Math 104: Homework 2 Solutions

The following problems are taken from the textbook, and listed according to the same numbering:

6.2 Show that if α and β are Dedekind cuts, then so is α + β = {r1 + r2 : r1 ∈ α and r2 ∈ β}. Solution: Let α, β ⊂ Q be Dedekind cuts. To show α + β is a Dedekind cut, we must verify three properties. First, we claim α + β is a nonempty proper subset of Q. To see this, note that each of α and β is nonempty, so we can find r1 ∈ α, r2 ∈ β, meaning r1 + r2 ∈ α + β, so α + β is nonempty. On the other hand, since α and β are Dedekind cuts we must have α 6= Q and β 6= Q, meaning we can find M1 ∈ Q with M1 6∈ α and M2 ∈ Q with M2 6∈ β. We claim that in fact r1 < M1 for all r1 ∈ α and r2 < M2 for all r2 ∈ β. This is because clearly we cannot have r1 = M1 or r2 = M2 by assumption that M1 6∈ α and M2 6∈ β. On the other hand, we also can’t have r1 > M1 because then property (ii) of Dedekind cuts would require M1 ∈ α. Similarly we cannot have r2 > M2. So since r1 < M1 for all r1 ∈ α and r2 < M2 for all r2 ∈ β we must have r1 + r2 < M1 + M2, and since r1 and r2 were arbitrary this shows that s < M1 + M2 for all s ∈ α + β. In particular, this means M1 + M2 6∈ α + β, so α + β is indeed a proper subset of Q.

The second property we must verify is that α + β is “closed downward,” i.e. if s1 ∈ α + β and s2 ∈ Q with s2 < s1 then s2 ∈ α + β as well. To see this, let s1 = r1 + r2 ∈ α + β (i.e. with r1 ∈ α and r2 ∈ β) and let s2 ∈ Q with s2 < s1. Then this means s2 < r1 + r2 so s2 − r2 < r1. Then since α is a Dedekind cut and therefore closed downward, we must have s2 − r2 ∈ α. Then we must have (s2 − r2) + r2 ∈ α + β, so indeed s2 ∈ α + β as desired. The final property we must verify is that α + β has no maximal element. To see this, let s1 ∈ α + β and we will show there exists some s2 ∈ α + β with s2 > s1. This is because, if s1 = r1 + r2 for some r1 ∈ α and r2 ∈ β then since α has no maximal element we can find 0 0 0 r1 ∈ α with r1 > r1. Then if we take s2 = r1 + r2 we indeed have s2 ∈ α + β with s2 > s1, so α + β has no maximal element.

