I am asking this question as I have not seen an answer to the question (that I can understand) as of yet (Credit for original question: Series expansion of infinite series raised to the $n$th power)
Can anyone explain how you can do it for an infinite $$\left( \sum_{n\in \mathbb N} \ f(x,n) \right)^n$$
Sorry for the bad explanation, linking and other stuff, I am not familiar with this site, but please explain the answer to this question as I am not quite familiar with infinite sums yet
Best Answer
Your example has three n's, the outer has to be another symbol
By definition (and if all terms exist and are unique)
$$\left( \sum_{n\in \mathbb N} \ f(x,n) \right)^m=\prod _{k=1}^m\left( \sum_{n_k\in \mathbb N} \ f(x,n_k) \right) $$
Any further simplification or expansion needs an ordering of all monomials in $f(x,k_j)$. This is simple by completely symmetric product structure of the f(x,n)f(x,m), eg powers, exponentials or the famous finite algebras.
For the classical complex Taylor series, one collects all products with constant total exponent
$$\prod _{k=1}^m\left( \sum_{n_k\in \mathbb N} \ a_{n_k} \ x^{n_k} \right) = \sum_{n\in \mathbb N} x^n \quad \left(\sum_{\sum _j k_j=n}\prod a_{k_j}\right) $$
Exercise: Power of Exponential series
$$\prod _{k=1}^m\left( \sum_{n_k\in \mathbb N} \ \frac{\ x^{n_k}}{n_k!} \right) = \sum_{n\in \mathbb N} x^n \quad \left(\sum_{\sum _j k_j=n}\prod \frac{1}{k_j!}\right) = \sum_{n\in \mathbb N} \frac{(m x)^n}{n!} $$