Evaluate:
Limit P_{n}(m)/m
m → ∞
where, P_{n}(m) denotes the n^{th}power mean of the m positive integers m+1, m+2, ....., 2m.
Next, evaluate this limit:
Limit L_{n}(m)/m
m → ∞
where, L_{n}(m) denotes the n^{th} Lehmer mean of the m positive integers m+1, m+2, ....., 2m.
Note: both answers will be formulas in terms of n.
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^(n1)), ones gets (if I did the calc and algebra correctly...)
(n/(n+1))*[2^(n+1)1]/[2^n1]

Posted by Kenny M
on 20100314 23:10:37 