bishnu Posted June 2, 2005 Posted June 2, 2005 can someone check my solution to the limit as n goes to infinity of n!^(1/n) L=n!^(1/n) Ln(L)=1/nSUM(Ln(n-i),i,0,n-1) use lhoptais rule Ln(L)=Sum(1/(n-i),i,0,n-1) Ln(L)=Sum(1/(1-i/n)*1/n,i,0,n-1) definiton of intergral LnL=int(1/(1-x),0,1) LnL=-Ln(1-x) from 0-1 LnL=Infinity-0 L=infinity
matt grime Posted June 2, 2005 Posted June 2, 2005 That's a bit messy and hard to follow, but it can be proved directly and easily. Let k be any positive integer, then n! > k^{n-k} for all n usfficiently large. Proof: Assume n>K, then the RHS is k multiplied by itself k-n times, but the LHS is n! = (k)!*(k+1)(k+2)....(k+(n-k)) > k!k^{n-k} >k^{n-k} as required Thus n!^{1/n} > k^{(n-k)/n} but the rhs tends to k as n tends to infinity thus n!^{1/n}> k-1 for n sufficiently large. But k was artbitrary thus n!^{1/n} must tend to infinity. Actually that is rather unnecessary since we know that the sum of k^n/n! is exp{k} it follows that k^n/n! tends to zero, and inparticular k^n/n! <1 for all n sufficiently large, but there's no harm in having two ways of proving something. As for your proof you are abusing the equals sign a lot. Yes [math] \frac{1}{n}\sum_{r=1}^{n}\log r[/math] does diverge, but I don't see how you can use l'hopital, which deals with real valued differentiable functions to conclude that.
uncool Posted June 2, 2005 Posted June 2, 2005 Hey guys, how do you do all those symbols (sum and product, etc.)? -Uncool-
Johnny5 Posted June 2, 2005 Posted June 2, 2005 Quote Hey guys' date=' how do you do all those symbols (sum and product, etc.)?-Uncool-[/quote'] Click on any particular latex image, and the code to write it should come up. for example, to write indefinite integral of x squared dx you would write: \int x^2 dx But you have to embed it in math tags like so: [ math] \int x^2 dx [ /math] (Just dont insert the extra spaces between the word math, and the [ brace. The printout is: [math] \int x^2 dx [/math] Regards PS: You can always read the Latex tutorial too.
uncool Posted June 3, 2005 Posted June 3, 2005 [math] L = \lim_{n \to \infty} n!^\frac{1}{n} ln[L] = ln[\lim_{n \to \infty} n!^\frac{1}{n}] = \lim_{n \to \infty} ln[n!^\frac{1}{n}] = \lim_{n \to \infty} \frac{ln[n!]}{n}[/math] Ratio of one to the next: [math]ln[n]\frac{n-1}{n} ln[n][/math] is always increasing, and will increase infinitely. [math]\frac{n-1}{n}[/math] is always decreasing, and will decrease towards 1. Therefore, the ratio will increase beyond one, and remain beyond one. Therefore, ln[L] does not exixt. Therefore, L does not exist. -Uncool- P.S. that was my first real LATEX thing. yay!
matt grime Posted June 3, 2005 Posted June 3, 2005 Quote [math]L = \lim_{n \to \infty} n!^\frac{1}{n}[/math] [math]\log [L] = \log[\lim_{n \to \infty} n!^\frac{1}{n}] = \lim_{n \to \infty} \log[n!^\frac{1}{n}] = \lim_{n \to \infty} \frac{\log[n!]}{n}[/math] Ratio of one to the next: [math]\log[n]\frac{n-1}{n} \log n[n][/math] Not sure I agree with that. the ratio of consective terms is [math]\frac{\log [( n+1)!](n)}{\log[n!] (n+1)}[/math] which is [math] (1- \frac{1}{n+1})(1+ \frac{\log(n+1)}{\log(n!)}[/math] which converges to 1 I believe so the ratio test doesn't show anything in this case.
uncool Posted June 5, 2005 Posted June 5, 2005 OK, matt, you are correct. I made a mistake (I divided two logs, and accidentally put the division inside). Shows wat happens when you skip a step... A few line break problems with what I had, too, though. The ln[n] is always decreasing... is supposed to be on a separate line. -Uncool-
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