Assignment #4 — Solutions Math 315.
Section 10.
10.7 Let t= sup Swith t6∈ S. Since every convergent sequence has a monotone subsequence
with the same limit, it will suffice to prove that there is some sequence (sn) in Sthat
converges to t. First note, though, that for each a < t there exists some s∈Swith
a < s < t. So for each n∈IN there exists some sn∈Swith t−1/n < sn< t. We claim
that lim sn=t. Indeed, t−1/n < sn< t for all n∈IN,and lim (t−1/n) = lim t=t,
so by squeezing, lim sn=t.
10.10 (a) 2/3,5/9,14/27;
(b) Let Pn:sn>1/2. Then (B) s1= 1 >1/2, so P1is true. (IS). Let n≥1 and
assume that Pnis true. Then sn+1 = (sn+ 1)/3(IH)
>(1/2 + 1)/3 = 1/2, so Pn+1 is true.
Thus, by induction sn>1/2 for all n∈IN.
(c) sn−sn+1 =sn−(sn+ 1)/3 = (2sn−1)/3>(1 −1)/3 = 0, so the sequence is
decreasing.
(d) Since (sn) is decreasing and bounded below (by 1/2), it converges. And if limsn=s,
then s= lim sn= lim sn+1 = lim(sn+ 1)/3 = (lim sn+ 1)/3 = (s+ 1)/3, so 3s=s+ 1
and s= 1/2.
Section 11.
11.2
(a) (b) (c) (d) (e)
lim inf lim sup
(an) (a2n){−1,1} −1 1 diverge bounded
(bn) (bn){0}0 0 converge bounded
(cn) (cn){+∞} +∞+∞diverge to +∞unbounded
(dn) (dn){7/6}6/7 6/7 converges bounded
Section 12.
12.1 Note that since sn≤tnfor all n∈IN, we have L(sn)⊆L(tn) and U(sn)⊇U(tn).
Then L(tn)+⊆L(sn)+and U(sn)−⊆U(tn)−whence
[lim inf tn,+∞) = L(tn)+⊆L(sn)+= [lim inf sn,+∞)
and
(−∞,lim sup sn] = U(sn)−⊆U(tn)−= (∞,lim sup tn]
so lim inf sn≤lim inf tnand lim sup sn≤lim sup tn.
12.3 (a) 0; (b) 1; (c) 2; (d) 3; (e) 4; (f) 0; (g) 2.