Evaluate:

Limit T(m)/m
m → ∞

where, T(m) is the geometric mean of m positive integers 1, 2, ..., m.

Bonus Problem

As a bonus, work out the following limit:

Limit U(m)/m
m → ∞

where, U(m) is the geometric mean of m positive integers 2m+1, 2m+2, ..., 3m.

Hi!

1.First limit

Let be a(m) = m!/m^m. So a(m+1) = (m+1)!/(m+1)^m+1

Cauchy - D'Alembert prove that the initial limit is the same with

lim a(m+1)/a(m) when m is going to infinit.

After calculus L(1) = lim 1/(1+1/m)^m = 1/e

2. Second limit

The same ideea!

Le be b(m) = (2m+1)(2m+2).....(2m+m)/m^m

So b(m+1) = (2m+3)(2m+4).....(3m)(3m+1)(3m+2)(3m+3)/(m+1)^(m+1)

The second limit is the same with lim b(m+1)/b(m) when m is going to infinit

After calculul L(2) = 27/4 * lim 1/(1+1/m)^m = 27/(4*e)

So i find

L(1) = 1/e

L(2) = 27/4*1/e

The problem can be generalise! 

Sorry that in the title i write Gauss - D'Alembert in fact is Caucy - D'Alembert 

The Cauchy - D'Alembert  criterion say that

lim square root(a(m)) = lim a(m+1)/a(m) where the square root is take m times. 

See the link below:

http://www.uel-pcsm.cines.fr/consultation/reference/mathematiques/serie/apprendre/chapitre4/partie1/titre3.htm

 

Edited on March 27, 2008, 3:31 am

Posted by Chesca Ciprian on 2008-03-27 03:17:52