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For that reason only the sufficiencyassertion is of interest.By hypothesis the set of values of the sequence {xn } is bounded above and hencehas a least upper bound s = supn∈N xn .By definition of the least upper bound, for every ε > 0 there exists an elementxN ∈ {xn } such that s − ε < xN ≤ s. Since the sequence {xn } is nondecreasing, wenow find that s − ε < xN ≤ xn ≤ s for all n > N . That is, |s − xn | = s − xn < ε.Thus we have proved that limn→∞ xn = s.Of course an analogous theorem can be stated and proved for a nonincreasingsequence that is bounded below.

In this case limn→∞ xn = infn∈N xn .Remark The boundedness from above (resp. below) of a nondecreasing (resp. nonincreasing) sequence is obviously equivalent to the boundedness of that sequence.Let us consider some useful examples.Example 11 limn→∞nqnProof Indeed, if xn =limn→∞ (1 +1 1n)q= 0 if q > 1.nqn ,then xn+1 == limn→∞ (1 +1n)·n+1nq xnfor n ∈ N. Since limn→∞limn→∞ q1=1·1q=1qn+1nq=< 1, there exists anindex N such that n+1nq < 1 for n > N . Thus we shall have xn+1 < xn for n > N , sothat the sequence will be monotonically decreasing from index N on.

As one cansee from the definition of a limit, a finite set of terms of a sequence has no effect theconvergence of a sequence or its limit, so that it now suffices to find the limit of thesequence xN +1 > xN +2 > · · · .The terms of this sequence are positive, that is, the sequence is bounded below.Therefore it has a limit.Let x = limn→∞ xn . It now follows from the relation xn+1 = n+1nq xn thatx = lim (xn+1 ) = limn→∞n→∞n+1n+11xn = lim· lim xn = x,n→∞ nqn→∞nqqfrom which we find (1 − q1 )x = 0, and so x = 0.Corollary 1limn→∞√nn = 1.Proof By what was just proved, for a given ε > 0 there exists N√ ∈ N such that1 ≤ n < (1 + ε)√n for all n > N . Then for n > N we obtain 1 ≤ n n < 1 + ε andhence limn→∞ n n = 1.3.1 The Limit of a Sequence89Corollary 2lim√nn→∞a = 1 for any a > 0.Proof Assume first that a ≥ 1.

For any ε > 0 √there exists N ∈ N such that 1 ≤ a <nalln>N,andwethenhave1≤a < 1 + ε for all n > N , which says(1 + ε)n for√limn→∞ n a = 1.For 0 < a < 1, we have 1 < a1 , and thenlimn→∞Example 12 limn→∞. . . · n.qnn!√1na = lim =n→∞n1a1limn→∞ = 1.n1a= 0; here q is any real number, n ∈ N, and n! := 1 · 2 ·Proof If q = 0, the assertion is obvious. Further, since | qn! | = |q|n! , it suffices toprove the assertion for q > 0. Reasoning as in Example 11, we remark that xn+1 =qn+1 xn . Since the set of natural numbers is not bounded above, there exists an indexq< 1 for all n > N . Then for n > N we shall have xn+1 < xn ,N such that 0 < n+1and since the terms of the sequence are positive, one can now guarantee that thelimit limn→∞ xn = x exists.

But thennx = lim xn+1 = limn→∞n→∞nqqxn = lim· lim xn = 0 · x = 0.n→∞ n + 1 n→∞n+1c. The Number eExample 13 Let us prove that the limit limn→∞ (1 + n1 )n exists.In this case the limit is a number denoted by the letter e, after Euler. This numberis just as central to analysis as the number 1 to arithmetic or π to geometry. We shallrevisit it many times for a wide variety of reasons.We begin by verifying the following inequality, sometimes called Jacob Bernoulli’s inequality:3(1 + α)n ≥ 1 + nαfor n ∈ N and α > −1.Proof The assertion is true for n = 1.

If it holds for n ∈ N, then it must also hold forn + 1, since we then have(1 + α)n+1 = (1 + α)(1 + α)n ≥ (1 + α)(1 + nα) == 1 + (n + 1)α + nα 2 ≥ 1 + (n + 1)α.3 Jacob (James) Bernoulli (1654–1705) – Swiss mathematician, a member of the famous Bernoullifamily of scholars. He was one of the founders of the calculus of variations and probability theory.903LimitsBy the principle of induction the assertion is true for all n ∈ N.Incidentally, the computation shows that strict inequality holds if α = 0 andn > 1.We now show that the sequence yn = (1 + n1 )n+1 is decreasing.Proof Let n ≥ 2.

Using Bernoulli’s inequality, we find thatn1 n)(1 + n−1yn−11nn2nn=1+≥==·12n2ynn −1 n+1(1 + n )n+1 (n − 1) n + 1n1nn≥ 1+ 2> 1+= 1.n n+1n −1 n+1Since the terms of the sequence are positive, the limit limn→∞ (1 + n1 )n+1 exists.But we then havelimn→∞11+nnn+1 1 −11+= lim=n→∞n1 n+11 n+111+= lim 1 +· lim=lim.n→∞n→∞ 1 + 1n→∞nnn11+nThus we make the following definition:Definition 101 n.e := lim 1 +n→∞nd. Subsequences and Partial Limits of a SequenceDefinition 11 If x1 , x2, .

. . , xn , . . . is a sequence and n1 < n2 < · · · < nk < · · · anincreasing sequence of natural numbers, then the sequence xn1 , xn2 , . . . , xnk , . . . iscalled a subsequence of the sequence {xn }.For example, the sequence 1, 3, 5, . . . of positive odd integers in their naturalorder is a subsequence of the sequence 1, 2, 3, . . . . But the sequence 3, 1, 5, 7, 9, . . .it not a subsequence of this sequence.Lemma 1 (Bolzano–Weierstrass) Every bounded sequence of real numbers contains a convergent subsequence.3.1 The Limit of a Sequence91Proof Let E be the set of values of the bounded sequence {xn }.

