generating function for $\frac{n!}{(2n)!}$

Question

How to construct generating function such that $$g(x) = \sum_{n=0}^\infty \frac{n!}{(2n)!} t^n $$

Answer 1

It might be better to consider the exponential generating function in this case. \begin{align} a_{n} &= \frac{n!}{(2n)!} \\ f(t) &= \sum_{n=0}^{\infty} a_{n} \, \frac{t^{n}}{n!} = \sum_{n=0}^{\infty} \frac{t^{n}}{(2n)!} = \cosh(\sqrt{t}). \end{align}

Linear generating function: \begin{align} g(t) &= \sum_{n=0}^{\infty} a_{n} \, t^{n} \\ &= \sum_{n=0}^{\infty} \frac{(1)_{n} \, (1)_{n} \, t^{n}}{n! \, (1)_{2n}} = \sum_{n=0}^{\infty} \frac{(1)_{n} \, t^{n}}{n! \, \left(\frac{1}{2}\right)_{n}} = {}_{1}F_{1}\left(1; \frac{1}{2}; \frac{t}{4}\right) \\ &= 1 + \frac{\sqrt{\pi \, t}}{2} \cdot e^{\frac{t}{4}} \cdot erf\left(\frac{\sqrt{t}}{2}\right). \end{align}

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Equation (21) of MathWorld is as follows: \begin{align} \sum_{n=0}^{\infty} \frac{t^{2n+1}}{(2n+1)!!} &= \frac{1}{t} \, \sum_{n=0}^{\infty} \frac{2^{n+1} \, (n+1)! \, t^{2n+2}}{(2n+2)!} \\ &= \frac{1}{t} \, \sum_{n=1}^{\infty} \frac{2^{n} \, n! \, t^{2n}}{(2n)!} \\ &= \frac{1}{t} \, \left[ -1 + g(2 t^{2}) \right] \\ &= \sqrt{\frac{\pi}{2}} \cdot e^{\frac{t^{2}}{2}} \cdot erf\left(\frac{t}{\sqrt{2}}\right) \end{align}

Answer 2

$$g(x)=1+e^{x/4}\frac{\sqrt{\pi x}}{2} \mbox{erf}(\sqrt{x}/2).$$

Answer 3

There's no easy way, I'm afraid... My suggestion would be to write the well-known Taylor series expansions for both the exponential and error function, where that of the latter can be found by integrating the former term by term, and then employ the dreadful Cauchy product formula for their multiplication.