Limits of Powers

Why the Exponent Must Be Fixed

The limit laws for sums, products, and quotients let us evaluate many expressions built from convergent sequences. A power with a fixed integer exponent fits into this framework, but it is useful to establish the result carefully: the exponent determines how many factors occur, and that number must not change as the index changes.

In this tutorial, \(a_n^m\) means that the term \(a_n\) is raised to a fixed integer power \(m\). We will prove that if \(a_n\to L\), then \(a_n^m\to L^m\) for every fixed nonnegative integer \(m\). We will then handle negative integer exponents when the terms and the limit are nonzero. A changing exponent, such as \(a_n^n\), is a different kind of problem and is not covered by this limit law.

Positive Integer Powers

For a positive integer \(m\), the difference between two \(m\)-th powers has a useful factorization. It turns the error in the powered sequence into the original error \(a_n-L\), multiplied by a sum that can be bounded. The boundedness of convergent sequences, established earlier in this course, supplies the bound we need.

Theorem (Limit of a Fixed Positive Integer Power): Let \(m\) be a fixed positive integer. If \(a_n\to L\), then \(a_n^m\to L^m\). More precisely, there is a constant \(C\geq0\) such that \[ |a_n^m-L^m|\leq C|a_n-L| \] for every \(n\).

Proof. Since \(a_n\to L\), the Theorem (Convergent Sequences Are Bounded) gives a number \(B\geq0\) such that \(|a_n|\leq B\) for every \(n\). Set

K=\max\{1,B,|L|\}.

Then \(|a_n|\leq K\), \(|L|\leq K\), and \(K\geq1\). For any real numbers \(x\) and \(y\), the difference-of-powers identity is

x^m-y^m=(x-y)\sum_{j=0}^{m-1}x^{m-1-j}y^j.

To see the cancellation in this identity, expand the product on the right:

(x-y)\sum_{j=0}^{m-1}x^{m-1-j}y^j =\sum_{j=0}^{m-1}\left(x^{m-j}y^j-x^{m-1-j}y^{j+1}\right) =x^m-y^m.

Apply the identity with \(x=a_n\) and \(y=L\). Each of the \(m\) terms in the sum has absolute value at most \(K^{m-1}\), because \(|a_n|\leq K\) and \(|L|\leq K\). Therefore,

|a_n^m-L^m| \leq |a_n-L|\sum_{j=0}^{m-1}|a_n|^{m-1-j}|L|^j \leq mK^{m-1}|a_n-L|.

This gives the claimed estimate with \(C=mK^{m-1}\). Let \(\varepsilon>0\). Since \(C>0\) and \(a_n\to L\), there is an \(N\) such that whenever \(n\geq N\),

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

For every such \(n\), the estimate gives \(|a_n^m-L^m|<\varepsilon\). This is the epsilon–N definition of \(a_n^m\to L^m\). \(\square\)

The exponent \(m\) is fixed in this proof: it determines the number of terms in the sum and the constant \(C\). The factorization also explains why a small change in the base produces a small change in a fixed power when the bases remain bounded.

Worked Example: A Square with a Positive Limit

Let

a_n=3+\frac{1}{n+2}, \qquad n\in\mathbb{N}_0.

Because \(1/(n+2)\to0\), the Limit of a Sum Theorem gives \(a_n\to3\). The exponent \(2\) is a fixed positive integer, so the Limit of a Fixed Positive Integer Power Theorem gives

a_n^2\longrightarrow 3^2=9.

For a direct check of the expression at \(n=0\), \(a_0=3+1/2=7/2\), and \(a_0^2=49/4\). The claimed limit concerns the terms as \(n\) grows; it does not assert that each term equals 9.

The Zeroth and Negative Integer Powers

The zeroth power is a separate, simple case. For every nonzero real number \(x\), \(x^0=1\). Thus, whenever the sequence terms are nonzero, \(a_n^0\) is the constant sequence with value 1. Its limit is 1, regardless of the limit of \(a_n\). We do not use \(0^0\) here; the zeroth power is being applied only to nonzero terms.

A negative integer power involves a reciprocal. The reciprocal limit theorem requires a nonzero limit, so this condition cannot be omitted.

Theorem (Limit of a Fixed Negative Integer Power): Let \(m\) be a fixed positive integer. Suppose \(a_n\neq0\) for every \(n\), \(a_n\to L\), and \(L\neq0\). Then \[ a_n^{-m}\longrightarrow L^{-m}. \]

Proof. By the Theorem (Limit of a Reciprocal), since \(a_n\to L\neq0\) and every \(a_n\neq0\),

\frac{1}{a_n}\longrightarrow\frac{1}{L}.

Apply the Limit of a Fixed Positive Integer Power Theorem to the sequence \(1/a_n\), with the same fixed positive integer \(m\). It follows that

\left(\frac{1}{a_n}\right)^m \longrightarrow \left(\frac{1}{L}\right)^m.

