# How to prove that $\sum\limits_{k = 0}^\infty\frac{(- 1)^k}{(2k + 1)^n}$ equals a certain infinite product

I'm trying to see how to get from

$$A = 1 - \frac{1}{3^n} + \frac{1}{5^n} - \frac{1}{7^n} + \frac{1}{9^n} - \cdots$$

to

$$A = \frac{3^n}{3^n + 1} \cdot \frac{5^n}{5^n - 1} \cdot \frac{7^n}{7^n + 1} \cdot \frac{11^n}{11^n + 1} \cdot \frac{13^n}{13^n - 1} \cdot \frac{17^n}{17^n - 1} \cdots$$

This equality stems from this post in Gaussianos, so if you can read Spanish it will help. I'm pretty clueless. This gives a proof a la Euler that we can write $\pi$ as

$$\pi = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} - \frac{1}{{10}} + \frac{1}{{11}} + \frac{1}{{12}} - \frac{1}{{13}} + \ldots$$

where the signs are as follows:

1. We leave a $+$ if the denominator is a prime of the form $4m-1$. We also leave a $+$ for $2$.
2. We change to $-$ if the denominator is a prime of the form $4m+1$.
3. If the number is composite, we calculate the product of the signs of the previous rules to obtain the corresponding sign.
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Define $\chi(m)$ to be equal to $1$ if $m\equiv1\pmod 4$, equal to $-1$ if $m\equiv3\pmod 4$, and $0$ if $m$ is even. Then $\chi$ is a completely multiplicative function. The general theory of Euler products then gives us the identity $\sum_{m=1}^\infty \chi(m)/m^z = \prod_p ( 1-\chi(p)/p^z)^{-1}$ that you want, as long as $\Re z > 1$. Convergence is certainly an issue, especially right at $z=1$, but you can find the details by looking up references on Dirichlet series and their Euler products. – Greg Martin May 2 '12 at 4:02
@GregMartin Now that seems awfully interesting! Thanks! – Pedro Tamaroff May 2 '12 at 19:59
@Michael Hardy You really can't stand bad $\LaTeX$, can't you? =) – Pedro Tamaroff May 2 '12 at 20:54
@PeterTamaroff : This one seemed like an attempt at parodying the things I mentioned earlier. – Michael Hardy May 2 '12 at 23:24
@MichaelHardy My comment or the question's formatting? The comment is meant with no harm Michael =), and the question, well, that's how I format. – Pedro Tamaroff May 2 '12 at 23:26

Note that we can write $\displaystyle A = 1 - \frac1{3^n} + \frac1{5^n} - \frac1{7^n} + \frac1{9^n} - \frac1{11^n} + \frac1{13^n} - \cdots$ as $$A = \left(1 - \frac1{3^n} + \left(\frac1{3^n} \right)^2 - \cdots \right) \left(1 + \frac1{5^n} + \left(\frac1{5^n} \right)^2 + \cdots \right)\left(1 - \frac1{7^n} + \left(\frac1{7^n} \right)^2 - \cdots \right) \times$$ $$\left(1 - \frac1{11^n} + \left(\frac1{11^n} \right)^2 - \cdots \right) \left(1 + \frac1{13^n} + \left(\frac1{13^n} \right)^2 + \cdots \right)\cdots$$ Since every odd number of the form $4k+3$ must have odd number of prime factors of the form $4k+3$ counted with multiplicity and every every odd number of the form $4k+1$ must have even number of prime factors of the form $4k+3$ counted with multiplicity.

EDIT

We will now prove what you have stated for $n>1$.

If we let $$F(n) = 1 - \frac1{3^n} + \frac1{5^n} - \frac1{7^n} + \frac1{9^n} - \frac1{11^n} + \frac1{13^n} - \frac1{15^n} + \frac1{17^n}\cdots,$$ then $$\frac{F(n)}{3^n} = \frac1{3^n} - \frac1{9^n} + \frac1{15^n} - \frac1{21^n} + \frac1{27^n} - \frac1{33^n} + \frac1{39^n} - \frac1{45^n} + \frac1{51^n}\cdots$$

Hence, $$F_3(n) = F(n) \left( 1 + \frac1{3^n} \right) = F(n) + \frac{F(n)}{3^n} = 1 + \frac1{5^n} - \frac1{7^n} - \frac1{11^n} + \frac1{13^n} + \frac1{17^n} - \frac1{19^n} - \cdots$$

By doing the above, we have now removed all the multiples of $3$. Note that while adding we are allowed to rearrange and add since we have assumed $n>1$ and we know that for $n>1$, $F(n)$ converges absolutely. Repeat the process for $F_3(n)$ using the rest of the primes by adding or subtracting depending on whether the prime is $\pm 1 \bmod 4$.

For instance, $$\frac{F_3(n)}{5^n} = \frac1{5^n} + \frac1{{25}^n} - \frac1{{35}^n} - \frac1{{55}^n} + \frac1{{65}^n} + \frac1{{85}^n} - \frac1{{95}^n} - \cdots$$

Subtract the above from $F_3(n)$ to get $$F_5(n) = F_3(n) \left(1 - \frac1{5^n} \right)$$ where $F_5(n)$ has all multiples of $3$ and $5$ removed.

This can be done over all the primes. Note the adding or subtracting depends on whether the prime is $\pm 1 \mod 4$. This is due to the following reason.

If $\displaystyle (4k+1) = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m} q_1^{\beta_1} q_2^{\beta_2} \cdots q_n^{\beta_n}$, where $p_i \equiv 1 \bmod 4$, $q_j \equiv 3 \bmod 4$, then $\beta_1 + \beta_2 + \cdots + \beta_n$ is even. (Why?)

If $\displaystyle (4k+3) = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m} q_1^{\beta_1} q_2^{\beta_2} \cdots q_n^{\beta_n}$, where $p_i \equiv 1 \bmod 4$, $q_j \equiv 3 \bmod 4$, then $\beta_1 + \beta_2 + \cdots + \beta_n$ is odd. (Why?)

Hence, $$\displaystyle \frac{1}{4k+1} = \frac1{p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m} q_1^{\beta_1} q_2^{\beta_2} \cdots q_n^{\beta_n}} = \frac1{p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m} (-q_1)^{\beta_1} (-q_2)^{\beta_2} \cdots (-q_n)^{\beta_n}}$$

Similarly, $$\displaystyle \frac{-1}{4k+3} = \frac1{p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m} q_1^{\beta_1} q_2^{\beta_2} \cdots q_n^{\beta_n}} = \frac1{p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m} (-q_1)^{\beta_1} (-q_2)^{\beta_2} \cdots (-q_n)^{\beta_n}}$$

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I don't follow why the equality follows. Maybe a small aisled example can help. – Pedro Tamaroff May 2 '12 at 0:01
@PeterTamaroff Each term within $()$ is a geometric progression which sums up to what you want. – user17762 May 2 '12 at 0:03
I know what goes on after, but not why the first equality follows. I see there is some number theory going on, so since I'm not illustrated in that, I need a little more information. – Pedro Tamaroff May 2 '12 at 0:09
@PeterTamaroff I will add some more details. – user17762 May 2 '12 at 0:14
@Peter Imagine expanding Marvis's product. Each term from an expansion would choose the "$1$" from all but finitely many factors. The non-$1$ selections from each factor would multiply together and correspond in a one-to-one manner with a terms from your original sum. (However this is not a rigorously proven equality without showing absolute convergence, which is not even true for $n=1$.) – alex.jordan May 2 '12 at 0:17