(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
Puzzle Solution
