# 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.

• Those sums are geometric series, so they converge absolutely wherever they converge (except possibly on the boundary of the disk of convergence); hence the same is true of the finite product. – Greg Martin Aug 28 '16 at 1:50
• Just a minor remark, using $i$ as a summation index with complex numbers is a really bad choice of variable name... – Lukas Geyer Sep 1 '16 at 19:22

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: $$\prod_{j=1}^2 \left(\sum_{k=1}^2 a_{jk} \right)= (a_{11}+a_{12})(a_{21} + a_{22}) = \sum_{k_1=1}^2 \sum_{k_2=1}^2 a_{1k_1} a_{2 k_2} = \sum_{k_1=1}^2 \sum_{k_2=1}^2 \left( \prod_{j=1}^2 a_{jk_j}\right)$$