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?
 A: 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*}
A: Note that in general
$$\sin x=\sum\limits_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots.$$
thus
$$\sin 2x =2x-\frac{8x^3}{6}+\frac{32x^3}{120}...$$
and
$$\frac{\sin 2x}{2x} =1-\frac{4x^2}{6}+\frac{16x^2}{120}...$$
A: Look at the Taylor expansion $$\sin t = t - \frac{t^3}{3!} +\frac{t^5}{5!} - \frac{t^7}{7!} + \cdots$$Now put $t = 2x$:$$\sin 2x = 2x - \frac{8x^3}{3!} +\frac{32x^5}{5!} - \frac{128x^7}{7!} + \cdots$$Divide by $2x$: $$\frac{\sin 2x}{2x} = 1 - \frac{4x^2}{3!} +\frac{16x^4}{5!} - \frac{64x^4}{7!} + \cdots$$Plugging $k=3/2$ for your $f_k(x)$ series gives precisely the above.