If E is finite, thereexists a point x ∈ E and a sequence n1 < n2 < · · · of indices such that xn1 = xn2 =· · · = x. The subsequence {xnk } is constant and hence converges.If E is infinite, then by the Bolzano–Weierstrass principle it has a limit point x.Since x is a limit point of E, one can choose n1 ∈ N such that |xn1 − x| < 1. Ifnk ∈ N have been chosen so that |xnk − x| < k1 , then, because x is a limit point of1.E, there exists nk+1 ∈ N such that nk < nk+1 and |xnk+1 − x| < k+11Since limk→∞ k = 0, the sequence xn1 , xn2 , .

. . , xnk , . . . so constructed converges to x.Definition 12 We shall write xn → +∞ and say that the sequence {xn } tends topositive infinity if for each number c there exists N ∈ N such that xn > c for alln > N.Let us write this and two analogous definitions in logical notation:(xn → +∞) := ∀c ∈ R ∃N ∈ N ∀n > N (c < xn ),(xn → −∞) := ∀c ∈ R ∃N ∈ N ∀n > N (xn < c),(xn → ∞) := ∀c ∈ R ∃N ∈ N ∀n > N c < |xn | .In the last two cases we say that the sequence {xn } tends to negative infinity andtends to infinity respectively.We remark that a sequence may be unbounded and yet not tend to positive infinnity, negative infinity, or infinity.

An example is xn = n(−1) .Sequences that tend to infinity will not be considered convergent.It is easy to see that these definitions enable us to supplement Lemma 1, statingit in a slightly different form.Lemma 2 From each sequence of real numbers one can extract either a convergentsubsequence or a subsequence that tends to infinity.Proof The new case here occurs when the sequence {xn } is not bounded.

Then foreach k ∈ N we can choose nk ∈ N such that |xnk | > k and nk < nk+1 . We then obtaina subsequence {xnk } that tends to infinity.Let {xk } be an arbitrary sequence of real numbers. If it is bounded below, onecan consider the sequence in = infk≥n xk (which we have already encountered inproving the Cauchy convergence criterion).

Since in ≤ in+1 for any n ∈ N, eitherthe sequence {in } has a finite limit limn→∞ in = l, or in → +∞.Definition 13 The number l = limn→∞ infk≥n xk is called the inferior limit of thesequence {xk } and denoted limk→∞ xk or lim infk→∞ xk . If in → +∞, it is said thatthe inferior limit of the sequence equals positive infinity, and we write limk→∞ xk =+∞ or lim infk→∞ xk = +∞. If the original sequence {xk } is not bounded below,923Limitsthen we shall have in = infk≥n xk = −∞ for all n. In that case we say that theinferior limit of the sequence equals negative infinity and write limk→∞ xk = −∞or lim infk→∞ xk = −∞.Thus, taking account of all the possibilities just enumerated, we can now writedown briefly the definition of the inferior limit of a sequence {xk }:lim xk := lim inf xk .n→∞ k≥nk→∞Similarly by considering the sequence sn = supk≥n xk , we arrive at the definitionof the superior limit of the sequence {xk }:Definition 14lim xk := lim sup xk .n→∞ k≥nk→∞We now give several examples:Example 14 xk = (−1)k , k ∈ N:lim xk = lim inf xk = lim inf (−1)k = lim (−1) = −1,n→∞ k≥nk→∞n→∞ k≥nn→∞lim xk = lim sup xk = lim sup(−1)k = lim 1 = 1.n→∞ k≥nk→∞n→∞ k≥nn→∞kExample 15 xk = k (−1) , k ∈ N:kklim k (−1) = lim inf k (−1) = lim 0 = 0,n→∞ k≥nk→∞kn→∞klim k (−1) = lim sup k (−1) = lim (+∞) = +∞.n→∞ k≥nk→∞n→∞Example 16 xk = k, k ∈ N:lim k = lim inf k = lim n = +∞,k→∞n→∞ k≥nn→∞lim k = lim sup k = lim (+∞) = +∞.k→∞Example 17 xk =(−1)kk ,n→∞ k≥nn→∞k ∈ N:!"− n1 ,if n = 2m + 1(−1)k(−1)k= lim inf= limlim= 0,n→∞ k≥nn→∞ − 1 , if n = 2mkkk→∞n+13.1 The Limit of a Sequence93(−1)k(−1)k= lim sup= limlimn→∞ k≥nn→∞k→∞kk!1n,1n+1 ,if n = 2m"if n = 2m + 1= 0.Example 18 xk = −k 2 , k ∈ N:lim −k 2 = lim inf −k 2 = −∞.n→∞ k≥nk→∞Example 19 xk = (−1)k k, k ∈ N:lim (−1)k k = lim inf (−1)k k = lim (−∞) = −∞,k→∞n→∞ k≥nn→∞lim (−1)k k = lim sup(−1)k k = lim (+∞) = +∞.k→∞n→∞ k≥nn→∞To explain the origin of the terms “superior” and “inferior” limit of a sequence,we make the following definition.Definition 15 A number (or the symbol −∞ or +∞) is called a partial limit of asequence, if the sequence contains a subsequence converging to that number.Proposition 1 The inferior and superior limits of a bounded sequence are respectively the smallest and largest partial limits of the sequence.4Proof Let us prove this, for example, for the inferior limit i = limk→∞ xk .

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