1/1

^{s}+ (1 + k_{1})/2^{s}+ (1 + k_{2})/3^{s}+ (1 + k_{1})/4^{s}+ (1 + k_{3})/5^{s}+ {(1 + k_{1})(1 + k_{2})}/6^{s}+ …
= 1/{1 – (1 + k

_{1})/(2^{s}+ k_{1})} * 1/{1 – (1 + k_{2})/(3^{s}+ k_{2})} * 1/{1 – (1 + k_{3})/(5^{s}+ k_{3})} * …
where k

So the standard equation represents the situation where the values for k

_{1},_{ }k_{2}, k_{3}, …_{ }are rational numbers which can be either positive or negative.So the standard equation represents the situation where the values for k

_{1},_{ }k_{2}, k_{3}, …_{ }= 0_{.}
It can be
seen here it is the distinct prime factors of a number that determines the
nature of the sum over the integers expression.

So for
example, 6 is the 1st composite natural number with two distinct prime factors.
Therefore as we can see from the new formula we must here thereby multiply 1 + k

_{1 }(associated with 2) by 1 + k_{2}(associated with 3) respectively.
One very
simple case with respect to this general formula is for k

_{1}= 1 and k_{2}, k_{3}, k_{4}, … = 0
Then,

1/1

^{s}+ 2/2^{s}+ 1/3^{s}+ 2/4^{s}+ 1/5^{s}+ … = 1/{1 – 2/(2^{s}+ 1)} * 1/(1 – 1/3^{s}) * 1/(1 – 1/5^{s}) * …
So when
example s = 2, then

1/1

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

^{2}/24.
And again
in general terms where s = 2, 4, 6, … and where only a finite number of k

_{1}, k_{2}, k_{3}, … terms are given a rational value (with all others = 0) then both the sum over the integers and product over the primes expressions have a value of the form t/π^{s}(where t is a rational number).
A very
interesting case arises when k

_{1 }= k_{2}= k_{3 }… = 1 for all terms. then
1/1

^{s}+ 2/2^{s}+ 2/3^{s}+ 2/4^{s}+ 2/5^{s}+ 4/6^{s}+ 2/7^{s}+ 2/8^{s}+ 2/9^{s}+ 4/10^{s}+ …
= 1/{1 – 2/(2

^{s}+ 1)}* 1/{1 – 2/(3^{s}+ 1)} * 1/{1 – 2/(5^{s}+ 1)} * …
So for
example again when s = 2,

1/1

^{2}+ 2/2^{2}+ 2/3^{2}+ 2/4^{2}+ 2/5^{2}+ 4/6^{2}+ 2/7^{2}+ 2/8^{2}+ 2/9^{2}+ 4/10^{2}+ …
= 1/{1 –
2/(2

^{2}+ 1)}* 1/{1 – 2/(3^{2}+ 1)} * 1/{1 – 2/(5^{2}+ 1)} * 1/{1 – 2/(7^{2}+ 1)} …
= 5/3 *
10/8 * 26/24 * 50/48 * …

= (5/4 *
4/3) * (10/9 * 9/8) * (26/25 * 25/24) * (50/49 * 49/48) * …

= (4/3 * 9/8
* 25/24 * 49/48 * …) * (5/4 * 10/9 * 26/25 * 50/49 * …)

= (π

^{2}/6) * (15/π^{2}) = 15/6 = 5/2.
So,
interestingly in this particular case where k

_{1 }= k_{2}= k_{3 }… = 1, the value of the sum over the integers and corresponding product over primes expressions is a rational number.
And again
this will always be the case where s is an even integer = 2, 4, 6, ...

For
example, again where k

_{1 }= k_{2}= k_{3 }… = 1 and s = 4,
1/1

^{4}+ 2/2^{4}+ 2/3^{4}+ 2/4^{4}+ 2/5^{4}+ 4/6^{4}+ 2/7^{4}+ 2/8^{4}+ 2/9^{4}+ 4/10^{4}+ …
= 1/{1 –
2/(2

^{4}+ 1)} * 1/{1 – 2/(3^{4}+ 1)} * 1/{1 – 2/(5^{4}+ 1)} * 1/{1 – 2/(7^{4}+ 1)} …
= 17/15 *
82/80 * 626/624 * 2402/2400 * …

= (17/16 *
16/15) * (82/81 * 81/80) * (626/625 * 625/624) * (2402/2401 * 2401/2400) * …

= (16/15 *
81/80 * 625/624 * 2401/2400 * …) * (17/16 * 17/16 * 626/625 * 2402/2401 * …)

= (π

^{4}/90) * (105/π^{4}) = 105/90 = 7/6.
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