I was playing on with the infinite product of the sine: $$\sin(x)=x\prod_{k=1}^\infty (1-\frac{x^2}{k^2\pi^2})$$
If we try expand this product, we will get the sum of reciprocals of squares of natural numbers and some others series: $$\sin(x)=x(1-\frac{x^2}{\pi^2}\sum_{n=1}^\infty \frac{1}{n^2}+junk)$$
I thought a scheme:
| Term | 1st factor | 2nd factor | 3rd factor | ... | Nth factor |
|---|---|---|---|---|---|
| 1st term | 1 | 1 | 1 | ... | 1 |
| 2nd term | $-\frac{x^2}{\pi^2}$ | $-\frac{x^2}{4\pi^2}$ | $-\frac{x^2}{9\pi^2}$ | ... | $-\frac{x^2}{N^2\pi^2}$ |
It's like this: first we do the product of all the columns of the first row. Then we need to get all the combinations which have ONE column of the second row, so we will have for the second term $-x^2\sum_{n=1}^N \frac{1}{n^2\pi^2}$. The third term will be the sum of all combinations, but with TWO columns of the second row, like this: $x^4\sum_{n=1, k=n+1}^\infty \frac{1}{n^2k^2\pi^4}$, and so on. So, aplying the limit as $N\rightarrow\infty$, the final product is like:
$$x\prod_{k=1}^\infty (1-\frac{x^2}{k^2\pi^2})=x-x^3\sum_{n=1}^\infty \frac{1}{n^2\pi^2}+x^5\sum_{n=1,k=n+1}^\infty \frac{1}{n^2k^2\pi^4}-...$$
But, I want to generalize a product like $\prod_{n=1}^\infty (1-\frac{x}{a_n})$, but I don't know how to manipulate the sum $\sum_{n_1 = 1, n_k = n_{k-1}+1, k = 2,3,...,m}^\infty\prod_{k=1}^m\frac{1}{n_k}$. So, if someone can help me, I'll be grateful.
AN APPROACH:
If we have a function $f$ defined by this way: $$f(x)=\prod_{n=1}^\infty (1-\frac{x}{a_n})$$ We have that the series of f at 0 equal this expression: $$f(x)=1-x\sum_{n=1}^\infty \frac{1}{a_n}+x^2\sum_{n=1}^\infty \sum_{k=n+1}^\infty \frac{1}{a_n a_k} -x^3\sum_{n=1}^\infty \sum_{k=n+1}^\infty \sum_{m=k+1}^\infty \frac{1}{a_n a_k a_m} +...$$ If we develop these functions, we can solve for all functions of the type above:
$$b_k(n)=\sum_{m=1}^n \frac{b_{k-1}(m)}{a_m}, b_1(n)=\sum_{m=1}^n \frac{1}{a_m}$$
For $a_n=\frac{1}{n^2\pi^2}$, we have that $f(x^2)=\frac{\sin(x)}{x}$, and $b_k(n)=\sum_{m=1}^n \frac{b_{k-1}(m)}{\pi^2 m^2}$