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.
The harmonic mean for the values in this problem is equal to:
m/[Ó (x=m+1 to 2m) 1/x. Divide this by m and you get
1/Ó (x=m+1 to 2m) 1/x.
If you take the limit, this starts to look a lot like a definite integral.
1 / {integral} m+1 to 2m of 1/x dx.
If you grant me this, then you get:
1/{ln(2m)  ln(m)} = 1/ln(2).
A spreadsheet check for m=1000 amd 10000 suggests strongly that this is the answer that is converged upon.
Sorry  my math is not any better than this. I trust that someone can be more rigorous than I.
Answer: 1/(ln(2)

Posted by Kenny M
on 20100307 22:30:59 