Sum over Countable Set Let $F$ be defined as the set of all non-negative integer sequences that only have a finite number of non-zero elements.
I am having difficulties in understanding the implications of the finiteness of number of non-zero elements.
First, I wonder wether it is mathematically sound to write $F$ as
$$F = \{{\bf k} \in \mathbb{N}_0^\infty: |{\bf k}|_1 < \infty \} \quad ?$$
Here ${\bf k} = (k_1,k_2,\ldots) \in \mathbb{N}_0^\infty$ denotes a multi-index of infinite length.
Second, I want to deal with sums of the form
$$G({\bf \gamma}) := \sum_{{\bf k} \in F} \exp\left(- \sum_{i=1}^\infty k_i \gamma_i  \right), $$
where $0 < \gamma_1 \leq \gamma_2 \leq ...$
Is it true, that 
$$G({\bf \gamma}) = \prod_{i=1}^\infty \left( \sum_{k=0}^\infty \exp(-k \gamma_i) \right) \, ?$$
If yes, where does the finiteness of the number of non-zero elements come into play? In other words -- when performing this summation - what is the difference between $F$ and the infinite-fold tensorprodut $\mathbb{N}_0^\infty = \bigotimes_{i=1}^\infty \mathbb{N}_0$? 
Maybe the countability of $F$ play a role?
Thank you! 
 A: Regarding your first question: Yes, or write $F=\mathbb N_0^\infty\cap \ell^1$.
Can we replace $\sum_{\mathbf k\in F}$ with $\sum_{\mathbf k\in \mathbb N_0^\infty}$ in the definition of $G(\gamma)$? First of all, even as is the $\Sigma$ does not denote a sum but a series: It could only be righteously be called a sum if almost all summands are $0$.
Similarly, if we extend to more than countably many summands, we must ensure that only countable many summands are nonzero (otherwise there exists $n\in\mathbb N$ such that infinitely many summands ae $>\frac 1n$ and produce divergence).
In the given situation we have $\sum k_i\gamma_i\ge \gamma_1|k|_1$, which is $+\infty$ if $\mathbb k\in\mathbf N_0^\infty\setminus  F$, hence $\exp(-\sum k_i\gamma_i)$ is not even defined for such $\mathbf k$. Then again, we could agree to extend the definition of the exponential function such that $\exp(-\infty)=0$. With this convention in mind we find that
$$ \sum_{\mathbf k\in F}\exp\left(-\sum_{i=0}^\infty k_i\gamma_i\right)=\sum_{\mathbf k\in \mathbb N_0^\infty}\exp\left(-\sum_{i=0}^\infty k_i\gamma_i\right).$$
Maybe it is just "uneconomic" to add so many extra summands which are just discarded in the end?
The transformation into an infinite product (with countably many factors having countably many summands) is indeed only allowed because we have only countably many summands in the defining series, each of which can be viewed as a product (of the $\exp(-k_i\gamma_i)$) where all but finitely many factors can be discarded (so that we do not have to bother with convergence of infinite products - they are justordinary finite products). Also, the convergence (if the series converges) of the series is absolute because all summands (in fact, all factors) are positive; otherwise the enourmeous resorting that is behind the transformation would not be allowed.
To see that the transformation is allowed, assume that the defining series converges. Then pick $\let\epsilon\varepsilon\epsilon>0$ and finitely many summands of the defining series such that the finite sum differs by less than $\epsilon$ from $G(\gamma)$.
Now comes a point that would be impossible if we used $\mathbb N_0^\infty$ in place of $F$: We find $n,m$ such that $k_i\le n$ for all $i$ and $k_i=0$ for $i>m$ holds for our finitely many summands. If we add more (but still finitely many) summands, namely all $\mathbf k$ with $k_i\le n$ for all $i$ and $k_i=0$ for $i>m$, then this larger finite sum approximates $G(\gamma)$ even better (because all summands are positive). Now we can use that
$$\begin{align}G(\gamma)&\approx\sum_{\mathbf k\in\{0,\ldots,n\}^m}\exp\left(-\sum_{i=0}^m k_i\gamma_i\right)\\ 
&=\sum_{\mathbf k\in\{0,\ldots,n\}^m}\prod_{i=0}^m\exp(-k_i\gamma_i)\\
&=\prod_{i=0}^m\sum_{k=0}^n\exp(-k\gamma_i)\end{align}$$ 
If we increase $m$ and $n$, this approximation can only get better.
If we first only increase $n$ we find that also
$$\left|G(\gamma)-\prod_{i=0}^m\sum_{k=0}^\infty\exp(-k\gamma_i)\right| <\epsilon$$
and this inequality still holds if we increase $m$. Hence the sequence of partial products converges to $G(\gamma)$ as desired. As certainly $G(\gamma)>0$, we conclude that the infinite product converges, hence
$$ G(\gamma)=\prod_{i=0}^\infty\sum_{k=0}^\infty\exp(-k\gamma_i)$$
which can be simplified to 
$$ G(\gamma)=\prod_{i=0}^\infty\sum_{k=0}^\infty\exp(-\gamma_i)^k=\prod_{i=0}^\infty\frac1{1-\exp(-\gamma_i)}.$$
