Steinberg´s formula for $A_{1}$

Question

I am trying to show, that Steinberg´s formula on the case $A_{1}$ yields the same result as the Clebsh-Gordan-formula, but I get a quite confusing result.
Let $V(\lambda)$, $V(\lambda´)$ and $V(\lambda´´)$ irreducible $L$-moduls for $L=sl(2,\mathbb{F})$. I am trying to compute the formula for $V(\lambda)$ in $V(\lambda´)\otimes V(\lambda´´)$. Let $W$ be the Weyl group.
First of all I know, that for $\lambda_{1}=\frac{1}{2}\alpha$ the fundamental dominat weight:
$\lambda=r\lambda_{1}$, $\lambda´=m\lambda_{1}$ and $\lambda´´=n\lambda_{1}$ $r, m, n\in \mathbb{Z}^{+}$ WLOG $n\leq m$.
$\sum\nolimits_{\sigma\in W} \sum\nolimits_{\tau\in W}sn(\sigma\tau)\rho(\lambda+ 2\lambda_{1}-\sigma(\lambda´+ \lambda_{1})-\tau(\lambda´´+\lambda_{1})) \\ =\rho(\lambda-\lambda´-\lambda´´)-\rho(\lambda - \lambda´+\lambda´´+2\lambda_{1})-\rho(\lambda+\lambda´-\lambda´´+2\lambda_{1})+\rho(\lambda+\lambda´+\lambda´´+4\lambda_{1}) \\ =\rho((r-m-n)\lambda_{1})-\rho((r-m+n+2)\lambda_{1})-\rho((r+m-n+2)\lambda_{1} + \rho((r+m+n+4)\lambda_{1})$
This showes, that $r=m+n-2i$ for some $i\in\mathbb{Z}^{+}$. So after applying:
$=\rho(-i\alpha)-\rho(((n+1)-i)\alpha)-\rho(((m+1)-i)\alpha)+\rho((m+n+2)-i)\alpha)$
Now I tried throught the possibilities of $i$:
For $0\leq i \leq n$ the formula becomes $1$, which is exactly what the Clebsh-Gordan-formula says.
For $n+1\leq i \leq m$ as well as $m+n+2\leq i$ the formula becomes $0$, which is also correct.
My problem lies with $m+1\leq i \leq m+n+1$ because the formula becomes $-1$, but I can´t find where I made a mistake.
Thank you for helping me.

Answer 1

I still don´t know exactly where the $-1$ comes from, but I succeeded in solving the problem:
Look at the formula before applying $r=m+n-2i$. The first summand is $1$ exactly when $r\leq m+n$,
the second when $r\leq m-n-2 \iff r-m+2\leq -n$,
the third when $r\leq -m+n-2 \iff r-n+2 \leq -m$ and
the forth when $r\leq -m-n-4$.
If only the first inequation is fulfilled, the sum is $1$. If the second is fulfilled, then so is the first and the sum is $0$.
If the third is fulfilled, then so are the first and the second, as well as:
$-m-n-4\geq r-n+2+r-m+2-4 = 2r-(m+n) \geq 2r-r=r$
So the forth is also fulfilled and the sum is $0$.
This showes that the sum is $1$ only for $m-n\leq r \leq m+n$ and has to be of the form $r=m+n-2i$ $i\in \mathbb{Z}^{+}$, what is exactly what the Clebsh-Gordan-formula says.