I have a question regarding interchanging the order of three nested summations. My expression looks like
\begin{align} \sum_{n=0}^\infty \sum_{k=0}^n \sum_{\nu=0}^{4n-2k}\frac{C_{nk\nu}}{k!(n-k)!}\binom{4n-2k}{\nu}, \end{align}
where $C_{nk\nu}$ is a term depending on the three indices which, for the purposes of this question is irrelevant. What I want to do (assuming convergence) is to put the $\nu$ sum to the far most left, however I am not sure about the proper summation indices after interchanging the sums. For instance, I know that due to the binomial coefficient $\binom{4n-2k}{\nu}$ and the $(n-k)!$ factor we have the following summation constraints
\begin{align} n-k\geq 0 \hspace{5mm} \Longrightarrow k \leq n, \end{align} \begin{align} 4n-2k-\nu\geq 0 \hspace{5mm} \Longrightarrow k \leq \dfrac{4n-\nu}{2}. \end{align}
From which I can "guess" the new summation indices after interchanging the $k \leftrightarrow \nu$ sums:
\begin{align} \sum_{n=0}^\infty \sum_{k=0}^n \sum_{\nu=0}^{4n-2k}\frac{C_{nk\nu}}{k!(n-k)!}\binom{4n-2k}{\nu}=\sum_{n=0}^\infty \sum_{\nu=0}^\infty \sum_{k=0}^{\max\left(\lfloor \frac{4n-\nu}{2}\rfloor,n\right)}\frac{C_{nk\nu}}{k!(n-k)!}\binom{4n-2k}{\nu}, \end{align}
However this is only true for the cases when $n=0$ and $\forall \ \nu$ but for instance when $n=1$ and $\nu=0$, $\lfloor \dfrac{4(1)-0}{2} \rfloor=\lfloor 2 \rfloor = 2$ so the $k$ sum will run up to $2$ but then the term $(n-k)!=(1-(2))!=(-1)!$ is undefined so the upper limit that I put is correct for some values but incorrect for others, I have been trying to do some modifications to my upper limit but so far I have failed. Do you have any suggestions to properly flip the summations?
Thanks in advance :)