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- From: clong@remus.rutgers.edu (Chris Long)
- Newsgroups: sci.math,rec.puzzles
- Subject: Re: HELP: limit (SPOILER)
- Message-ID: <Jan.12.04.30.20.1993.8013@remus.rutgers.edu>
- Date: 12 Jan 93 09:30:21 GMT
- References: <C0pAC9.CtH@news.cso.uiuc.edu>
- Followup-To: sci.math
- Organization: Rutgers Univ., New Brunswick, N.J.
- Lines: 34
-
-
- A correction:
-
- In article <C0pAC9.CtH@news.cso.uiuc.edu>, Jon Hamkins writes:
-
- > k+q
- > --- i
- > -k \ k
- > lim e / ----- ,
- > k->infty --- i!
- > i=0
-
- > where q is a positive integer constant.
-
- The limit is 1/2. Using the fact that P = \lim_{k \to \infnty}
- \sum_{i=0}^{k} k^i/i! ~ 1/2*e^{k} (*), we see that limit =
- \lim_{k \to \infnty} e^{-k}*(P + \sum_{i=k+1}^{k+q} k^i/i!) =
- \lim_{k \to \infnty} e^{-k}*P +
- \lim_{k \to \infnty} e^{-k}*\sum_{i=k+1}^{k+q} k^i/i!). The first
- term is easily seen to go to 1/2 by (*), and the last term is seen
- to be 0 by the fact that e^{-k}*k^{k+c}/{k+c}! ~
- e^{-k}*e^{k+c}*(1-c/(k+c))^{k+c}/sqrt(2*pi*(k+c)) ~ 1/sqrt(2*pi*(k+c)),
- where ~ is used throughout in the asymptotic sense.
-
- (*) This is proven in Newman's _A Problem Seminar_. I can't seem
- to find my copy of it to verify that it's there, but my memory
- is pretty good about things like this.
-
- Challenge: For what finite constant c does \sum_{i=0}^{p*k} k^i/i!
- ~ c*e^{k} as k \to \infnty?
- --
- Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618
-
- Score: 0, Diff: 1, clong killed by a Harvard Math Team on 1
-