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.

For positive integers AM ≥ GM
Lt (m→inf) T(m)/m < Lt (m→inf) (1+2+3+....+m)/mē
=> Lt (m→inf) T(m)/m < Lt (m→inf) mē+m/2mē = 1/2
As m→inf, the given limit may approach 1/2.

Second Part:
Lt (m→inf) T(m)/m < Lt (m→inf) (2m+1+2m+2+....+3m)/mē
=> Lt (m→inf) T(m)/m < Lt (m→inf) (m/2)*(2m+1+3m)/mē=5/2
As m→inf, the given limit may approach 5/2.
 

Edited on March 27, 2008, 3:05 am

Posted by Praneeth on 2008-03-27 01:19:45