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Limiting Mean (Posted on 2010-03-07) Difficulty: 2 of 5
Evaluate:

Limit H(m)/m
m → ∞

where, H(m) denotes the harmonic mean of the m positive integers m+1, m+2, ....., 2m.

As a bonus, evaluate this limit:

Limit R(m)/m
m → ∞

where, R(m) denotes the root mean square (RMS) of the m positive integers m+1, m+2, ....., 2m.

No Solution Yet Submitted by K Sengupta    
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Solution Solution Comment 3 of 3 |
Harmonic case:

First a note that as n approaches infinity the sum 1 + 1/2 + 1/3 + ... + 1/n approaches ln(n) + euler-macheroni constant.

What is sought is m/(1/(m+1)+1/(m+2)...+1/(2m))/m
= 1/(1/(m+1)+1/(m+2)+...+1/(2m))
Which can be rewritten as the difference
=1/([1+1/2+1/3+...+1/(2m)] - [1+1/2+1/3+...+1/m])
As m approaches infinity the constants cancel and we get
=1/(ln(2m)-ln(m)
=1/(ln(2m/m))
=1/ln(2)
or about 1.4427


RMS case.

First a note that the sum of 1^2 + 2^2 + 3^2 + ... +n^2 = (2n^3+3m^2+m)/6

What is sought is sqrt(((m+1)^2+(m+2)^2+...+(2m)^2)/m)/m
=sqrt(((m+1)^2+(m+2)^2+...+(2m)^2)/m^3)
which can we written as the difference
=sqrt(([1^2+2^2+3^2+...+(2m)^2]-[1^2+2^2+3^2+...+n^2])/m^3)
=sqrt(([2(2m)^3+3(2m)^2+2m]/6-[2m^3+3m^2+m]/6)/m^3)
=sqrt((14m^3+9m^2+m)/(6m^3))
as m approaches infinity this approaches
sqrt(14/6) = sqrt(7/3)
or about 1.5275

  Posted by Jer on 2010-03-08 15:21:12
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