wu :: forums « wu :: forums - Limit of a Combinatorial Sum » Welcome, Guest. Please Login or Register. Apr 12th, 2024, 4:01am RIDDLES SITE WRITE MATH! Home Help Search Members Login Register
 wu :: forums    riddles    putnam exam (pure math) (Moderators: Grimbal, Icarus, william wu, Eigenray, towr, SMQ)    Limit of a Combinatorial Sum « Previous topic | Next topic »
 Pages: 1 Reply Notify of replies Send Topic Print
 Author Topic: Limit of a Combinatorial Sum  (Read 788 times)
Michael Dagg
Senior Riddler

Gender:
Posts: 500
 Limit of a Combinatorial Sum   « on: Aug 1st, 2008, 11:51am » Quote Modify

Suppose

G(m) = \sum_{i=1}^m  \sum_{j=1}^m C(m,i) C(m,j) i^{m-j} j^{m-i}  .

Show that

lim m->oo [ (G(m))^{1/(2m)}  ln m ]/m = 1/e   .
 « Last Edit: Aug 3rd, 2008, 8:47am by Michael Dagg » IP Logged

Regards,
Michael Dagg
Obob
Senior Riddler

Gender:
Posts: 489
 Re: Limit of a Combinatorial Sum   « Reply #1 on: Aug 1st, 2008, 1:54pm » Quote Modify

Should that be j^{m-1} or j^{m-i}?
 IP Logged
Michael Dagg
Senior Riddler

Gender:
Posts: 500
 Re: Limit of a Combinatorial Sum   « Reply #2 on: Aug 2nd, 2008, 8:38pm » Quote Modify

Sorry, I made typo --   j^{m-i}   is correct.  Good observation, however, I hope
no one spent any time on it -- as written, the sum (whose terms are all positive)
exceeds the term when   i=j=m   and that term is   m^{m-1}  .
 IP Logged

Regards,
Michael Dagg
 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 »