|
|
Re: trying to get an upper bound on 1/log(n/m) when (n,m) = 1 and n
>m > 0 , n & m integers
Posted:
Apr 13, 2012 5:42 PM
|
|
On Apr 13, 8:53 am, David Bernier <david...@videotron.ca> wrote:
> (a) n > m > 0
> (b) n and m are positive integers
> (c) (n, m) = 1.
The problem statement says "upper bound".
I interpret this to mean "upper bound over all possible m for
a given n such that (n,m) = 1". This will happen when n/m is as
close to 1 as possible;
i.e. when m = n-1.
estimating 1/log( n/(n-1)) should not be hard.
It is just 1/log(1 + 1/n - 1/n^2 + ....) , just use the Taylor
series for log(1+epsilon)....
|
|
