At 09:40 PM 5/11/00 EDT, [EMAIL PROTECTED] wrote:
>This is totally weird, and I think I messed up in getting this, but it is
>true. Could someone with a lot of time and a lot of patience and a big
>calculator check this?
You can't prove it with a calculator; you need to do some algebra.
>1 = 1/(2^2- 1) + 1/(2^3 - 1) + 1/(2^4 - 1) + 1/(2^5 - 1) + ... + 1/(3^2 -
1) +
>1/(3^3 - 1) + 1/(3^4 - 1) + ... + 1/(5^2 - 1) + 1/(5^3 - 1) + 1/(5^4 - 1)
+ ...
>In other words, if set x contains integral powers, (2^3, 3^3, 2^2,
etc....) and
>x' is the set of integers minus set x. Then this says:
>1 = sum(n=2 to n=infinity (sum( i=1 to i = infinity, 1/((x'i)^n - 1))))
[I've replaced the non-ASCII characters used for exponents]
I think you simply mean 1 = sum 1/n, where n is in x'.
This result is a theorem due to a man better known for a conjecture:
Christian Goldbach. The proof that follows is based on one given in
_Concrete Mathematics_, by Graham, Knuth & Patashnik, where Goldbach's
theorem is exercise 2.53.
Let P be the set of perfect powers greater than 1, i.e.,
P = {m^n | m,n > 1}. Let Q be the integers greater than 1 not in P;
Q = {n>1 | n not in P}. The crucial fact we need is that every element
n of P can be expressed uniquely in the form q^k, where q is in Q and
k > 1, and, conversely, every such q^k is in P. (If you write n as a
product of prime powers p_i^a_i, then k is the GCD of the a_i.)
The sum in question is
Sum 1/(m - 1).
m in P
Expanding each term as a geometric series yields
Sum m^(-n).
m in P
n >= 1
Separating out the terms with n = 1, we get
Sum m^(-n) + Sum 1/m.
m in P m in P
n >= 2
But, from the fact above, this is the same as
Sum m^(-n) + Sum m^(-n).
m in P m in Q
n >= 2 n >= 2
Combining the sums again, we have
Sum m^(-n).
m >= 2
n >= 2
The terms involving a fixed m are a geometric series, so this is
Sum 1/(m*(m-1)).
m >= 2
Splitting into partial fractions yields a telescoping series:
Sum (1/(m-1) - 1/m) = (1 - 1/2) + (1/2 - 1/3) + ... = 1.
m >= 2
--
Fred W. Helenius <[EMAIL PROTECTED]>
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