Interchanging summations with complicated, nested indices

Question

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 :)

Answer 1

Since by definition $\binom pq=0$ whenever $q>p,$ your sum (assuming absolute convergence) can simply be written $$\sum_{n,k,\nu\ge0}\frac{C_{n,k,\nu}}{n!}\binom nk\binom{4n-2k}\nu$$ or, if you "want to put the $ν$ sum to the far most left": $$\sum_{\nu\ge0}\sum_{n,k\ge0}\frac{C_{n,k,\nu}}{n!}\binom nk\binom{4n-2k}\nu$$ $$=\sum_{\nu\ge0}\sum_{k\ge0}\sum_{n\ge\max(k,(2k+\nu)/4)}\frac{C_{n,k,\nu}}{k!(n-k)!}\binom{4n-2k}\nu$$ $$=\sum_{\nu\ge0}\sum_{n\ge\nu/4}\sum_{0\le k\le\min(n,(4n-\nu)/2)}\frac{C_{n,k,\nu}}{k!(n-k)!}\binom{4n-2k}\nu.$$ Note the $\min$ in this last line, instead of your annoying $\max.$ This is because $k\le a$ and $k\le b$ is not equivalent to $k\le\max(a,b)$ but to $k\le\min(a,b).$

Edit, after seeing epi163sqrt's answer:

The expression can be slightly improved by the change of index $m=2n-k:$ $$\sum_{0\le k\le n,\;0\le\nu\le4n-2k}\frac{C_{n,k,\nu}}{k!(n-k)!}\binom{4n-2k}\nu$$ $$=\sum_{0\le n\le m\le 2n,\;0\le\nu\le2m}\frac{C_{n,2n-m,\nu}}{(2n-m)!(m-n)!}\binom{2m}\nu$$ $$=\sum_{\nu\ge0}\sum_{m\ge\nu/2}\sum_{m/2\le n\le m}\frac{C_{n,2n-m,\nu}}{(2n-m)!(m-n)!}\binom{2m}\nu.$$

Answer 2

Since we are interested in a transformation of the index regions only we consider wlog the sum \begin{align*} \sum_{n=0}^{\infty}\sum_{k=0}^n\sum_{\nu=0}^{4n-2k} a_{n,k,\nu} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{n=0}^{\infty}\sum_{k=0}^{n}\sum_{\nu=0}^{4n-2k}a_{n,k,\nu}} &=\sum_{n=0}^{\infty}\sum_{k=0}^n\sum_{\nu=0}^{2n+2k}a_{n,n-k,\nu}\tag{1}\\ &=\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}\sum_{\nu=0}^{2n+2k}a_{n,n-k,\nu}\tag{2}\\ &=\sum_{k=0}^{\infty}\sum_{n=2k}^{\infty}\sum_{\nu=0}^{2n}a_{n-k,n-2k,\nu}\tag{3}\\ &=\sum_{n=0}^{\infty}\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\sum_{\nu=0}^{2n}a_{n-k,n-2k,\nu}\tag{4}\\ &=\sum_{n=0}^{\infty}\sum_{\nu=0}^{2n}\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}a_{n-k,n-2k,\nu}\tag{5}\\ &\,\,\color{blue}{=\sum_{\nu=0}^{\infty}\sum_{n=\left\lfloor\frac{\nu}{2}\right\rfloor}^{\infty} \sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}a_{n-k,n-2k,\nu}}\tag{6} \end{align*}

Comment:

• In (1) we change the order of summation of the middle sum $k\to n-k$.

• In (2) we exchange left and middle sum noting \begin{align*} \sum_{n=0}^{\infty}\sum_{k=0}^n \Box=\sum_{0\leq k\leq n<\infty}\Box=\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}\Box \end{align*}

• In (3) we shift the index of the middle sum by $k$ to get rid of $k$ in the upper limit of the right sum.

• In (4) we exchange left and middle sum noting the equivalence of \begin{align*} &0\leq k<\infty&&0\leq n<\infty\\ &2k\leq n<\infty&&0\leq k\leq \left\lfloor\frac{n}{2}\right\rfloor \end{align*}

• In (5) we exchange middle and right sum.

• In (6) we exchange left and middle sum noting the equivalence of \begin{align*} &0\leq n<\infty&&0\leq \nu<\infty\\ &0\leq \nu\leq 2n&&n\leq \left\lfloor\frac{\nu}{2}\right\rfloor<\infty \end{align*}