[math-fun] funny little limit on central binomial coefficients
it seems to be 3, but why?: (2n choose n) /( Sum(k=0..n-1, (2k choose k) ) or (2n choose n) /(1+Sum(k=0..n-1, (2k choose k) ) hmmm, Mathematica 4.0.1.0 doesn't seem to handle it. Wouter. email: wouter.meeussen@vandemoortele.com =============================== This email is confidential and intended solely for the use of the individual to whom it is addressed. If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. You are explicitly requested to notify the sender of this email that the intended recipient was not reached.
??? makelist(choose(2*n,n)/sum(choose(2*k,k),k,0,n-1),n,1,11); [2, 2, (20/9), (70/29), (28/11), (308/117), (1144/425), (1430/523), (48620/17577), (184756/66197), (235144/83651)] At 06:40 PM 11/18/2002 +0100, you wrote:
it seems to be 3, but why?:
(2n choose n) /( Sum(k=0..n-1, (2k choose k) ) or (2n choose n) /(1+Sum(k=0..n-1, (2k choose k) )
hmmm, Mathematica 4.0.1.0 doesn't seem to handle it.
Wouter.
email: wouter.meeussen@vandemoortele.com
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At 06:40 PM 11/18/2002 +0100, Wouter wrote:
it seems to be 3, but why?:
(2n choose n) /( Sum(k=0..n-1, (2k choose k) ) or (2n choose n) /(1+Sum(k=0..n-1, (2k choose k) )
hmmm, Mathematica 4.0.1.0 doesn't seem to handle it.
Wouter.when n gets large enough.
It's easy to get that lim n->oo (2n choose n) / (2(n-1) choose (n-1)) = 4 directly from the definition. If you take the reciprocal of your quotient (the first one), you get Sum(k=0.n-1) ((2k choose k) / (2n choose n)) which as n gets sufficiently large looks a lot like Sum(k=1..n) 1/4^k = 1/3. (it's the latter terms the sum in your denominator that make up the first terms of the power series, so the more significant terms are as close as you like to powers of 1/4). I'm sure I'm violating some rules or other by doing it this way... I imagine this could also be gotten directly from a combinatoric argument involving growing a binary word which is half 1's and half 0's by two bits (one 1 and one 0), but that seems harder since you'd have throw out duplicates.
participants (3)
-
Marc LeBrun -
Meeussen Wouter (bkarnd) -
Shel Kaphan