6.3 (a) Show α + 0∗ = α for all Dedekind cuts α. Solution: Let α ⊂ Q be an arbitary Dedekind cut. By definition, the Dedekind cut ∗ ∗ 0 ⊂ Q is the set consisting of all negative rational numbers. Now let s ∈ α + 0 , say ∗ s = r1 + r2 with r1 ∈ α and r2 ∈ 0 . Then since r2 < 0 we have r1 + r2 < r1 so since α ∗ is closed downward we have s = r1 + r2 ∈ α, showing that α + 0 ⊂ α. On the other hand, let r1 ∈ α. Then since α has no maximal element we can find ∗ r2 ∈ α with r2 > r1. This means r1 − r2 < 0 and therefore r1 − r2 ∈ 0 . So then ∗ ∗ ∗ r2 + (r1 − r2) ∈ α + 0 , so r1 ∈ α + 0 , meaning α ⊂ α + 0 as well. Since containment holds in both directions, we have the equality of sets α = α + 0∗. (b) We claimed, without proof, that addition of Dedekind cuts satisfies property A4. Thus if α is a Dedekind cut, there is a Dedekind cut −α such that α + (−α) = 0∗. How would you define −α? Solution: The “naive” approach would be to attempt the definition −α = {−r : r ∈ α}. However, a quick sketch of what this would look like on the number line should be enough to convince one that this will not work, since Dedekind cuts are supposed to essentially resemble intervals of the form (−∞, x) and this definition would produce something of the form (−x, ∞). However, there is a simple solution to this issue, since (again, as we can see intuitively from a sketch) the naive definition actually happens to produce what 1 looks like the complement of the set we are aiming for – i.e. it produces the set of all rational numbers greater than the place where we are “cutting” the rational line. So we should expect the definition −α = Q \ {−r : r ∈ α} = {r ∈ Q : −r 6∈ α} to work. There is still one technical problem, however. If the original Dedekind cut α happened to represent a rational number, then its complement Q \ α will have a minimum element (equal to the rational number α represents). Under negation, this will unfortunately become a maximal element of −α. However, −α is easy to define in this case, because ∗ in this case we have α = s = {r ∈ Q : r < s} for some s ∈ Q, so we can just directly ∗ define −α = (−s) = {r ∈ Q : r < −s}. While the problem does not require you to verify that this definition is satisfactory, I’ll do so here anyway for your edification. Our first task is to show that our definition in fact produces a Dedekind cut. Clearly since we defined −α = (−s)∗ in ∗ the case α = s for some s ∈ Q, we have that −α is a Dedekind cut because we already ∗ know all sets of the form q for some q ∈ Q are Dedekind cuts, so we focus on the case where α is not rational. To see that −α is nonempty, note that since α is not all of Q we can find r ∈ Q with r 6∈ α. Then we see (since α is not rational), −r ∈ −α since r 6∈ α. On the other hand, we cannot have −α = Q because α is nonempty, and for any r ∈ α we have −r 6∈ −α. To see that −α is closed downward, let r ∈ −α and s < r. Then that means −r 6∈ α, and we also have −s > −r. This means we cannot have −s ∈ α because then since α is closed downward that would mean −r ∈ α as well. So since −s 6∈ α we in fact have s ∈ −α as desired. To see −α has no maximal element, suppose for the sake of contradiction that it does, say s = max(−α). Then we claim −s = min(Q \ α). To see this, let r ∈ Q \ α. Then since r 6∈ α we have −r ∈ −α, which means −r < s, meaning r > −s, so −s is a lower bound of Q \ α. On the other hand we also have since s ∈ −α that −s 6∈ α, meaning −s ∈ Q \ α, so indeed −s = min(Q \ α). However, this means s is the least upper bound for α, i.e. s = sup α, contradicting the assumption that α was not rational. Now that we have verified that our definition of −α indeed produces a Dedekind cut, we still must verify that it satisfies the desired property of being the additive inverse of α, i.e. that α + (−α) = 0∗ as we have defined it. First we will handle the case α = s∗. First ∗ ∗ ∗ ∗ let r ∈ α + (−α) = s + (−s) . Then we have r = r1 + r2 where r1 ∈ s and r2 ∈ (−s) . ∗ So r1 < s and r2 < −s. But then r1 + r2 < s + −s = 0, so r1 + r2 ∈ 0 , showing s∗ + (−s)∗ ⊆ 0∗. On the other hand, let r ∈ 0∗. Then s + r/2 < s and −s + r/2 < −s so s + r/2 ∈ s∗ and −s + r/2 ∈ (−s)∗. So then r = (s + r/2) + (−s + r/2) ∈ s∗ + (−s)∗. So we have shown s∗ + (−s)∗ ⊆ 0∗, completing the proof that s∗ + (−s)∗ = 0∗.

Now suppose α is irrational, and let r ∈ α + (−α), meaning r = r1 + r2 where r1 ∈ α and r2 ∈ −α. Then we claim −r2 > r1. This is because r2 ∈ −α means −r2 ∈ Q \ α, but if −r2 ≤ r1 we would have −r2 ∈ α since α is closed downward. So since −r2 > r1 ∗ ∗ we have r1 + r2 < 0, meaning r = r1 + r2 ∈ 0 , showing α + −α ⊆ 0 . On the other hand, let r ∈ 0∗. Now we claim we can find s ∈ α with s − r/2 6∈ α. This is because if we take an arbitrary q ∈ α we can consider the set {q − nr/2 : n ∈ N}, which is unbounded above (by the Archimedean property) and therefore contains some element which is not in α since α itself is bounded above. Then by taking the smallest n for which q − nr/2 6∈ α we see that q − (n − 1)r/2 gives the desired choice of s. Now if we take r1 = s + r/2 we clearly have r1 ∈ α since s + r/2 < s. Moreover, if we take r2 = −(s − r/2) = −s + r/2, we have r2 ∈ −α since −r2 6∈ α by our choice of s. So then 2 ∗ r1 + r2 = s + r/2 + −s + r/2 = r ∈ α + (−α), showing 0 ⊂ α + (−α), completing the proof that α + (−α) = 0∗ in the case α is irrational.

