simplifying triple summations from series expansion

Question

I have the following equation I would like to extract $x^k$ out of $$ \sum^\infty_{k=0}x^k \Gamma \bigg(l,\frac{ax-1}{a^2}\bigg) $$ I start by expanding the incomplete gamma function, and then the ensuing exponential $$ \begin{align} &= \sum^\infty_{k=0}x^k\Gamma(l)e^{-\frac{ax-1}{a^2}}\sum^{l-1}_{m=0}\bigg(\frac{ax-1}{a^2}\bigg)^m\frac{1}{m!}\\ &=\Gamma(l)\sum^\infty_{k=0}x^k\sum^\infty_{n=0}\bigg(\frac{1-ax}{a^2}\bigg)^n\frac{1}{n!}\sum^{l-1}_{m=0}\bigg(-\frac{1-ax}{a^2}\bigg)^m\frac 1{m!} \end{align} $$ I'm then not sure how to proceed.

My attempt is to pull the terminal sum out front and change the starting index of $k$ $$ =\Gamma(l)\sum^{l-1}_{m=0}\sum^\infty_{k=m}x^k\bigg(-\frac{1-ax}{a^2}\bigg)^m\frac 1{m!}\sum^\infty_{n=0}\bigg(\frac{1-ax}{a^2}\bigg)^n\frac{1}{n!} $$ I think then perhaps a binomial expansion of $1-ax$ to isolate $x$ and recombine as $$ =\Gamma(l)\sum^{l-1}_{m=0}\sum^\infty_{k=m}\sum^m_{j=0}\begin{pmatrix} m \\ j \end{pmatrix} (a^{-2})^j(a^{-1})^{m-j}\frac 1{m!}\sum^\infty_{n=0}\sum^n_{q=0}\begin{pmatrix} n \\ q \end{pmatrix} \left(a^{-2}\right)^{q}\left(-a^{-1}\right)^{n-q}(x)^{m-j+k+n-q}\frac{1}{n!} $$ I now have $x$ on its own, but have no idea how to pull it out front. I then want to reuse the exponential and incomplete gamma functions to tidy the resulting formula up. I'd imagine a quotient containing the incomplete gamma function as a final result.

** Edit **

In a similar (and simpler) example we would have $$ \sum^\infty_{k=0} x^k \sum^n_{m=0}\frac{(nx)^m}{m!} = \sum^n_{m=0}\sum^\infty_{k=m}x^k\frac{n^m}{m!} = \sum^\infty_{k=0}x^k\sum^{(k,n)}_{m=0}\frac{n^m}{m!} = e^n\sum^\infty_{k=0}x^k\frac{\Gamma((k,n)+1,n)}{\Gamma((k,n)+1)} $$ where $(k,n)$ denotes the lesser of $k$ or $n$. I wish to retain the sum over k and pull $x^k$ outside (if possible).

Answer 1

It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$.

We obtain for $j\geq 0$: \begin{align*} [x^j]\sum^\infty_{k=0}&x^k\sum^\infty_{n=0}\left(\frac{1-ax}{a^2}\right)^n\frac{1}{n!}\sum^{l-1}_{m=0}\left(-\frac{1-ax}{a^2}\right)^m\frac 1{m!}\\ &=\sum_{k=0}^j[x^{j-k}]\sum^\infty_{n=0}\left(\frac{1-ax}{a^2}\right)^n\frac{1}{n!}\sum^{l-1}_{m=0}\left(-\frac{1-ax}{a^2}\right)^m\frac 1{m!}\tag{1}\\ &=\sum_{k=0}^j\sum_{n=0}^\infty \sum_{m=0}^{l-1}\frac{(-1)^m}{n!m!a^{2(n+m)}}[x^k](1-ax)^{n+m}\tag{2}\\ &=\sum_{k=0}^j\sum_{n=0}^\infty\sum_{m=0}^{l-1}\frac{(-1)^m}{n!m!a^{2^{(n+m)}}}[x^k]\sum_{r=0}^{n+m}\binom{n+m}{r}(-ax)^r\tag{3}\\ &\,\,\color{blue}{=\sum_{k=0}^j\sum_{n=0}^\infty \sum_{m=0}^{l-1}\binom{n+m}{k}\frac{(-1)^{m+j}}{n!m!a^{2(n+m)-k}}}\tag{4} \end{align*}

Comment:

• In (1) we select the coefficient of $x^j$ and apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$. We set the upper limit of the outer sum to $j$ since higher indices do not contribute to $[x^j]$.

• In (2) we change the order of summation of the outer sum by letting $k\to j-k$ and do some rearrangements.

• In (3) we expand the binomial as preparation for the next step.

• In (4) we select the coefficient of $x^k$. Note that the double sum is zero as long as $n+m<k$.