3

UPPER AND LOWER LIMITS OF SEQUENCES OF REAL NUMBERS

3.1GENERALIZATION OF THE LIMIT CONCEPT

In calculus it was often necessary to compute the limit of a particular sequence, and occasionally a sequence having no limit at all was encountered. For example, consider the sequence

The sequence {x_n} has no limit, but the subsequence x₁, x₄, x₇, … converges to 1, the subsequence x₂, x₅, x₈,… converges to 0, and the subsequence x₃, x₆, x₉, … converges to –1. When we encounter sequences having no limit, it will be useful for us to work with the largest of all limits of subsequences (+1 in the above example) and the smallest of all subsequential limits (–1 above).

3.1.1Definitions and Comments

Let {x_n} be a sequence of real numbers; by Theorem 2.3.4, {x_n} has at least one convergent subsequence, if +∞ and –∞ are allowed as limits. Thus the set S of all subsequential limits is a nonempty subset of the extended reals . It follows that S has a sup and an inf. For clearly ∞ is an upper bound of S; if ∞ is not the least upper bound, then S has a real upper bound, and consequently S has a sup by Theorem 2.4.2. The argument that S has an inf is similar.

We define the upper limit of the sequence {x_n} as sup S and the lower limit of {x_n} as inf S.

The standard notation for the upper limit is

and the lower limit is denoted by

In fact we can show that there are subsequences converging to sup S and inf S, and therefore sup S and inf S actually belong to S. Thus, lim sup x_n is the largest element of S, in other words, the largest subsequential limit, and lim inf x_n is the smallest element of S, the smallest subsequential limit.

To see this, let s = sup S; we must show that s ∈ S. We know that there is a sequence of points s₁, s₂,… in S such that s_k → s as k → ∞ (see Theorem 2.4.3). Since s_k ∈ S we have a subsequence x_k1, x_k2, … of {x_n} converging to s_k. We summarize the relevant information as follows:

First assume that s is finite. For each n pick m = m_n such that |x_nj – s_n| < 1/n for all j ≥ m_n. Then consider the “diagonal sequence” (a useful description, although we do not necessarily go down the diagonal of the above array) x_1m1 x_2m2, x_3m3,.… We adjust the indices m_n so as to obtain a true subsequence; in other words, if, for example, x_1m1 = x₇, x_2m2 = x₆, we increase m₂ so that x_2m2 becomes x_r for some r > 7. Now

hence,

Therefore, x_nmn → s, so s ∈ S. Since s = sup S, it follows that s is the largest element of S.

Now assume s = ∞. (If s = –∞, then S must consist of –∞ alone; therefore s ∈ S. Alternatively, a direct argument similar to the s = ∞ case can be made). If ∞ ∈ S we are finished, so assume ∞ ∉ S. As in Theorem 2.4.3, we can obtain a sequence of finite numbers s_n ∈ S with s_n → ∞. If the subsequence {x_nmn} is constructed as above, then x_nmn > s_n – 1/n, so that x_nmn → ∞. Thus ∞ ∈ S, as desired.

The above argument shows that if s₁, s₂, … ∈ S and s_n → t then t ∈ S; thus S is a closed set in .

Upper and lower limits form a natural generalization of the basic limit concept, as the next result shows.

3.1.2THEOREM. Suppose {x_n} is a sequence of real numbers. If lim_n→∞ x_n = x, then lim sup x_n = lim inf x_n = x. Conversely, if lim sup x_n = lim inf x_n = x, then x_n → x.

    Proof. If x_n → x, then all subsequences converge to x also, and hence S = {x}. Therefore, lim sup x_n = lim inf x_n = x. Conversely, suppose x_n does not converge to x. Assume first that x is finite. The statement x_n → x means that for every ∈ > 0 there is a positive integer N such that for all n ≥ N we have |x_n – x| < ∈. Symbolically,

We know how to find the negation of statement of this type (see Section 1.4.5).

The statement x_n x means

In other words, there is an ∈ > 0 such that for all N we can find an n ≥ N with |x_n – x| ≥ ∈. Set N = 1 and find n = n₁ such that |x_n1 – x | ≥ ∈. Then set N = n₁ + 1 and find n = n₂ ≥ N such that |x_n2 – x| ≥ ∈. Proceeding in this fashion, we obtain a sequence x_n1, x_n2,… with |x_nj – x| ≥ ∈ for all j. This sequence has a convergent subsequence by Theorem 2.3.4, and the limit must be outside the interval (x – ∈, x + ∈). But if lim sup x_n = lim inf x_n = x, we must have S = {x }, a contradiction.

