lin n!^(1/n)

[bishnu]

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]

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]

Hey guys, how do you do all those symbols (sum and product, etc.)?

-Uncool-

[Johnny5]

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]

[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]

[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]

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-