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?

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.