7.3 For each sequence below, determine whether it converges and, if it converges, give its limit. No proofs are required.

1/n (f) sn = 2 1 1/n 0 Solution: The intuition here is that n → 0, so 2 → 2 = 1. (−1)n (i) n n n Solution: We have (−1) = 1 → 0, so we expect (−1) → 0 as well. n n n 1
2 (r) 1 + n 1 1
2 2 Solution: Since n → 0, we expect 1 + n → (1 + 0) = 1. 3n (q) n! Solution: Though we haven’t proved the “ratio test” for sequences yet, we can use it here because the problem does not require rigorous proof. We see that in this case n |sn+1/sn| = 3/(n + 1) → 0, so we expect 3 /n! → 0. 7.4 Give examples of

(a) A sequence (x ) of irrational numbers having a limit lim x that is a rational number. n √ n Solution: We can take a sequence like 2/n, which converges to zero.

(b) A sequence (rn) of rational numbers having a limit lim rn that is an irrational number. Solution: This is a little bit trickier. The problem does not explicitly ask for a proof that the sequence you exhibit actually converges to an irrational number, so I think simply describing the example should suffice. In fact, there are many readily available to us. One straightforward example would be the sequence 1, 1√.4, 1.41, 1.414, 1.4142,... whose nth term is the truncation of the decimal expansion of 2 to the first n digits. Another famous example would be the sequence (1 + 1/n)n which converges to e, or −1 the sequence defined recursively√ by s1 = 1 and sn+1 = (sn + 2sn )/2 for n ≥ 1. This sequence also converges to 2 (this is known as “Newton’s Method” for approximating −1 √ square roots – if the sequence is defined by sn+1 = (sn + asn )/2 then sn → a, at least provided a > 0 and s1 > 0 as well).

1 8.2 (e) Determine the limit of the sequence sn = n sin n and prove your claim. Solution: Our intuition is that, even though sin n behaves somewhat unpredictably, it is at least bounded between ±1, so that asymptotically the growth of n in the denominator will render the oscillation of sin n irrelevant and the sequence should converge to zero. Indeed, showing the limit is zero entails choosing N so that |(1/n) sin n| < ε for n > N, and we immediately see that we can directly exploit the bound |(1/n) sin n| ≤ |1/n| = 1/n (since | sin n| ≤ 1), and we have 1/n < ε so long as n > 1/ε. So our proof might go as follows: Let ε > 0. Choose N > 1/ε. Then for all n > N we have n > 1/ε, meaning 1/n < ε. But then this means |(1/n) sin n| ≤ 1/n < ε for all n > N, so we have shown that (1/n) sin n → 0.

8.5 (a) Consider three sequences (an), (bn) and (sn) such that an ≤ sn ≤ bn for all n and lim an = lim bn = s. Prove lim sn = s. This is called the “squeeze lemma.”

3 Solution: Let (an), (bn) and (sn) be sequences as in the problem statement. As usual with general problems of this sort, we must exploit the convergence of (an) and (bn) to control the behavior of (sn). To do this, we consider the question of the manner in which the “error bounds” on an and bn constrain sn. The answer in this case (which is not difficult to see if one makes a rudimentary sketch of the situation) turns out to be particularly simple – if |an − s| < ε and |bn − s| < ε then an and bn are both in the interval (s − ε, s + ε) and since sn is between them then sn must be as well. So if we let ε > 0 and simply choose N1 so that |an − s| < ε for n > N1 and N2 so that |bn − s| < ε for n > N2, then taking N = max{N1,N2} guarantees that for n > N we have both |an − s| < ε and |bn − s| < ε. In particular, this means s − ε < an and bn < s + ε, so we have s − ε < an ≤ sn ≤ bn < s + ε, i.e. s − ε < sn < s + ε, meaning |sn − s| < ε for all n > N as well. This shows sn → s.

