TITLE: How to prove that $\sum\limits_{k = 0}^\infty\frac{(- 1)^k}{(2k + 1)^n}$ equals a certain infinite product
QUESTION [5 upvotes]: 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:

We leave a $+$ if the denominator is a prime of the form $4m-1$. We also leave a $+$ for $2$.
We change to $-$ if the denominator is a prime of the form $4m+1$.
If the number is composite, we calculate the product of the signs of the previous rules to obtain the corresponding sign.

REPLY [5 votes]: 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,
$$\displaystyle \frac{-1}{4k+3} = \frac1{p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m} q_1^{\beta_1} q_2^{\beta_2} \cdots q_n^{\beta_n}} = \frac1{p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m} (-q_1)^{\beta_1} (-q_2)^{\beta_2} \cdots (-q_n)^{\beta_n}}$$