I am reading Abate and Whihtt "1996An Operational Calculus for Probability Distributions via Laplace Transforms" and meet one summation question.
In Section 8, authors mentioned that the Laplace transform of a particular Inverse Gaussian distribution is $$\tilde{f}(s;1,v)=exp\left({-\frac{\sqrt{1+ 2 v s}-1}{ v}}\right),$$ and the moment is $$m_{n+1}=\sum _{k=0}^n \frac{(n+k)!}{k!(n-k)!}\left(\frac{v}{2}\right)^k.$$
Then, by the relationship between the Laplace transform and the moments, we should have $$\tilde{f}(s;1,v)=exp\left({-\frac{\sqrt{1+ 2 v s}-1}{ v}}\right)=\sum _{n=0}^{\infty } m_n\frac{(-s)^n}{n!}=1+\sum _{n=0}^{\infty } m_{n+1}\frac{(-s)^{n+1}}{(n+1)!}.$$
But I do not know how to simplify the summation to get the exp function. Exchange the order of summation does not seem to work here: \begin{align*} &1+\sum _{n=0}^{\infty } m_{n+1}\frac{(-s)^{n+1}}{(n+1)!}\\ =&1+\sum _{n=0}^{\infty } \sum _{k=0}^n \frac{(n+k)!}{k!(n-k)!}\left(\frac{v}{2}\right)^k\frac{(-s)^{n+1}}{(n+1)!}\\ =&1+\sum _{k=0}^{\infty } \frac{1}{k!}\left(\frac{v}{2}\right)^k\sum _{n=k}^{\infty } \frac{(n+k)!}{(n-k)!}\frac{(-s)^{n+1}}{(n+1)!}\\ =&1+\sum _{k=0}^{\infty } \frac{1}{k!}\left(\frac{v}{2}\right)^k\sum _{n=0}^{\infty } \frac{(n+2k)!}{(n+k+1)!}\frac{(-s)^{n+1}}{n!}. \end{align*}
Thanks in advance.
For convenience, denote the left hand side as a real-valued function of $s$ where $t$ is a parameter: $$g(s) \equiv \exp\left({-\frac{\sqrt{1+ 2 v s}-1}{ v}}\right)$$
Compare the MacLaurin series (Taylor expansion at $s = 0$) of $g$ with the target right hand side $$g(s) = \sum_{n=0}^{\infty} g^{(n)}(0) \frac{ s^n }{n!} \qquad \text{v.s.} \qquad \sum _{n=0}^{\infty } m_n\frac{(-s)^n}{n!}$$ one sees that the objective is to show $g^{(n)}(0) = -m_n~,$ where $g^{(n)}(0) = \dfrac{d^ng(s)}{ds^n} \Bigg|_{s=0}$ is the $n$-th derivative evaluated at zero.
Let's manually compute the first few terms of $g^{(n)}(s)$ for general $s$ (before evaluated at $s = 0$) to examine the pattern of the functional form. Use the shorthand $g \equiv g(s)$ for the zero-th order at $n = 0$.
\begin{align} g'(s) &= \frac{-g}{ \sqrt{1 + 2vs} } \\ g''(s) &= \frac{ g }{ 1 + 2vs } \left( 1 + \frac{ v }{ \sqrt{1 + 2vs} } \right) \\ g'''(s) &= \frac{ -g }{ (1 + 2vs)^{3/2} } \left( 1 + \frac{ 3v }{ \sqrt{1 + 2vs} }+ \frac{ 3v^2 }{ 1 + 2vs } \right) \\ g^{(4)}(s) &= \frac{ g }{ (1 + 2vs)^2 } \left( 1 + \color{blue}{ \frac{ 6v }{ \sqrt{1 + 2vs} } } + \color{blue}{ \frac{ 15v^2 }{ 1 + 2vs } } + \frac{ 15v^3 }{ (1 + 2vs)^{3/2} }\right) \\ g^{(5)}(s) &= \frac{ -g }{ (1 + 2vs)^{5/2} } \left( 1 + \frac{ 10v }{ \sqrt{1 + 2vs} } + \color{magenta}{ \frac{ 45v^2 }{ 1 + 2vs } } + \frac{ 105v^3 }{ (1 + 2vs)^{3/2} } + \frac{ 105v^4 }{ (1 + 2vs)^2 } \right) \end{align} No trick was used (nor needed) in the calculation above. Just be patient and carry out each step. The point is that through this process one obtains a concrete sense of how one order $g^{(n-1)}(s)$ yields the next order $g^{(n)}(s)$.
