?
Can convergence of \(0a_n\) imply that \(a_n\) converges? Give a reason.
What additional issue arises if the scalar is \(c_n\) and varies with the index?