(In reply to

Answer by K Sengupta)

We know that the nth harmonic number is given asymptotically by:

H(n) ~ ln n + gamma + (2n)^-1 - 1/12*(n^-2) + (n^-4)/120 - (n^-6)/252 + ......, where gamma represents Euler-Mascheroni constant

(Reference:

http://mathworld.wolfram.com/HarmonicNumber.html )

Then, it follows that:

H(2n) - H(n)

-> ln (2n) - ln(n) = ln 2, as n -> infinity

or, 1/(n+1)+1/(n+2)+1/(n+3)+...+1/(2n) -> ln 2 as n -> infinity

*Edited on ***August 28, 2008, 1:32 pm**