If you have actually done the differentiation above, you should find it reasonable to conjecture that $$g^{(n)}(s) = (-1)^n \frac{g}{ (1 + 2vs)^{n/2} } \sum_{k = 0}^{n-1} \frac{ v^k \cdot A_{n,k} }{ (1 + 2vs)^{k/2} } \tag*{Eq.(1a)} $$ where $A_{n,k}$ are the constant coefficients that satisfy the following recurrence relation $$A_{n,k} = A_{n-1,k} + (n+k-2) A_{n-1,k-1} \tag*{Eq.(1b)}$$ accompanied by the initial and boundary conditions $$\begin{aligned} A_{n,0} &= 1 & A_{n,n} &= 0 & \forall~ n &\geq 0 \end{aligned} \tag*{Eq.(1c)}$$ For example, at $n = 5$ and $k = 2$ the magenta $\color{magenta}{45}$ comes from the $n = 4$ blue $\color{blue}{6}$ and $\color{blue}{15}~$, as in $45 = 15 + (n+k-2)\cdot 6 = 15 + 5 \cdot 6$ $$\begin{array}{c|ccccccccc} & k = 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline n = 1 & 1 & \color{gray}{0} & \\ n = 2 & 1 & 1 & \color{gray}{0} \\ n = 3 & 1 & 3 & 3 & \color{gray}{0} \\ n = 4 & 1 & \color{blue}{6} & \color{blue}{15} & 15 & \color{gray}{0} \\ n = 5 & 1 & 10 & \color{magenta}{45} & 105 & 105 & \color{gray}{0} & \\ n = 6 & 1 & 15 & 105 & 420 & 945 & 945 & \color{gray}{0} \\ n = 7 & 1 & 21 & 210 & 1260 & 4725 & 10395 & 10395 & \color{gray}{0} \\ \end{array}$$
The proof of this "conjecture" $\text{Eq}.(1)$ by induction is literally summarizing/repeating what happens in the chain rule during the differentiation:
$A_{n,k}$ is the coefficient of $\displaystyle \frac{ g \cdot v^k }{ (1 + 2vs)^{(n+k)/2} }$, which from order $n - 1$ can only come from two places: \begin{align} \text{one:}& & &\frac{ v^k }{ (1 + 2vs)^{( n - 1 + k )/2} } \cdot \frac{d}{ds} g & &\because g' = \frac{-g}{ \sqrt{1 + 2vs} } \\ \text{two:}& & & g \cdot \frac{d}{ds} \frac{ v^{k-1} }{ (1 + 2vs)^{(n - 1 + k - 1)/2} } & &\because \left( \frac1{ (1 + 2vs)^{(n - 1 + k - 1)/2} } \right)' = \frac{ (n+k-2) \cdot v }{ (1 + 2vs)^{(n + k )/2} } \end{align} The minus sign is easy so was omitted in the above, and the recurrence holds trivially for the initial $n = 1$ to $n = 2$ for all their relevant $k$.$\quad Q.E.D$
The recurrence relation Eq.($1$b) or its equivalent alternate expression (that defines the "triangle" displayed above) is well-studied and known as the coefficients of the Bessel polynomial (for those who don't like wiki, see e.g. Wolfram MathWorld).
In the OEIS (Online Encyclopedia of Integer Sequences) it is listed as A001498, where one can find and related formulae and various combinatorial interpretations. Note: the index $n$ differs from here by one in OEIS.
Besides close connections to the several famous polynomials from ODE, Bessel polynomial is also related to: matchings of the complete graph, enumeration of unordered forests composed of ordered rooted trees, and the differential operator $D \equiv \frac1{t} \frac{d}{dt}$.
This is well-documented in all the three sources provided above (wiki, Wolfram, OEIS), in addition to the often-cited 1949 paper by H.L. Krall that is somewhat friendly to all levels of readers. In particular, one should check out Eq.(21) for the recurrence relations and Eq.(25) in the paper for the generating function (or the section in wiki, or Eq.(11) in Wolfram) that states exactly the identity of the posted question.