I would like to know wether it is possible to represent $$f(x)=\sum_{n=1}^\infty \frac{\prod_{k=1}^n (1/4-k)}{n!} a_n x^n$$ in terms of $$g(x)=\sum_{n=0}^\infty a_n x^n$$
I have tried to represent the product and the $n!$ as a binomial coefficient, but once there I do not know what to do and how to continue (both differentiation and integration seem useless). In other words, I do not know where to start from.
Any help/idea will be appreciated. Thank you.
We assume that the series $$g(x)=\sum^\infty_{n=0}a_n\,x^n$$ has convergence radius $R$. This means that it converges for some positive $x=r<R,$ so that $\lim_{n\rightarrow\infty}a_n\,r^n=0$ and thus $|an\,r^n\le C$ for some $C>0$ and $n\ge0.$ Then, the series is absolutely convergent for $|x|<r$: $$\sum^\infty_{n=0}|a_n\,x^n|\le\sum^\infty_{n=0}C\,r^{-n}\,|x^n|=\frac{C}{1-\frac{|x|}{r}}.\tag1$$ Let $$c_n=\frac{\displaystyle{\prod_{k=1}^n (k-1/4})}{n!}.$$ Obviously, $c_n=O(n^{-1/4})$, so the radius of convergence of both series is equal. The sequence $c_n$ can be represented by the beta function, $$B(x,y)=\int^1_0t^{x-1}(1-t)^{y-1}\,dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}:$$ From the well-known recursion $$B(x+1,y)=B(x,y)\cdot\frac{x}{x+y}$$ we get immediately $$c_n=\frac{B(n+3/4,1/4)}{B(3/4,1/4)},$$ where by Euler's reflection formula $$B(3/4,1/4)=\frac{\Gamma(3/4)\Gamma(1/4)}{\Gamma(1)}=\frac{\pi}{\sin\frac\pi4}=\pi\sqrt{2}.$$ So $$f(x)=\frac1{\pi\sqrt{2}}\sum^\infty_{n=0}(-1)^n\left(\int^1_0t^{n-1/4}(1-t)^{-3/4}\,dt\right)\,a_nx^n.$$ The series on the RHS is absolutely convergent for $|x|<r$ because of (1), so we can interchange the order of summation and integration.
This means $$f(x)=\frac1{\pi\sqrt{2}}\int^1_0t^{-1/4}(1-t)^{-3/4}g(-tx)\,dt,$$ and this derivation is valid in every closed subinterval of the interval of convergence, since the series is absolutely convergent, there.