= Solution
Use the corrected <generating series of the Kostant partition function> and the permitted <Weyl character formula>:
$$
\chi_\lambda
=\left(\sum_{w\in W}\varepsilon(w)e^{w(\lambda+\rho)-\rho}\right)
\left(\sum_\nu p(\nu)e^{-\nu}\right).
$$
For a fixed $w$, the coefficient of $e^\mu$ arises exactly when
$$
w(\lambda+\rho)-\rho-\nu=\mu,
\qquad\text{that is,}\qquad
\nu=w(\lambda+\rho)-(\mu+\rho).
$$
There are only finitely many Weyl-group terms, and the <partition function> vanishes outside the cone of nonnegative <root> combinations. <Coefficient extraction> is therefore legitimate in the formal completion used in the preceding part. Summing the signed contributions proves
$$
\boxed{m_\lambda(\mu)=\sum_{w\in W}\varepsilon(w)\,p\bigl(w(\lambda+\rho)-(\mu+\rho)\bigr).}
$$
This proves the <Kostant multiplicity formula> for every <compact> <connected> <Lie group>, not just the <unitary groups>. In the presence of a central <torus>, <roots> have zero central component, so the <partition function> automatically gives zero for any incompatible central <weight>. Also $\rho$ need not itself be an integral <weight> for the given global <group>: $w\rho-\rho$ always belongs to the <root lattice>, so every argument in the formula is nevertheless integral whenever $\lambda$ and $\mu$ are integral <group> <weights>.
As a normalization check, for $SU(2)$ write a <highest weight> as the <nonnegative integer> $\ell$, with <positive root> $2$ and $\rho=1$. The two Weyl-group terms give
$$
m_\ell(\mu)=p(\ell-\mu)-p(-\ell-\mu-2).
$$
Since $p(a)=1$ precisely when $a$ is even and nonnegative, this is one for $\mu=\ell,\ell-2,\ldots,-\ell$ and zero otherwise, including the cancellation below the lowest <weight>. This recovers the full familiar <weight string>.
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