--- In

primenumbers@yahoogroups.com, Peter Kosinar <goober@...> wrote:

>

> > Fine trick, but what makes you sure the integral is not e.g. =

0.008?

> > Didn't you only shift the problem upon the integral?

>

> The integral can be evaluated explicitly, exactly as I did in my

original

> post. I just didn't find this method "nice trick"-y enugh, as it

required

> adding 168 terms first (as opposed to the two terms in the other

sum).

> It's easy to check that the integral int(log(t)/t^2, t=A..infinity)

is

> equal to (ln(A)+1)/A.

>

> Adam first added all the terms corresponding to primes smaller than

1000,

> which resulted in roughly 0.4921003 and then bounded the remaining

terms

> by the value of the integral corresponding to p=1009 (which means

> A = p-1 = 1008). It's easy to see that (ln(1008)+1)/1008 < 0.007853.

> These two values add up to 0.4999533, which is strictly less than

0.5.

>

> Peter

You are quite right. I integrated only numerically. Everything clear.

Thanks to you both.

Werner

>