The identity
1 1 1 1
zeta(x) = 1 + ---- + ---- + ---- + ---- + ... =
x x x x
2 3 4 5
1 1 1 1
------------ * ------------ * ------------ * ------------ * ...
1 1 1 1
( 1 - ---- ) ( 1 - ---- ) ( 1 - ---- ) ( 1 - ---- )
x x x x
2 3 5 7
is very important, because it relates one series based on all the positive integers in order to another series, this time a series product instead of a sum, which involves only the sequence of prime numbers.
The prime numbers have long been a fascinating part of mathematics, and they are chiefly studied in the part of mathematics known as number theory.
Having the above equation available to us lets us learn things about number theory in general, and the prime numbers in particular, using techniques borrowed from the very rich area of mathematics known as analysis, which includes calculus and differential equations, and which is usually used for very ordinary and practical concerns.
We will not need to voyage through a large area of mathematics to see the truth of the equation above. Using the fact that every integer greater than or equal to 2 can be expressed as the product of primes in exactly one way, and some elementary algebra, this beautiful result can be shown to be true.
First, however, we do need one result that could be obtained by using a Taylor series. However, we can also obtain that result a simpler algebraic result, known as the binomial theorem, as well.
Can we express 1/(1-x) as a power series in x?
The decimal expansion of 1/9 is .111111.... . Thus, 1/9 = 1/10 + 1/100 + 1/1000 + ... . Why is this?
What is 1/9 - 1/10? It is 10/90 - 9/90, or 1/90. What is 1/90 - 1/100? Clearly, it must be 1/900, since we're dealing with the same equation, but with everything divided by ten.
In general, since
1/(n-k) - 1/n
equals
(n - (n-k))/((n-k)n)
by the usual rule for cross-multiplying when adding and subtracting fractions, which equals
k/((n-k)n)
when we add together the variables in the numerator, which also equals
k/(n^2-nk)
when we multiply out the denominator, and which finally gives us
(k/n)(1/(n-k))
by dividing n out of n^2-nk, since
1/(n-k) - 1/n = (k/n)(1/(n-k))
is now seen to be true, then
1/(n-k) = 1/n + (k/n)(1/(n-k)).
This equation is important, because 1/(n-k) appears on both sides of it. So we can use that equation over and over again, like this:
1/(n-k) = 1/n + (k/n)(1/n + (k/n)(1/(n-k)))
1/(n-k) = 1/n + (k/n)(1/n + (k/n)(1/n + (k/n)(1/(n-k))))
and so on and so forth, which means that we have an infinite series in powers of k/n for 1/(n-k):
1/(n-k) = 1/n * (1 + (k/n) + (k/n)^2 + (k/n)^3 + (k/n)^4 + ... )
Since this means we never have n-k in the denominator, but only n, if we apply this formula to all the factors in the infinite product for the Riemann Zeta Function, we appear, in one way, to approach more closely to the infinite sum for the Riemann Zeta Function.
That is,
1 1 1 1
------------ * ------------ * ------------ * ------------ * ...
1 1 1 1
( 1 - ---- ) ( 1 - ---- ) ( 1 - ---- ) ( 1 - ---- )
x x x x
2 3 5 7
becomes
1 1 2 1 3 1 4 1 5
( 1 + ---- + (----) + (----) + (----) + (----) + ... ) *
x x x x x
2 2 2 2 2
1 1 2 1 3 1 4 1 5
( 1 + ---- + (----) + (----) + (----) + (----) + ... ) *
x x x x x
3 3 3 3 3
1 1 2 1 3 1 4 1 5
( 1 + ---- + (----) + (----) + (----) + (----) + ... ) *
x x x x x
5 5 5 5 5
1 1 2 1 3 1 4 1 5
( 1 + ---- + (----) + (----) + (----) + (----) + ... ) * ...
x x x x x
7 7 7 7 7
But this colossal expression is now infinite in two directions! Can we deal with this? Yes, we can. When we multiply two expressions, like a+b+c and d+e+f, the result is the sum of every possible combination of one term in the first expression and one term in the second expression. So, here we have nine terms.
Similarly, then, the above expression can be changed from an infinite product of infinite sums to an infinite sum of infinite products.
Thus, by choosing to take products from it in the right order, we get:
(1*1*1*1...)+
1
(----*1*1*1...)+
x
2
1
(----*1*1*1...)+
x
3
1 2
((----) *1*1*1...)+
x
2
1
(----*1*1*1...)+
x
5
1 1
(----*----*1*1...)+...
x x
2 3
and since every integer can be expressed as the product of primes in one and only one way, this will indeed work out neatly, becoming exactly equal to our first expression for the Riemann Zeta Function,
1 1 1 1 1
1 + ---- + ---- + ---- + ---- + ---- + ...
x x x x x
2 3 4 5 6
Now, if you happened to look at my page on Infinite Cardinals, you may think that I have pulled a fast one here. This sum corresponds only to those infinite products that, after a finite number of factors, become nothing but 1*1*1*1... . What about the much larger number of infinite products that never do this?
Well, those products, since they are the products of an infinite number of factors, each of which is less than 1 by a definite amount, along with other stuff that is either 1 or also less than 1, if not by as much, must all be equal to zero.