Now if x = ∞, we write the definition of x_n → ∞:

thus, x_n ∞ means

In other words, there is a positive real number M such that for every positive integer N we can find n ≥ N with x_n ≤ M. As above, we find a subsequence with a limit that is less than or equal to M, contradicting the fact that S = {∞}. If x = –∞, the argument is similar. ■

Problems for Section 3.1

    1.Let S = {–3, 2, 4, 10}. Construct a sequence {x_n} of real numbers such that the set of subsequential limits of {x_n} is precisely S.

    2.Suppose x_n+1 = f(n)x_n, where f(n) → 3 as n → ∞. What are the possible values of lim sup x_n and lim inf x_n ? Does {x_n} always converge?

    3.Let c be a positive real number. Show that lim inf(cx_n) = c lim inf x_n and lim sup(cx_n) = c lim sup x_n.

    4.Let c be a negative real number. Show that lim inf(cx_n) = c lim sup x_n and lim sup(cx_n) = c lim inf x_n.

3.2SOME PROPERTIES OF UPPER AND LOWER LIMITS

Before looking at further properties of upper and lower limits, we need some additional terminology.

3.2.1Definitions and Comments

Let P₁, P₂,… be a sequence of statements (for example, P_n might be “x_n < 3,” where {x_n} is a sequence of real numbers). We say that P_n holds eventually (sometimes abreviated ev.) if P_n holds for all but finitely many n. For example, if P_n holds for all n ≥ 100, then P_n holds ev. For n < 100, P_n might be true for some n and false for others, but it doesn’t matter; once we get to n = 100, P_n is true from that point on. We say that P_n holds infinitely often (sometimes abbreviated i .o.) if P_n is true for infinitely many n. For example, if P_n is true whenever n is even, then P_n holds i.o., regardless of what happens when n is odd. Note that if P_n holds eventually, then P_n holds infinitely often, but not conversely.

If is the negation of P_n, then the negation of “P_n holds infinitely often” is “P_n holds for only finitely many n”; that is, “ holds for all but finitely many n”; in other words, “ holds eventually.” If we replace P_n by , we see that the negation of “P_n holds eventually” is “ holds infinitely often.” These ideas will now be used in establishing basic properties of upper and lower limits.

3.2.2THEOREM. Let {x_n} be a sequence of real numbers, and assume that

lim sup x_n = x, lim inf x_n = y.

        (a)If z > x, then x_n < z eventually

        (b)If z < x, then x_n > z infinitely often.

Furthermore, (a) and (b) characterize the lim sup; that is, if x′ is a number in satisfying (a) and (b), then x′ = x. Similarly:

        (c)If z < y, then x_n > z eventually

        (d)If z > y, then x_n < z infinitely often.

If y′ is a number in satisfying (c) and (d), then y′ = y.

        Proof

        (a)If “x_n < z ev.” is false, then by Section 3.2.1, x_n ≥ z i.o., and, hence, by Theorem 2.3.5 there is a subsequence coverging to a limit that is at least z. Thus, x = lim sup x_n ≥ z.

        (b)If “x_n > z i.o.” is false, then by Section 3.2.1, x_n ≤ z ev., and therefore all subsequential limits are less than or equal to z. Thus, x = lim sup x_n ≤ z. (Alternatively, there is a subsequence x_nk → x so that if z < x then, for all sufficiently large k, x_nk > z.)

        (c)If “x_n > z ev.” is false, then x_n ≤ z i.o., and it follows as in (a) that y ≤ z.

        (d)If “x_n < z i.o.” is false, then x_n ≥ z ev., and therefore, as in (b), y ≥ z. (Alternatively, there is a subsequence x_nk → y so that if z > y then x_nk < z for all sufficiently large k.)

Now suppose x′ satisfies (a) and (b). Assume (without loss of generality) that x < t < x′. Since t < x′, (b) gives x_n > t i.o.; since t > x, (a) gives x_n < t ev. This is a contradiction. Similarly, if y′ satisfies (c) and (d) and y < u < y′, then by (c), x_n > u ev., and by (d), x_n < u i.o., a contradiction. ■

The last part of the proof is a typical “uniqueness argument.”

