Let $G$ be a compact Lie group and $a\in\mathfrak{g}^*$ (dual of Lie algebra of Lie group $G$). Then let $\mathcal O_a$ be a coadjoint orbit. Then every co-adjoint orbit is Kähler manifold and also projective variety. How can we compute the Kodaira dimension of co-adjoint orbit as projective variety? 

In fact I am looking for 

$$\kappa(\mathcal O_a)=\limsup_{m\to \infty}\frac{\log\text{dim}H^0(\mathcal O_a, K_{\mathcal O_a}^{\otimes m})}{\log m}$$

Note : By Borel-Weil theorem $h^{0,0}$ is only non-zero case. So normaly $\kappa(\mathcal O_a)$ must vanishes.  Is it correct?

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I am not sure my method of solution is correct or not, so commentes are welcomed:

We have the following formula by using Kirillov's dimension formula or Wyel dimension formula, (Maybe my following formula is not computed correctly, so take care, please)

 $$\dim H^0(\mathcal O_a, K_{\mathcal O_a}^{\otimes m})=\frac{m^n}{n!}\prod_{\alpha>0}\frac{<\alpha,ma+\rho>}{<\alpha,\rho>}$$ where $\dim \mathcal O_a=n$ and $\rho=\frac{1}{2}\sum_{\alpha>0}\alpha$ and $\alpha$ is the positive roots of Lie Group $G$, So 

$$\kappa(\mathcal O_a)=\limsup_{m\to \infty}\frac{\log\text{dim}H^0(\mathcal O_a, K_{\mathcal O_a}^{\otimes m})}{\log m}=\limsup_{m\to \infty}\frac{\log \frac{m^n}{n!}\prod_{\alpha>0}\frac{<\alpha,ma+\rho>}{<\alpha,\rho>} }{\log m}$$

> So, the Question is again , what is the $\kappa(\mathcal O_a)$?