Answers & Hints for Selected Exercises 645- Use induction to show that the sequence is monotone increasing and bounded
above by 2; thus, it converges, say to L. Then, since Xn+l = v2 + Xn, we have
L^2 = 2 + L, from which we conclude L = 2 (supply reas~ns).
EXERCISE SET 2.61.- (a) (1 + ~)2n = [(1 + ~rJ2 -->'e2. (b) (1 + 3~)2n = [(1 + 3~)3n]3----> e~
since { (1 + 3 ~)
3
n} is a subsequence of {(1 + ~r}.( C) e ( d) ( n~ 1 r = ( ~ r = ( H
1
* r ----> ~ · ( e) ~- (a) False. If 3n 1 EN 3 n :::'.'. n 1 ==> Xn EA and 3n2 EN 3 n :::'.'. n2 ==> Xn EB,
then n :::'.'. max { n 1 , n2} ==> Xn E A n B = 0, a contradiction.
(b) True. {(-l)n} is frequently in (-2,0) and frequently in (0,2).
- (a) Suppose Xn----> oo and { xnk} is a subsequence of {xn}· Let M > 0. Then
3no EN 3 n :::'.'.no==> Xn > M. So, k :::'.'.no==> nk :::'.'. k :::'.'.no==> Xnk > M.
(b) Same as (a), with "> M" replaced by "< -M." - (a)
(g)
(k)0, +oo; diverges ( c) -oo, 0, +oo; diverges ( e) -5, 5; diverges
0 , 2>^1 2 v'3 >^1 , -2^1 , -2, v'3 -1· , d " 1verges (") 1 0 ; converges
-1, 1; diverges- {1,2,1,2,3,1,2,3,4, 1, 2,3, 4, 5, 1, 2,3, 4,5,6, 1,2,3,4,5,6, 7,1, ... }
- {xn} not bounded above==> Vk EN, 3nk EN 3 Xnk > k ==> Xnk ----> +oo.
Similarly for {xn} not bounded below. - ( ~) Suppose every subsequence of { Xn} has a subsequence converging to
real number L. Then Lis a cluster point of {xn}· If L' is another, then { Xn} has
a subsequence converging to L', and this in turn has a subsequence converging
to L. By Thm. 2.6.8 and the uniqueness of limits, L = L'. Thus, { Xn} has one
and only one cluster point. By Thm. 2.6.17, Xn ----> L.
i9. Redo Case 1, using (L, +oo), changing inequalities, etc.
- Suppose { xn} is bounded and all its convergent subsequences converge to
L. If { Xnk} is any subsequence, it is bounded, so by Bolzano-Weierstrass it has
a convergent subsequence, which must converge to L. So, every subsequence of
{ xn} has a subsequence converging to L. By Exercise 2.6.17, Xn ----> L.