Evaluate:

Limit T(m)/m
m → ∞

where, T(m) is the geometric mean of m positive integers 1, 2, ..., m.

Bonus Problem

As a bonus, work out the following limit:

Limit U(m)/m
m → ∞

where, U(m) is the geometric mean of m positive integers 2m+1, 2m+2, ..., 3m.

We can write T(m) as (m!)^(1/m)

Call the limit L. Then the natural log of the limit is the limit of the natural log of the expression.

ln(L) = lim (m->infiinity) ln(T(m)/m) = -ln(m) + (1/m)ln(m!)

as m-> infinity, Sterling's approximation for ln(m!) is increasingly accurate. (NOTE: making this substitution may not be legitimate and I offer it without proof, although I'm fairly confident it's correct.)

If so,

ln(L) = lim -ln(m) + (1/m)(m ln(m) - m) = -1

if ln(L) = -1 then L = 1/e


The same approach should work for the U(m) case, once we write U(m) = ((3m)! / (2m)!)^(1/m)

then letting the limit be represented by L and taking the log of both sides gives:

ln(L) = lim -ln(m) + (1/m)(3m ln(m) + 3m ln(3) - 3m -2m ln (m) - 2m ln(2) +2m)

ln(L) = lim -ln(m) + 3ln(m) + 3ln(3) - 3 -2ln(m) - 2ln(2) + 2
ln(L) = 3ln(3) - 3 - 2ln(2) + 2 = 3ln(3) - 2ln(2)  - 1

e^(a+b) = e^a*e^b and e^(a ln(b)) = b^a, so the limit is:

L = 3^3 / 2^2 * 1/e = 27/(4e)

In general, we can solve this kind of problem for and F(m) that's the geometric mean of m positive integers starting with km+1 (k>=0), and the result will be (k+1)^(k+1)/(e*k^k).

k=0 corresponds to T(m) and (since 0^0 = 1) gives the expected 1/e. k= 2 corresponds to U(m) and gives the previously stated 27/(4e).

Of course, all of this is for naught if Sterlings approximation cannot be used to substitute for ln(m!) but I'll let some braver reader than I prove or disprove that assumption.

Posted by Paul on 2008-03-27 02:56:37