So, 1/1

^{s }+ 1/2

^{s }+ 1/3

^{s }

^{ }+.... = 1/(1 – 1/2

^{s}) * 1/(1 – 1/3

^{s}) * 1/(1 – 1/5

^{s}) *....

Thus for example in the simple case, where s = 2,

1/1

^{2 }+ 1/2^{2 }+ 1/3^{2 }^{ }+... = 1/(1 – 1/2^{2}) * 1/(1 – 1/3^{2}) * 1/(1 – 1/5^{2}) *...., so that,
1 + 1/4 + 1/9 + ... = 4/3 * 9/8 * 25/24 * ... = π

However when we remove the even terms (in the sum over natural numbers) we get a new expression that is equally a product over primes (excluding the 1st prime, i.e. 2)

Thus, 1/1

1 + 1/9 + 1/25 + ... = 9/8 * 25/24 * 49/48 * ... = π

It only recently struck me that this in fact represents a specific case of a more general phenomenon.

So just as we can derive a new (more truncated) expression for the natural numbers over the primes through the omission of the 1st term (i.e. 4/3) in the product over primes, equally we can achieve a corresponding (truncated) sum over the natural numbers, when we omit any of the other terms in the product over primes expression.

The key here is that when we remove a term for a particular value of p (in the product over primes) we then remove in corresponding fashion each (ordinal) term in the sum over natural numbers expression that is exactly divisible by p.

Thus when p = 2, as we have seen, we omit the even terms (in the sum over natural numbers expression) which - by definition - are all divisible by 2!

Therefore to obtain the corresponding expression (in the sum over natural numbers) that corresponds to the product over primes formula (with 9/8 - relating to p = 3 omitted) we remove all ordinal terms that are divisible by 3.

In other words we systematically omit each 3rd terms (i.e. 3rd, 6th, 9th etc).

Therefore 1/1

And because the number of terms in the product over primes is potentially unlimited, this means that the corresponding expressions for the (truncated) sums over natural numbers are likewise potentially unlimited.

Of course, it should be apparent that now that we have expressions for the sum over all natural numbers and equally the reduced sum of natural terms (with terms divisible by 3 omitted) that we we can now likewise provide through subtraction of both results, an expression here for terms that are divisible by 3.

So 1/3

In this case the product over primes expression = 4/3 * 1/8 * 25/24 * .....

So we have the same terms (as in the full product over primes expression with the exception of the terms involving p = 3 (where 9/8 is now replaced by 1/8).

And though we have illustrated here exclusively for value of s = 2, this can then be extended to all other values of s (real and complex) except 1.

And for all even integer values of s, both the resulting (reduced) product over primes and sum over natural numbers expressions can be expressed as π

And then as well as omitting terms (in the product over primes expression) in an individual manner, we can omit terms in any combination we choose, while preserving a corresponding fashion a unique reduced expression for the sum over the natural numbers.

Thus for example (again with reference to s = 2) when we now omit the first two terms i.e. 4/3 and 9/8 in the product over primes (which relate to p = 2 and p = 3 respectively), we then in corresponding fashion, omit all those terms in the sum over natural numbers expression that are divisible by either 2 or 3.

In fact this means that up to 50 besides all the prime numbered terms, we would only include the 25th, 35th and 49 terms (which are neither divisible by 2 or 3)!

So 1 + 1/5

And once again these reduced results involving the omission of selected combinations (in the product over primes expression) give rise to corresponding unique reduced expressions (for the sum over natural numbers) for all values of s (except 1).

Also through subtraction, we can provide a corresponding expression for terms divisible by both 2 and 3.

So 1/2

^{2}/6 (= 1.6449...)However when we remove the even terms (in the sum over natural numbers) we get a new expression that is equally a product over primes (excluding the 1st prime, i.e. 2)

Thus, 1/1

^{2 }+ 1/3^{2 }+ 1/5^{2 }^{ }+... = 1/(1 – 1/3^{2}) * 1/(1 – 1/5^{2}) * 1/(1 – 1/7^{2}) * ... so that,1 + 1/9 + 1/25 + ... = 9/8 * 25/24 * 49/48 * ... = π

^{2}/8 (= 1.2337...)It only recently struck me that this in fact represents a specific case of a more general phenomenon.

So just as we can derive a new (more truncated) expression for the natural numbers over the primes through the omission of the 1st term (i.e. 4/3) in the product over primes, equally we can achieve a corresponding (truncated) sum over the natural numbers, when we omit any of the other terms in the product over primes expression.

The key here is that when we remove a term for a particular value of p (in the product over primes) we then remove in corresponding fashion each (ordinal) term in the sum over natural numbers expression that is exactly divisible by p.

Thus when p = 2, as we have seen, we omit the even terms (in the sum over natural numbers expression) which - by definition - are all divisible by 2!

Therefore to obtain the corresponding expression (in the sum over natural numbers) that corresponds to the product over primes formula (with 9/8 - relating to p = 3 omitted) we remove all ordinal terms that are divisible by 3.

In other words we systematically omit each 3rd terms (i.e. 3rd, 6th, 9th etc).

Therefore 1/1

^{2 }+ 1/2^{2 }+ 1/4^{2 }+ .... = 4/3 * 25/24 * 49/47 * ... = 4π^{2}/27 (= 1.4621...).And because the number of terms in the product over primes is potentially unlimited, this means that the corresponding expressions for the (truncated) sums over natural numbers are likewise potentially unlimited.

Of course, it should be apparent that now that we have expressions for the sum over all natural numbers and equally the reduced sum of natural terms (with terms divisible by 3 omitted) that we we can now likewise provide through subtraction of both results, an expression here for terms that are divisible by 3.

So 1/3

^{2 }+ 1/6^{2 }+ 1/9^{2 }+... = π^{2}/6 – 4π^{2}/27 = π^{2}/54.In this case the product over primes expression = 4/3 * 1/8 * 25/24 * .....

So we have the same terms (as in the full product over primes expression with the exception of the terms involving p = 3 (where 9/8 is now replaced by 1/8).

And though we have illustrated here exclusively for value of s = 2, this can then be extended to all other values of s (real and complex) except 1.

And for all even integer values of s, both the resulting (reduced) product over primes and sum over natural numbers expressions can be expressed as π

^{s}/k (where k is a natural number).And then as well as omitting terms (in the product over primes expression) in an individual manner, we can omit terms in any combination we choose, while preserving a corresponding fashion a unique reduced expression for the sum over the natural numbers.

Thus for example (again with reference to s = 2) when we now omit the first two terms i.e. 4/3 and 9/8 in the product over primes (which relate to p = 2 and p = 3 respectively), we then in corresponding fashion, omit all those terms in the sum over natural numbers expression that are divisible by either 2 or 3.

In fact this means that up to 50 besides all the prime numbered terms, we would only include the 25th, 35th and 49 terms (which are neither divisible by 2 or 3)!

So 1 + 1/5

^{2}+ 1/7^{2}+ ...... = 25/24 * 49/47 * 121/120 * .... = π^{2}/9 (= 1.0966...).And once again these reduced results involving the omission of selected combinations (in the product over primes expression) give rise to corresponding unique reduced expressions (for the sum over natural numbers) for all values of s (except 1).

Also through subtraction, we can provide a corresponding expression for terms divisible by both 2 and 3.

So 1/2

^{2}+ 1/3^{2}+ 1//4^{2 }+ 1//6^{2}... = π^{2}/6 – π^{2}/9 = π^{2}/18,
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