Verify that for $k=3/2 ,$ $\quad f_{3/2}(x)=\frac{\sin(2x)}{2x}$ This is a part of a proof I am studying on:
Let  $f_k(x) = 1 - \frac{x^2}k+\frac{x^4}{2! k(k+1)}-\frac{x^6}{3! k(k+1)(k+2)} + \cdots \qquad (k\notin\{0,-1,-2,\ldots\}) $
For $k=3/2 ,$ it's shown that $\quad f_{3/2}(x)=\frac{\sin(2x)}{2x}$  
How is it concluded this way?

Note that (using double factorial notation)
\begin{eqnarray*}
\frac{3}{2} \frac{5}{2} \cdots \frac{2n+1}{2} = \frac{(2n+1)!!}{2^n}.
\end{eqnarray*}
So the $n^{th}$ term can be written as
\begin{eqnarray*}
\frac{x^{2n}}{n! \frac{3}{2} \frac{5}{2} \cdots \frac{2n+1}{2}} =\frac{( 2^n)^2 x^{2n}}{(2n)!! (2n+1)!!} =\frac{(2x)^{2n}} {(2n+1)!}. 
\end{eqnarray*}