Theorem 3.2.2 may be used to show that lim sup x_n is really the “limit of a sup,” and similarly for lim inf x_n.

3.2.3COROLLARY

    Proof. As n increases, we are taking the sup of a smaller set in evaluating sup_{k ≥ n} x_k, so by Theorem 2.4.6, sup_{k ≥ n} x_k decreases to a limit x. (For example, if , then

Similarly, inf_{k ≥ n} x_k will increase to a limit y. If z > x, then eventually, sup_{k ≥ n} x_k < z, so x_k < z for all k ≥ n. Thus, x_n < z ev. If z < x, then for all n, sup_{k ≥ n} x_k > z, so for all n there is, by Theorem 2.4.3, a positive integer k ≥ n such that x_k > z. In other words, x_n > z i.o. By Theorem 3.2.2, x must be lim sup x_n.

If z < y, then inf_{k ≥ n}x_k > z ev., so x_k > z for all k ≥ n. Thus, x_n > z ev. If z > y, then inf_{k ≥ n} x_k < z for all n, so for all n there is, by Theorem 2.4.3, a positive integer k ≥ n such that x_n < z. Thus, x_n < z i.o. By Theorem 3.2.2, y must be lim inf x_n. ■

3.2.4Remark

In proving Corollary 3.2.3, we saw that sup_{k ≤ n} x_k decreases and inf_{k ≥n} x_k increases as n increases. Thus, we may write

Problems for Section 3.2

    1.Let {x_n} and {y_n} be sequences of real numbers. Show that if x_n ≤ y_n for all n then lim inf x_n ≤ lim inf y_n and lim sup x_n ≤ lim sup y_n.

    2.Show that lim sup(x_n + y_n) ≤ lim sup x_n + lim sup y_n and

            (assuming in each case that the right-hand side is not of the form +∞ – ∞ or – ∞ + ∞).

In Section 3.3, we study convergence of power series, so it will be useful to look at a few problems involving finite and infinite series of real numbers.

    3.Let a₁,…, a_n, b₁,…, b_n be real numbers. Prove the Cauchy-Schwarz inequality

            [Hint:

    4.Use Problem 3 to show that if a_n ≥ 0 for all n and converges, then converges. Can you obtain this result without using Problem 3?

    5.In R^ρ, define , where x = (x₁,…, x_n). Then the Euclidean metric is given by d(x, y) = |x – y|. Use Problem 3 to show that d is actually a metric.

    6.Express the statement “x_n > 3 eventually” using existential and universal quantifiers. Then use the technique of Section 1.4.5 to take the negation and verify that the result is “x_n ≤ 3 infinitely often.”

3.3CONVERGENCE OF POWER SERIES

An important application of upper and lower limits is to the problem of convergence of power series. (A power series is an expression of the form .) The natural domain for the discussion is the complex plane C rather than the set R of real numbers. One reason is that, given any differentiable function ƒ defined on an open subset V of C and a point z₀ ∈ V, there is a power series representation valid in a neighborhood of z₀· The corresponding result for real-valued functions is false. However, no knowledge of complex variable theory is needed here; we only use the absolute value (magnitude) of a complex number; that is, |a + ib| = (a² + b²)^1//2. You can assume if you like that all numbers in the discussion to follow are real, and no harm will be done. Also, for convenience, we assume z₀ = 0; this amounts to making a change of variable w = z – z₀, so no generality is lost.

Let’s begin with an example. Suppose we are given the power series ; for which values of z does the series converge? (Converge of the series means convergence of the sequence of partial sums . Also, in the discussion to follow, “convergence” always means “convergence to a finite limit”.) A technique from calculus called the ratio test works in most “practical” examples. We find the limit of the absolute value of the ratio of the (n + l)st term of the series to the nth term:

If the limit is less than 1, that is, |z| < 2, the series converges. If the limit is greater than 1, that is, |z| > 2, the series diverges. If the limit is 1, that is, |z| = 2, the test gives no information.

In general, the limit involved in the ratio test might not exist, so a different test is used.

