Need some help expressing this as a series, if possible Okay, so I'm working on a question, and after a bunch of simplification and manipulations, I get the following expression. Just wanted to ask if there's any way I can get this in the form of a series, with $n$ going from 0 to $\infty$. a is a random positive integer here.
$$
U(x) = Q\bigg[1 - \frac{a(a+1)}{2!}x^2 + \frac{a(a+1)(a+2)(a+3)}{4!}x^4 + ...\bigg] 
+ R\bigg[x - \frac{(a-1)(a+2)}{3!}x^3 + \frac{(a-1)(a+2)(a-3)(a+4)}{5!}x^5 + ...\bigg]
$$
I'm having trouble trying to figure out a way to simplify this further, so I'm just curious if it's possible. If not, then that's okay too.
Thanks for taking the time to look at this!!
Editing to add the DE I was working on:
$$(1-x^2)U''(x) -2xU'(x) +a(a+1)U(x) = 0$$

This is just for your curiosity
$U(x)$ is already expressed as a linear combination of two Taylor series (for which you give only the first terms but in which the patterns of the coefficients is quite clear). I suppose that this is the power series solution of a second oreder differential equation which could be intresting to know.
So, what I think is that the problem would better be to find the analytic functions to which correspond these Taylor series.
The first one is "amazing" since it seems that we have
$$f(x)=\sum_{n=0}^\infty (-1)^n \frac {c_n}{(2n)!} x^{2n}\quad \text{with} \quad c_n=\prod_{i=0}^{2n-1} (a+i)=a \,(a+1)_{2 n-1}$$ where appear Pochhammer symbols. This makes
$$f(x)=\, _2F_1\left(\frac{a}{2},\frac{a+1}{2};\frac{1}{2};-x^2\right)$$ where appears the gaussian hypergeometric function. But, guess what ?
$$\color{blue}{f(x)=\left(1+x^2\right)^{-\frac a2} \cos \left(a \tan ^{-1}(x)\right)}$$
For the second series, it seems to be much more complex since we have
$$g(x)=\sum_{n=0}^\infty (-1)^n \frac {d_n}{(2n+1)!} x^{2n+1}$$ with
$$d_n=\Bigg[\prod_{i=0}^{n} (a+2i)\Bigg]\,\,  \Bigg[\prod_{i=0}^{n-1} (a+2i-1)\Bigg]=(-1)^{n+1}(a-1)  2^{2 n-1} \left(\frac{3-a}{2}\right)_{n-1}   \left(\frac{a+2}{2}\right)_n$$
$$\color{blue}{g(x)=x \, _2F_1\left(\frac{1-a}{2},\frac{a+2}{2};\frac{3}{2};x^2\right)}$$ which, unfortunately, does not simplify but explains why was required a power series solution.
For integer values of $a$, we have quite nice and simple expressions
$$\left(
\begin{array}{cc}
 a & \left(x^2+1\right)^a\,f(x) \\
 1 & 1 \\
 2 & 1-x^2 \\
 3 & 1-3 x^2 \\
 4 & x^4-6 x^2+1 \\
 5 & 5 x^4-10 x^2+1 \\
 6 & -x^6+15 x^4-15 x^2+1 
\end{array}
\right)$$
For $g(x)$, it is a bit more complex but it write
$$g(x)= x Q(x)+ R(x) \tanh^{-1}(x)$$
$$\left(
\begin{array}{ccc}
a & Q(x) & R(x) \\
 1 & 1 & 0 \\
 2 & \frac{3}{4} & -\frac{3 x^2}{4}+\frac{1}{4} \\
 3 & -\frac{5 x^2}{3}+1 & 0 \\
 4 & -\frac{105 x^2}{64}+\frac{55}{64} & \frac{105 x^4}{64}-\frac{45
   x^2}{32}+\frac{9}{64} \\
 5 & \frac{21 x^4}{5}-\frac{14 x^2}{3}+1 & 0 \\
 6 & \frac{1155 x^4}{256}-\frac{595 x^2}{128}+\frac{231}{256} & -\frac{1155
   x^6}{256}+\frac{1575 x^4}{256}-\frac{525 x^2}{256}+\frac{25}{256}
\end{array}
\right)$$
