Convergence of a power series and interchanging order of summation and integration. If we have the expression $$f(x) = \frac{1}{(2\pi i)^n} \int_{\gamma_1} \int_{\gamma_2} \cdots \int_{\gamma_n} f(z) \prod_{j=1}^n \left( \sum_{i=0}^{\infty} \frac{x_j^i}{z_j^{i+1}} \right)dz_j.$$ How would I show that the series $\prod_{j=1}^{n} \sum_{i=0}^{\infty} \frac{x_j^i}{z_j^{i+1}}dz_j$ converges absolutely, such that I can write $f(x)$ as $$f(x) = \frac{1}{(2\pi i)^n} \sum_{i=0}^{\infty} \prod_{j=1}^n \int_{\gamma_j} \frac{f(z)}{z_j^{i+1}}dz_j?$$
Note that this is a small argument that makes up a larger proof that I'm working on.
 A: Assuming we are working on disks, that $|x_j|\leq r_j < |z_j|=R_j$
and $|f(z)|\leq M$ on the product disk,
it suffices to show absolute convergence. We have
$$ \sum_{k\geq 0} \left|  \frac{x_j^k}{z_j^{k+1}} \right| \leq
\sum_{k\geq 0} \frac{r_j^k}{R_j^{k+1}}  = \frac{1}{R_j-r_j} $$
and then
$$  |f(z)| \prod_{j=1}^n \sum_{k\geq 0} \left|\frac{x_j^k}{z_j^{k+1}} \right| \leq M \prod_{j=1}^n \frac{1}{R_j-r_j} $$
Being absolutely convergent the product sum is well-defined and you may permute the order of taking sums/products as it pleases you most. The last expression you wrote, however, is wrong as the index $i$ must depend upon $j$. More precisely we have:
$$  \int_{\gamma_1} \cdots \int_{\gamma_n} f(z)\prod_{j=1}^n \left( \sum_{k\geq 0}  \frac{x_j^k}{z_j^{k+1}} \right)\frac{dz_j}{2\pi i}
 = \sum_{k_1\geq 0} \cdots \sum_{k_n\geq 0} 
\int_{\gamma_1} \cdots \int_{\gamma_n}
 f(z) \left( \prod_{j=1}^n 
 \frac{x_j^{k_j}}{z_j^{{k_j}+1}} \frac{dz_j}{2\pi i}\right) $$ 
You may compare with:
$$ \prod_{j=1}^2 \left(\sum_{k=1}^2 a_{jk} \right)= (a_{11}+a_{12})(a_{21} + a_{22})
 = \sum_{k_1=1}^2 \sum_{k_2=1}^2 a_{1k_1} a_{2 k_2} =
\sum_{k_1=1}^2 \sum_{k_2=1}^2 \left( \prod_{j=1}^2 a_{jk_j}\right)$$
