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 Limiting Mean part 3 (Posted on 2010-03-14)
Evaluate:

Limit Pn(m)/m
m → ∞

where, Pn(m) denotes the nthpower mean of the m positive integers m+1, m+2, ....., 2m.

Next, evaluate this limit:

Limit Ln(m)/m
m → ∞

where, Ln(m) denotes the nth Lehmer mean of the m positive integers m+1, m+2, ....., 2m.

Note: both answers will be formulas in terms of n.

 No Solution Yet Submitted by Jer No Rating

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 Soution (non-rigorous?) Comment 1 of 1

Part 1 - the nth power mean is equal to [1/n *{(Ó m+1 to 2m) An^n}]^(1/n).  Realizing that the limit of the sum defines a definitie integral, one needs to solve the following:

Integral (m to 2m) x^n dx = [x^(n+1)]/(n+1) evaluated over 2m to m.  Perform this, divide by n, and take the nth root and the answer becomes (the m's cancel out):

{[2^(n+1) - 1]/[n+1] }^(1/n) (answer to part 1)  Note that for n=2, the nth power mean equals the RMS (see Limiting Mean 2) and the answer abover reverts to that answer = sqrt(7/3).

Part 2 - using similar methods, and realizing that the nth Lehmer Mean is equal to (Ó m+1 to 2m of m^n) /(Ó m+1 to 2m m^(n-1)), ones gets (if I did the calc and algebra correctly...)

(n/(n+1))*[2^(n+1)-1]/[2^n-1]

 Posted by Kenny M on 2010-03-14 23:10:37

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