Prove that you can write:
sum(n=1 to infinity) 1/n^s = (a/b)π^s,
where a/b is some rational number.
sum(n=1 to infinity) 1/n^s = (a/b)π^s,
where a/b is some rational number.
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This is possibly false in general (though I know of no *proof* of this), and certainly an open question at best: consider s = 3. As you may be aware, this series serves as the definition of the Riemann zeta function [1] z(s) = sum[n=1:inf] n^(s) for Re(s) > 1 [1]. [1] offers that z(3) ≈ 1.20207, a value known as Apéry's constant [2]. Though Apéry's constant is known to be irrational, it is *not* known to be transcendental [3], as any rational multiple of π^3 must be!
For even integers s, this is true. The expression of z(2n) in forms of Bernoulli numbers [4] establishes this, since it is clear from the explicit definition [5] that each B(n) is a sum of products of rational numbers and consequently rational, and further, we may write:
z(2n) = (-1)^(n+1) B(2n)(2π)^(2n)/[2(2n)!]
= (-1)^(n+1) B(2n) * 2^(2n-1)/(2n)! * π^(2n)
where (-1)^(n+1)B(2n)*2^(2n-1)/(2n)! is a product of rationals and thus rational.
For more information, you'll need to tease it out of the basic references I've provided. I'm no number theorist, as I'm sure I've pointed out to you before. And the ideas you're probing here are a bit...more complicated...than the elementary calculus-type approach you may be looking for.
For even integers s, this is true. The expression of z(2n) in forms of Bernoulli numbers [4] establishes this, since it is clear from the explicit definition [5] that each B(n) is a sum of products of rational numbers and consequently rational, and further, we may write:
z(2n) = (-1)^(n+1) B(2n)(2π)^(2n)/[2(2n)!]
= (-1)^(n+1) B(2n) * 2^(2n-1)/(2n)! * π^(2n)
where (-1)^(n+1)B(2n)*2^(2n-1)/(2n)! is a product of rationals and thus rational.
For more information, you'll need to tease it out of the basic references I've provided. I'm no number theorist, as I'm sure I've pointed out to you before. And the ideas you're probing here are a bit...more complicated...than the elementary calculus-type approach you may be looking for.