This proof is extremely important, because it can be used to show that
2=1. It works this way:
log 2 = 0, from the proof, above.
I got an email questioning the statement that log 2 = 1-1/2+1/3-1/4+..., so I
found these references to support this fact, which, by the way, has nothing to
do with the error in the proof, above.
http://numbers.computation.free.fr/Constants/Log2/log2Formulas.html
attributes log(2)=1-1/2+1/3-1/4 to Nicolaus Mercator and gives dozens of other formulas for log(2).
http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/harmonic-series
gives a sketch of the proof that this series is the integral of dy/y from 1 to 2, which is log(2).
H(n) = 1 + 1/2 + 1/3 + ... + 1/n.
H(2n) - H(n) = 1/(n+1) + 1/(n+2) + ... + 1/(2n).
Prop A: H(2n) - H(n) = 1 - 1/2 + 1/3 - 1/4 ... + 1/(2n-1) - 1/(2n)
Follows from Induction: n=0 : H(0) - H(0) = 0
n->n+1: H(2(n+1))-H(n+1) =
H(2n) + 1/(2n+1) + 1/(2n+2) - H(n) - 1/(n+1) =
H(2n) - H(n) + 1/(2n+1) - 1/(2n+2) =
1 - 1/2 + 1/3 - 1/4 ... + 1/(2n+1) - 1/(2n+2).
Prop B: H(2n) - H(n) is monoton increasing and converges to log(2).
The limit of 1 - 1/2 + 1/3 - 1/4 ... was first calculated by Nicolaus Mercator [Logarithmotechnia,
1667]. He found the series
log(1+x) = x - x^2/2 + x^3/3 - x^4/4 ... But you can do that without using the
Taylor series.
Adding the estimates 1/(n+1) < Integral[1/x, x, {n,n+1}] < 1/n gives
H(2n)-H(n) < Integral[1/x, x, {n,2n}] < H(2n-1)-H(n-1) =
H(2n)-H(n)+1/(2n).
Now Integral[1/x, x, {n,2n}] = Integral[1/y, y, {1,2}] via y = n x. (An
interpretation of this transform gives [Yegorov].)
So we get log(2) - 1/(2n) < H(2n) - H(n) < log(2).
As log(t) is defined as the Integral[1/y, y, {1,t}] see [Yegorov].
Regrouping the
Harmonic Series, by Dan Teague, 2001, Department of Mathematics and Computer Science,
North Carolina School of Science and Mathematics
