Goldbach–Euler theorem

In mathematics, the Goldbach–Euler theorem (also known as Goldbach's theorem), states that the sum of 1/(p − 1) over the set of perfect powers p, excluding 1 and omitting repetitions, converges to 1:

\sum_{p}\frac{1}{p-1}= {\frac{1}{3} %2B \frac{1}{7} %2B \frac{1}{8}%2B \frac{1}{15} %2B \frac{1}{24} %2B \frac{1}{26}%2B \frac{1}{31}}%2B \cdots = 1.

This result was first published in Euler's 1737 paper "Variae observationes circa series infinitas". Euler attributed the result to a letter (now lost) from Goldbach.

Proof

Goldbach's original proof to Euler involved assigning a constant to the harmonic series: \textstyle x = \sum_{n=1}^\infty \frac{1}{n}\ , which is divergent. Such a proof is not considered rigorous by modern standards. It is also interesting to note that there is a strong resemblance between the method of sieving out powers employed in his proof and the method of factorization used to derive Euler's product formula for the Riemann zeta function.

Let x be given by

x = 1 %2B \frac{1}{2} %2B \frac{1}{3} %2B \frac{1}{4} %2B \frac{1}{5} %2B \frac{1}{6} %2B \frac{1}{7} %2B \frac{1}{8} \cdots

Since the sum of the reciprocal of every power of two is \textstyle 1 = \frac{1}{2} %2B \frac{1}{4} %2B \frac{1}{8} %2B \frac{1}{16} %2B \cdots, subtracting the terms with powers of two from x gives

x - 1 = 1 %2B \frac{1}{3} %2B \frac{1}{5} %2B \frac{1}{6} %2B \frac{1}{7} %2B \frac{1}{9} %2B \frac{1}{10} %2B \frac{1}{11} %2B \cdots

Repeat the process with the terms with the powers of three: \textstyle \frac{1}{2} = \frac{1}{3} %2B \frac{1}{9} %2B \frac{1}{27} %2B \frac{1}{81} %2B \cdots

x - 1 - \frac{1}{2} = 1 %2B \frac{1}{5} %2B \frac{1}{6} %2B \frac{1}{7} %2B \frac{1}{10} %2B \frac{1}{11} %2B \frac{1}{12} %2B \cdots

Absent from the above sum are now all terms with powers of two and three. Continue by removing terms with powers of 5, 6 and so on until the right side is exhausted to the value of 1. Eventually, we obtain the equation

x - 1 - \frac{1}{2} - \frac{1}{4} - \frac{1}{5} - \frac{1}{6} - \frac{1}{9} - \cdots = 1

which we rearrange into

x - 1 = 1 %2B \frac{1}{2} %2B \frac{1}{4} %2B \frac{1}{5} %2B \frac{1}{6} %2B \frac{1}{9} %2B \cdots

where the denominators consist of all positive integers that are the non-powers minus one. By subtracting the previous equation from the definition of x given above, we obtain

1 = \frac{1}{3} %2B \frac{1}{7} %2B \frac{1}{8}%2B \frac{1}{15} %2B \frac{1}{24} %2B \frac{1}{26}%2B \frac{1}{31} %2B \cdots

where the denominators now consist only of perfect powers minus one.

While mathematically incorrect, Goldbach's proof provides a reasonably intuitive visualization of the problem. Rigorous proofs require proper and more careful treatment of the divergent terms of the harmonic series. Other proofs make use of the fact that the sum of 1/p over the set of perfect powers p, excluding 1 but including repetitions, converges to 1 by demonstrating the equivalence:

\sum_{p}\frac{1}{p - 1} = \sum_{m=2}^\infty \sum_{n=2}^\infty \frac{1}{m^n} = 1.

See also

References