wu :: forums « wu :: forums - A sequence » Welcome, Guest. Please Login or Register. Oct 15th, 2024, 11:56pm

RIDDLES SITE WRITE MATH! Home Help Search Members Login Register

wu :: forums    riddles    putnam exam (pure math) (Moderators: Icarus, towr, william wu, Grimbal, Eigenray, SMQ)    A sequence « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: A sequence  (Read 732 times) wmat1234 Newbie     Posts: 1 A sequence   « on: Aug 31st, 2007, 7:18am » Quote Modify Sequence:   x_1 = 1   x_n = x_{n-1} + sqrt(x_{n-1})   what is lim x_n/n^2 IP Logged Eigenray wu::riddles Moderator Uberpuzzler     Gender: Posts: 1948 Re: A sequence   « Reply #1 on: Aug 31st, 2007, 9:21am » Quote Modify Let yn = 2xn; then   yn+12 = yn2 + 2yn.   On the one hand,   yn+12 < (yn+1)2,   so yn < n, and limsup yn/n 1.   On the other hand,   Clearly xn xn-1 + 1, so xn (hence yn) goes to infinity.  Now, fix 0<a<1.  Then there exists an N such that for n>N, yn > C = a2/[2(1-a)], and therefore   yn+12 = yn2 + 2yn > yn2 + 2ayn + a2 = (yn + a)2,   so for n>N, yn > C + a(n-N).   Therefore liminf yn/n a.  Since this holds for all a<1, we have liminf yn/n 1.  Since also yn < n, it follows lim yn/n = 1, and thus lim xn/n2 = 1/4. IP Logged Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------  - easy   - medium   - hard   - what am i   - what happened   - microsoft   - cs => putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board