[seqfan] (sum lnp) / (sum lnp {lnm/lnp}) limit

[Leroy Quet]

(I posted this over at sci.math, but no one there answered it.)

I am just wondering... 
Let {x} = x - floor(x). Let lnx = natural logarithm of x. 
Let A(m) = sum(p=primes <= m)  lnp * {lnm/lnp}. 
And B(m) = sum(p=primes <= m)  lnp. 
Then does the limit (as m approaches infinity) A(m)/B(m) exist? 
If so, what is it? 
0 <= A(m)/B(m) <= 1 for all m. But I don't know if the ratio converges 
to a constant. 
By the way, I know that A(m) = ln(m^pi(m) /LCM(1,2,3,...,m)), where pi 
(m) is the number of primes <= m. 
Thanks, 
Leroy Quet 

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