We know that pi(k) is roughly equal to k/ln(k). Are there any related functions that lead to a new appreciation of the number of primes up to some given number? An approximation of ln(k) is the harmonic sum: (1/1)+(1/2)+(1/3)+(1/4)+ . . .

The harmonic sum is equivalent to zeta(n) for n=1, which is (again): (1/1)+(1/2)+(1/3)+(1/4)+ . . .+(1/k) as k approaches infinity. Note that this is not a convergent sum. But also note that the difference (1/1)+(1/2)+(1/3)+ . . . +(1/k)-ln(k) (the harmonic sum minus the natural log) approaches gamma as k approaches infinity, a constant often associated with Euler. In Riemann zeta function I mention another form of zeta(n), i.e.:

           | |     1
zeta(n)=   | | --------
           | | 1-p^(-n)
for p in P, the set of prime numbers. Thus, ln(k) is also related to the multiplication:

   1       1       1            1
1-(1/2) 1-(1/3) 1-(1/5)     1-(1/p(k))
where p(k) is the largest prime less than or equal to k. So another approximation for the number of primes up to some number k might be:

  2-1 3-1 5-1     p(k)-1
   2   3   5       p(k)
In fact, this is always an underestimate. It may still be an underestimate for all k if the above expression is multiplied by k. The multiplication seems to be intimately related to the Sieve of Eratosthenes, since the following process seems to apply:

Begin with k (or k2) consecutive integers, the smallest of which is 1.
Remove one from every two numbers (remove one from 1,2, another from 3,4, another from 5,6, etc.).
Remove one from every three numbers remaining.
Remove one from every five numbers remaining.
Etc. up to p(k).

The first step is like multiplying by 1/2. The second step is like multiplying by 2/3, the third like multiplying by 4/5, and so on. In the limit as k approaches infinity, k will have gone to infinity but the other terms will have multiplied together to get zero (since 1/zeta(1)=0). The number of primes *is* infinite, so this multiplication should be always increasing without bound.