3.3.1THEOREM (Root Test)

Let {a_n} be a sequence of real numbers, and define

If a < 1, the series ∑a_n converges; in fact, it converges absolutely (that is, the series ∑|a_n| converges). If a > 1, the series diverges; if a = 1, the test gives no information

    Proof. If a < 1, pick b such that a < b < 1. By Theorem 3.2.2(a), |a_n|^1/n < b eventually, so |a_n| < bⁿ eventually. Thus, ∑|a_n| converges by comparison with a geometric series. Now absolute convergence implies convergence, for if then as n, m → ∞. (Recall from calculus that convergence does not imply absolute convergence; the standard example is ).

If a > 1, pick b such that a > b > 1. By Theorem 3.2.2(b), |a_n|^1/n > b infinitely often, or |a_n| > bⁿ infinitely often; consequently, a_n cannot approach 0 as n → ∞. The series therefore diverges, for if Σ_na_n converges to s, then .

To show that a = 1 provides no information, consider the divergent series ∑1/n and the convergent series ∑ 1/n². Since (l/n^r)^1/n → 1 as n → ∞ for any r > 0 (take logarithms to verify this), a = 1 in both cases. (The root test works for complex numbers also; the proof is the same.) ■

Now consider the power series . We look at the series of absolute values , which brings the problem back to real variables. We have the following result.

3.3.2THEOREM. Let a₀ = lim sup |c_n |^1/n, and let r = 1/a₀ (take r = ∞ if a₀ = 0). If |z| < r, the power series Σc_nzⁿ converges absolutely, and if |z| > r the series diverges.

    Proof. We apply the root test to Σ|c_n||z|ⁿ; we compute

The result follows from Theorem 3.3.1. ■

The number r defined in Theorem 3.3.2 is called the radius of convergence of the series. The series converges inside the circle of radius r and center at 0 (in the real case, we have instead an interval of length r) and diverges outside the circle. The case r = ∞ is not uncommon; for example, the familiar expansions , cos z = l – z²/2! + z⁴/4! – …, and sin z = z – z³/3! + z⁵/5! – … converge for all z.

Problems for Section 3.3

    1.Consider the series

            Verify that the ratio test cannot be applied, but the root test yields convergence.

    2.Find the radius of convergence of each of the following power series:

        (a)Σn^kzⁿ, k a positive integer

        (b)

    3.Suppose the power series coefficients c_n have the property that |c_n|^1/n ≥ 3 infinitely often. What can be said about the radius of convergence of the series?

    4.A series that is convergent but not absolutely convergent can be rearranged so as not to converge to the same sum. For example, if

            show that

    5.Continuing Problem 4, a conditionally but not absolutely convergent series can be rearranged to converge to any given real number r, or to diverge to + ∞, or to diverge to –∞, or simply to diverge (not to ±∞). Can you describe a strategy for accomplishing this?

    6.Suppose converges to f(x) for –1 < x < 1, and let s_n = c₀ + …+ c_n, s_–1 = 0. Assume s_n → s (finite) as n → ∞.

        (a)Show that for |x| < 1.

        (b)Given ∈ > 0, choose N so that |s – s_n| < ∈/2 for all n > N. Show that

        (c)Prove Abel’s Theorem: If , where converges, then

REVIEW PROBLEMS FOR CHAPTER 3

    1.Give an example of

        (a)a sequence of real numbers such that lim sup x_n ≠ lim inf x_n.

        (b)a sequence {x_n} of real numbers and an open interval I such that x_n ∈ I infinitely often, but x_n is not eventually in I.

    2.Give an example of a power series whose radius of convergence is 2.

    3.True or false: the radius of convergence of a power series is always strictly greater than zero.

    4.True or false, and explain briefly ({x_n } is a sequence of real numbers throughout).

        (a)If lim sup x_n = 5, then x_n > 4 eventually.

        (b)If lim sup x_n = 5, then x_n > 4 infinitely often.

        (c)If lim inf x_n = 5, thenjt„ > 4 eventually.

        (d)If x_n > 4 eventually, then lim inf x_n > 4.

    5.Give an example of a sequence of real numbers whose set of subsequential limits is S = {–∞, ∞}.

    6.Let

Let A be the set of all real numbers x such that x ∈ A_n infinitely often, and let B be the set of real numbers x such that x ∈ A_n eventually. Find A and B.