Quantum mechanical spin coherent states I'm really struggling with how to derive the right hand side of this equation. It's the Spin operator on coherent states of the harmonic oscillator. I've done it for the $S_+$ operator but can't seem to get this result:

$<\mu|$ here is $(1+|\mu|^2)^{-2s}$$\Sigma$ $\mu^*$ $(\frac{(2S)!}{p!(2S-p)!})^{1/2}$$<p|$
My attempt at a solution was to use $S_-|p>$ = $((2S-p)(p+1))^{1/2}|p+1>$ and input that which i've shown below, but unfortunately it still isn't getting me anywhere.

I apologise if I haven't formatted this correctly or anything, this is my first post here. Any help would be greatly appreciated!
 A: Well... part of the problem is that you have used $p$ twice as a summation index and so I suspect your first (leftmost) sum is actually over $q$.  This will require some reorganization on your part but otherwise 
you need to move your $\langle q\vert $ inside your rightmost summation over $p$ to eliminate this sum through $\langle q\vert p+1\rangle=\delta_{p,q-1}$.  
If I didn't make some typo your rightmost sum on the right will become something like:
\begin{align}
&\langle q \vert \sum_{p=0}^{2s} \mu^p
\sqrt{\frac{(2S)!(2S-p)!(p+1)}{p!(2S-p)!}}\vert p+1\rangle \\
&= \sum_{p=0}^{2s} \mu^p
\sqrt{\frac{(2S)!(2S-p)!(p+1)}{p!(2S-p)!}}\langle q\vert p+1\rangle\, ,\\
&= \sum_{p=0}^{2s} \mu^p
\sqrt{\frac{(2S)!(2S-p)!(p+1)}{p!(2S-p)!}}\delta_{p,q-1}\, ,\\
&=\mu^{q-1}
\sqrt{\frac{(2S)!(2S-q+1)!(q)}{(q-1)!(2S-q+1)!}}\, .
\end{align}
You can already see how you have $(\mu^*)^q$ but $\mu^{q-1}$, suggesting a leftover $\mu^*$ as per your answer.  
There is still some work to do but I think this should put you on the right track.
