Assignment #4 — Solutions. Math 315.

11.4. (a) Monotone subsequences:

(w2n), (x2n), (y2n), (z4n).

(c) lim sup wn = +∞, lim inf wn = −∞;lim sup xn = 1/5, lim inf xn = 5lim sup yn = 2, lim inf yn = 0lim sup zn = +∞, lim inf zn = −∞

(d) None of the four sequences converges or diverges to ±∞.

(e) Only the two sequences (xn) and (yn) is are bounded.

11.6. Let σ, τ be strictly increasing functions from N to N. Then for each n ∈ N, τ(n) <τ(n + 1) =⇒ σ(τ(n)) < σ(τ(n + 1)), so σ ◦ τ is strictly increasing. Thus, for eachsequence (sn), the subsequence (s(στ)(n)) = (sσ(τ(n))) of (sσ(n)) is a subsequence of (sn).

12.1. Let Sm = {sn : n > m} and Tm = {tn : n > m} be the terms in the mth tailsof (sn) and (tn). Then for every m > N0, ∀ n > m, inf Sm ≤ sn ≤ tn ≤ sup Tm =⇒sup Sm ≤ sup Tm, and inf Sm ≤ inf Tm. Thus, by Exercise 9.9, lim sup sn ≤ lim sup tn,and lim inf sn ≤ lim inf tn.

12.2. lim sn = 0 ⇐⇒ lim |sn| = 0 ⇐⇒ lim inf |sn| = lim sup |sn| = 0. So lim sn = 0 =⇒lim sup |sn| = 0. Conversely, by 12.1, 0 ≤ lim inf |sn| ≤ lim sup |sn| so lim sup |sn| =0 =⇒ lim inf |sn| = 0 and lim sn = 0.

12.3. lim inf sn = lim sup tn = 0 and lim sup sn = lim sup tn = 2, so

(a) lim inf sn + lim inf tn = 0 (b) lim inf(sn + tn) = 1;

(c) lim inf sn + lim sup tn = 2 (d) lim sup(sn + tn) = 3;

(e) lim sup sn + lim sup tn = 4; (f) lim inf sntn = 0;

(g) lim sup sntn = 2.

12.4. Let Sm = {sn : n > m} and Tm = {tn : n > m} be the terms in the mth tailsof (sn) and (tn). Then for each m ∈ N, n > m =⇒ sn + tn ≤ sup Sm + sup Tm =⇒sup{sn + tn : n > m} ≤ sup Sm + sup Tm so by Exercise 9.9

lim sup(sn + tn) ≤ lim(sup Sm + sup Tm) ≤ lim(sup Sm) + lim(sup Tm)

≤ lim sup(sn) + lim sup(tn).