slightly OT: harmonic series revisited...

[Max]

santi_spadaro at virgilio.it wrote:
> Anybody knows an answer (and a neat way to show that the answer is 
> true)?
> 
> "Define a_n = 1/n if n is composite and a_n = -(1/n) if n is 
> prime. Does the series of a_n (sum from n to infinity of a_n) diverges?"
> 

Sure it does.

Let's consider a sum of the first pk terms of an where pk is the k-th prime
and write this sum as
Sum[n=1..pk] an = S(k) + T(k) + C(k)

where

S(k) = - 1/p1 - 1/p2 - ... - 1/pk

T(k) = 1/(p1+1) + 1/(p2+1) + ... + 1/(pk+1)

C(k) = the rest

Claims:

1) S(k)+T(k) has a finite limit as k goes to infinity
Since
-1/p + 1/(p+1) ~= -1/p^2
and sum of reciprocals of squares converges.


2) C(k) >= H(pk) - H(2k)
where H() stands for harmonic numbers
We know that H(m) ~= ln(m) and that k = pi(pk) ~= pk/ln(pk), hence,
C(k)~ >= ln(pk) - ln(2k) ~= ln(pk) - ln(2pk/ln(pk)) = ln(ln(pk)) - ln(2)
which tends to infinity.

Therefore,
Sum[n=1..m] an >= ln(ln(m)) + C
where C is some constant.

This sketch is very informal but it can be formalized up to a complete proof.

Max