4.12. EXERCISES 81

12. Suppose limn→∞ xn = x. Show that then limn→∞1n ∑

nk=1 xk = x. Give an example

where limn→∞ xn does not exist but limn→∞1n ∑

nk=1 xk does.

13. Suppose r ∈ (0,1) . Show that limn→∞ rn = 0. Hint: Use the binomial theorem. r =1

1+δwhere δ > 0. Therefore, rn = 1

(1+δ )n < 11+nδ

, etc.

14. Prove limn→∞n√

n = 1. Hint: Let en ≡ n√

n− 1 so that (1+ en)n = n. Now observe

that en > 0 and use the binomial theorem to conclude 1+ nen +n(n−1)

2 e2n ≤ n. This

nice approach to establishing this limit using only elementary algebra is in Rudin[24].

15. Find limn→∞ (xn +5)1/n for x ≥ 0. There are two cases here, x ≤ 1 and x > 1. Showthat if x > 1, the limit is x while if x≤ 1 the limit equals 1. Hint: Use the argumentof Problem 14. This interesting example is in [11].

16. Find limsupn→∞ (−1)n and liminfn→∞ (−1)n . Explain your conclusions.

17. Give a careful proof of Theorem 4.10.12.

18. Let {an} be a sequence in (−∞,∞). Let Ak ≡ sup{an : n≥ k} so that

λ ≡ lim supn→∞

an = limn→∞

An,

the An being a decreasing sequence.

(a) Show that in all cases, there exists Bn < An such that Bn is increasing andlimn→∞ Bn = λ .

(b) Explain why, in all cases there are infinitely many k such that ak ∈ [Bn,An].Hint: If for all k≥m> n, ak ≤Bn, then ak <Bm also and so sup{ak : k ≥ m}≤Bm < Am contrary to the definition of Am.

(c) Explain why there exists a subsequence{

ank

}such that limk→∞ ank = λ .

(d) Show that if γ ∈ [−∞,∞] and there is a subsequence{

ank

}which has the prop-

erty that limk→∞ ank = γ, then γ ≤ λ .

This shows that limsupn→∞ an is the largest in [−∞,∞] such that some subsequenceconverges to it. Would it all work if you only assumed that {an} is not −∞ forinfinitely many n? What if an = −∞ for all n large enough? Isn’t this case fairlyeasy? The next few problems are similar.

19. Let λ = limsupn→∞ an. Show there exists a subsequence,{

ank

}such that

limk→∞

ank = λ .

Now consider the set S of all points in [−∞,∞] such that for s∈ S, some subsequenceof {an} converges to s. Show that S has a largest point and this point is limsupn→∞ an.

20. Let {an} ⊆ R and suppose it is bounded above. Let

S≡ {x ∈ R such that x≤ an for infinitely many n}

Show that for each n, sup(S) ≤ sup{ak : k ≥ n} . Why is sup(S) ≤ limsupn→∞ ak?Next explain why the two numbers are actually equal. Explain why such a sequencehas a convergent subsequence. For the last part, see Problem 19 above.