Limit P_n(m)/m
m → ∞

where, P_n(m) denotes the n^thpower 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^(n-1)), ones gets (if I did the calc and algebra correctly...)

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