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Limiting Mean part 2 (Posted on 2010-03-10) Difficulty: 1 of 5
Evaluate:

Limit A(m)/m
m → ∞ where A(m) denotes the arithmetic mean of the m positive integers m+1, m+2, ....., 2m.

Next evaluate:

Limit G(m)/m
m → ∞ where, G(m) denotes the geometric mean of the m positive integers m+1, m+2, ....., 2m.

No Solution Yet Submitted by Jer    
Rating: 1.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
analytical solutions | Comment 1 of 2
part 1.
A(m) = (3m+1)/2
A(m)/m = (3m+1)/(2m) = (3/2) + (1/2m)
thus the limit is 3/2 = 1.5

part 2.
G(m) = [(2m)!/m!]^(1/m)
now
n! is approximately
sqrt(2pi*n)*n^n/e^n thus we have an approximation for
G(m) as
(sqrt(4*pi*m)*(2m)^(2m)*e^m /[ sqrt(2*pi*m)*m^m*e^(2m)])^(1/m)
simplifying we get
[sqrt(2)*4^m*m^m/e^m]^(1/m) =
4m/e*(sqrt(2))^1/m
thus
G(m)/m is approximately
4/e*(sqrt(2))^(1/m)
and thus the limit is 4/e

  Posted by Daniel on 2010-03-10 11:33:54
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