p.86 - 3 (1, .5, 3, .25, 5, .17, ...); (1, 2, .33, 4, .2, 6, ...) [more decimal places will be needed] 8a lim (4 + 2/n + 1/(n^2)); by lilmit of sums, limit is 4 12d We are looking at (8/9)^n; the ratio of adjacent terms is manifestly 8/9 LT 1, hence the llimit is 0. p.93 - 1 Note first that for 1 LT x_n LT 2, 1 LT x_(n+1) LT 2; this shows bounded. Monotone can be demonstrated by showing that x_n - (2 - 1/x_n) is positive, hence x_(n+1) is less than x_n The limit is found by setting x_(n+1) = x_n and solving. (lim = 1) 7 By definiton of sup, there exists a member of A in every V_(1/n)(u) subscript those eements as x_n corresponding to the 1/n neighborhoods. Note that it may be necessary to use some elements of A more than once (e.g., if A is the set of all negative integers) p.99 - 6d Convergence can be established by appropriately altering the arguments on pp. 92, 93. Then by noting that the subsequence obtained by taking the square root for even values of n converges to e, we can show that the sequence converges to e^2 8 Since x_n GEQ 0, The subsequence for odd n converges to a limit LEQ 0, The subsequence for even n converges to a limit GEQ 0 [The subsequences converge because all subsequences of a sequence converge to the limit of the sequence] Since the subsequences converge to the same limit, that limit must be 0, and x_n will converge to it. 10 Let's start with the interval [-1, 1]. I shall use the positive half, hence start with 1/2, then to be in side 1/2, x_2 =1/4, then to be in side 1/4, x_3 =1/6, then to be inside 1/8, x_4=1/10, etc.

Sun Sep 28 1997