(b) Suppose (sn) and (tn) are sequences such that |sn| ≤ tn for all n and lim tn = 0. Prove lim sn = 0. Solution: Note that we have 0 ≤ |sn| ≤ tn, so, since tn → 0, applying the squeeze lemma from (a) with an = 0 and bn = tn shows that |sn| → 0. Now, we also have that −|sn| ≤ sn ≤ |sn|, so a second application of the squeeze lemma shows that sn → 0 as well. (Technically we are using the fact that |sn| → 0 implies −|sn| → 0 which we would either have to prove on its own or justify using a theorem from section 9, but let’s not worry about that.) √ 2 1 8.8 (b) Prove that lim[ n + n − n] = 2 . Hint: Try rewriting the sequence as √ √ p ( n2 + n − n)( n2 + n + n) (n2 + n) − n2 n n2 + n − n = √ = √ = √ . n2 + n + n n2 + n + n n2 + n + n

Solution: As in the hint, we will work with the expression n sn = √ . n2 + n + n

This means we would like to show that with a suitable choice of N we can ensure |sn −(1/2)| < ε (for any ε > 0). With a little further manipulation, we see √ n 1 2n − ( n2 + n + n)

|sn − s| = √ − = √ n2 + n + n 2 2( n2 + n + n) √ n − n2 + n

= √ 2 n2 + n + 2n q 1 − 1 + 1 n = q 1 2 1 + n + 2

q 1 1 + n − 1 = . q 1 2 + 2 1 + n

To bound this last expression above by ε, it would be nice to first bound it above by some expression that doesn’t involve radicals. The denominator can simply be bounded below 4 by 4, since p1 + 1/n > 1. On the other hand, we also have 1 + 1/n > p1 + 1/n since p1 + 1/n > 1, so we see the numerator can be bounded above by 1 + 1/n − 1 = 1/n. In all, this means we have |sn − s| < 1/(4n), so a choice like N > 1/(4ε) should suffice, as then we will have 1/(4n) < ε, which in turn guarantees |s − sn| < ε for n > N.

8.9 Let (sn) be a sequence that converges.

(a) Show that if sn ≥ a for all but finitely many n, then lim sn ≥ a. Solution: Suppose for the sake of contradiction that s = lim sn < a. Then pick ε = (a − s)/2. Since sn → s, we can find N such that |sn − s| < ε (for all n > N) for this choice of ε, meaning sn < s + (a − s)/2 = (a + s)/2 < (a + a)/2 = a since s < a. But since this is true for all n > N, we have shown that sn < a for infinitely many n, contradicting our assumption that sn ≥ a for all but finitely many n.

(b) Show that if sn ≤ b for all but finitely many n, then lim sn ≤ b. Solution: The argument is very similar to (a). If we suppose s = lim sn > b then by taking ε = (s − b)/2 we ultimately find we can choose N so that sn > (s + b)/2 > b for all n > N, contradicting that sn ≤ b for all but finitely many n.

(c) Conclude that if all but finitely many sn belong to [a, b], then lim sn belongs to [a, b]. Solution: This follows by invoking both (a) and (b) above. The assumption that all but finitely many sn belong to [a, b] directly implies that all but finitely many sn satisfy a ≤ sn ≤ b, so by (a) the first inequality means lim sn ≥ a and by (b) the second implies lim sn ≤ b, so since a ≤ lim sn ≤ b we indeed have lim sn ∈ [a, b].

3 3 an+4an a +4a 9.3 Suppose lim an = a and lim bn = b, and sn = 2 . Prove lim sn = 2 carefully, using bn+1 b +1 the limit theorems.