By the definition of negative integer powers, the left side is \(a_n^{-m}\), and the right side is \(L^{-m}\). This proves the result. \(\square\)

The assumption \(L\neq0\) is essential for this argument and for the conclusion in general. For instance, if \(a_n=1/(n+1)\), then \(a_n\to0\), but \(a_n^{-1}=n+1\) does not converge to a finite real number. The condition \(a_n\neq0\) also matters because otherwise the negative power may not be defined at every index.

Worked Example: A Negative Power Near a Nonzero Limit

Consider

d_n=2+\frac{1}{n+1}, \qquad n\in\mathbb{N}_0.

Since \(1/(n+1)\to0\), we have \(d_n\to2\). Also \(d_n\geq2>0\) for every \(n\), so the terms are nonzero and the limit is nonzero. The negative-power theorem, with \(m=2\), gives

d_n^{-2}\longrightarrow2^{-2}=\frac14.

At the initial index, \(d_0=2+1=3\), so \(d_0^{-2}=1/9\), in agreement with the definition \(d_0^{-2}=1/(d_0^2)\). As the index increases, \(d_n\) approaches 2 and its reciprocal square approaches \(1/4\).

Examples with Zero and Negative Limits

The fixed-power result does not require the limit \(L\) to be positive. Positive integer powers of negative terms are handled by the same theorem, with the usual algebraic signs. It also applies when \(L=0\): every fixed positive integer power then converges to zero.

Worked Example: An Odd Power with a Negative Limit

Let

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

The term \(1/(n+1)\) tends to zero, so the Limit of a Sum Theorem gives \(b_n\to-2\). Applying the fixed-power theorem with \(m=3\),

b_n^3\longrightarrow(-2)^3=-8.

For \(n=0\), \(b_0=-2+1=-1\), and \(b_0^3=-1\). This does not conflict with the limit \(-8\); the theorem says that the cubes approach \(-8\), not that they equal \(-8\) at every index.

Worked Example: A Positive Power of a Sequence Converging to Zero

Set

c_n=\frac{1}{n+1}.

The sequence \(c_n\to0\). Using the fixed positive integer power theorem with \(m=4\), we obtain

c_n^4=\frac{1}{(n+1)^4}\longrightarrow0^4=0.

The initial term is \(c_0^4=1\), while the later terms decrease toward zero. The exponent remains 4 throughout; this is precisely the setting of the theorem.

Why a Changing Exponent Is Different

It is tempting to use the fixed-power theorem when both the base and exponent depend on \(n\). That theorem does not justify this step: it applies to one fixed integer \(m\), not to a different exponent at each index. A simple example shows that the distinction matters.

Worked Example: A Base Approaching One with a Changing Exponent

For \(n\in\mathbb{N}_0\), define

x_n=1+\frac{1}{n+1}, \qquad y_n=x_n^{n+1}.

The base satisfies \(x_n\to1\). However, the exponent \(n+1\) changes with \(n\), so the fixed-power theorem cannot be applied to conclude that \(y_n\to1\). In fact, the Lemma (A Linear Lower Bound for Powers) gives \((1+h)^k\geq1+kh\) whenever \(h>0\) and \(k\in\mathbb{N}_0\). Take \(h=1/(n+1)\) and \(k=n+1\). Then

y_n =\left(1+\frac{1}{n+1}\right)^{n+1} \geq 1+(n+1)\frac{1}{n+1} =2

for every \(n\in\mathbb{N}_0\). Thus the terms \(y_n\) are never close to 1, even though the bases \(x_n\) approach 1. This estimate does not determine the limit of \(y_n\); it shows why the fixed-power result cannot be used as though the exponent were constant.

There is a related distinction between fixed integer powers and powers with non-integer exponents. This tutorial has established results for integer exponents, for which powers are defined by multiplication and reciprocals. Powers involving roots require additional results and will be considered next. When evaluating a limit, first identify whether the exponent is fixed and whether it is an integer; those details determine which result is available.

Check the exponent.
A fixed positive integer exponent is covered by the factorization estimate; a changing exponent is not.

Check the type of power.
The zeroth power is 1 for nonzero terms. A negative integer power requires nonzero terms and a nonzero limit.

Apply the limit to the base.
For a fixed positive integer \(m\), if \(a_n\to L\), then \(a_n^m\to L^m\), whether \(L\) is positive, negative, or zero.

Check Your Understanding

Use the fixed-power results and their hypotheses to answer these questions.

If \(a_n\to-3\), what is the limit of \(a_n^4\), and which theorem justifies the conclusion?
Why is boundedness of \(a_n\) useful in the factorization proof for positive integer powers?
What additional conditions are needed to conclude \(a_n^{-2}\to L^{-2}\)?
If \(a_n\to0\), what is the limit of \(a_n^5\)? Does the fixed-power theorem apply to \(a_n^{-5}\) under these conditions?
Why can the fixed-power theorem not be applied directly to \(\left(1+\frac{1}{n+1}\right)^{n+1}\)?

What You'll Learn