
Using this proof, you can double any number without changing its value!

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.
log 1 = 0, because e^{0}=1
log 2 = log 1, because both logs are equal to 0
e^{log 2} = e^{log 1}, taking the exponent of both sides
2 = 1, because e^{log a} = a
I got an email questioning the statement that log 2 = 11/2+1/31/4+..., 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)=11/2+1/31/4 to Nicolaus Mercator and gives dozens of other formulas for log(2).
http://www.mathematik.unibielefeld.de/~sillke/PUZZLES/harmonicseries 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/(2n1)  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).
Or Prop A directly:
H(2n) = 1 + 1/2 + 1/3 + 1/4 + ... + 1/(2n1) + 1/(2n)
H(n) = 0/1 + 2/2 + 0/3 + 2/4 + ... + 0/(2n1) + 2/(2n)
H(2n)H(n) = 1  1/2 + 1/3  1/4 + ... + 1/(2n1)  1/(2n)Prop B: H(2n)  H(n) is monotonically 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(2n1)H(n1) = H(2n)H(n)+1/(2n).
Now Integral[1/x, x, {n,2n}] = Integral[1/y, y, {1,2}] via y = nx. (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, 2003, Department of Mathematics and Computer Science, North Carolina School of Science and Mathematics
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