Solution: Since lim an = a and lim bn = b, we have by several applications of Theorem 9.4 3 3 2 2 that lim an = a and lim bn = b . We also have by 9.2 that lim 4an = 4a, so using 9.3 we 3 3 find lim(an + 4an) = a + 4a. On the other hand clearly lim 1 = 1 so by 9.3 again we have 2 2 2 2 lim(bn + 1) = b + 1. Now clearly bn + 1 > 0 and b + 1 > 0, so we have by 9.6 that

3 3 an + 4an a + 4a lim 2 = 2 , bn + 1 b + 1 as desired.

9.9 Suppose there exists N0 such that sn ≤ tn for all n > N0.

(a) Prove that if lim sn = ∞ then lim tn = ∞. Solution: Suppose lim sn = ∞. Let M ∈ R. Then we can find N1 so that sn > M for n > N1. Take N = max{N0,N1}. Then for n > N we have tn ≥ sn > M. Since M was arbitrary, we see tn → ∞ as well.

(b) Prove that if lim tn = −∞ then lim sn = −∞. Solution: Suppose lim tn = −∞. Let M ∈ R. Then we can find N1 so that tn < M for n > N1. Take N = max{N0,N1}. Then for n > N we have sn ≤ tn < M. Since M was arbitrary, we see sn → −∞ as well.

(c) Prove that if lim sn and lim tn exist, then lim sn ≤ lim tn. Solution: Suppose lim sn = s and lim tn = t. Suppose for the sake of contradiction t < s. Then take ε = (s − t)/2 and choose N1 so that |sn − s| < ε for n > N1 and N2 5 so |tn − t| < ε for n > N2, and put N = max{N0,N1,N2}. Then for n > N we have sn > s − ε = (s + t)/2, while on the other hand tn < t + ε = (s + t)/2, so we see that tn < sn for n > N, contradicting the assumption that sn ≤ tn for n > N0.

sn+1 9.12 Assume all sn 6= 0 and that the limit L = lim exists. sn

(a) Show that if L < 1, then lim sn = 0. Hint: Select a so that L < a < 1 and obtain N so n−N that |sn+1| < a|sn| for n ≥ N. Then show |sn| < a |sN | for n > N. Solution: Let ε > 0. As in the hint, we choose a with L < a < 1. Then if we take

ε0 = a − L, we can use the assumption lim sn+1 = L to find N for which sn

sn+1 0 − L < ε , sn which means in particular we have

sn+1 0 < L + ε = a, sn

and after multiplying on both sides by |sn| we see we indeed have |sn+1| < a|sn| for n−N n > N. We now argue by induction that this means |sn| < a |sN | for n > N. To see this, note that the argument we just made shows the base case |sN+1| < a|sN |. k−N Now suppose we know |sk| < a |sN | for some particular k > N. Then we have |sk+1| < a|sk| by our previous argument, and by the inductive hypothesis this means k−N k+1−N |sk+1| < a(a |sN |) = a |sN |, completing the induction. Now since we know an → 0 (by Theorem 9.7(b)), we can find some N 0 for which |an−0| = n 0 00 0 00 a < ε/|sN | for n > N . Then if we take N = N + N we see that for n > N we have n−N ε |sn| < a |sN | < |sN | = ε, |sN | since for n > N 00 we have n − N > N 0. (b) Show that if L > 1, then lim |s | = ∞. Hint: Apply (a) to the sequence t = 1 ; see n n |sn| Theorem 9.10. Solution: As in the hint, we let t = 1 . Then we see n |sn| −1 −1 tn+1 |sn+1| sn+1 lim = lim −1 = lim . tn |sn| sn

Since we know lim sn+1 = L > 0, we can apply Lemma 9.5 to say sn

tn+1 1 lim = < 1. tn L

Then we see by part (a) that we must have tn → 0. But then by 9.10 (since |sn| > 0 for all n) we see this implies lim |sn| = ∞. an 9.15 Show limn→∞ n! = 0 for all a ∈ R. n Solution: Say sn = a /n!. Note that n+1 sn+1 a /(n + 1)! a lim = lim n = lim = 0, sn a /n! n + 1

so by part (a) of problem 9.12 we have that lim sn = 0 as well.

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