With
[tex]\bar X=\displaystyle\sum_{i=1}^n\frac{X_i}n[/tex]
we find the mean is
[tex]\mathrm E[\bar X]=\displaystyle\frac1n\sum_{i=1}^n\mathrm E[X_i]=\frac{n\mu}n=\mu[/tex]
and the variance is
[tex]\mathrm V[\bar X]=\displaystyle\frac1{n^2}\mathrm V\left[\sum_{i=1}^nX_i\right][/tex]
[tex]\mathrm V[\bar X]=\displaystyle\frac1{n^2}\left(\sum_{i=1}^n\mathrm V[X_i]+2\sum_{1\le i<j\le n}\mathrm{Cov}[X_i,X_j]\right)[/tex]
[tex]\mathrm V[\bar X]=\displaystyle\frac1{n^2}\left(n\sigma^2+2\frac{n(n-1)}2\rho\sigma\right)[/tex]
[tex]\mathrm V[\bar X]=\dfrac{\sigma^2}n+\dfrac{n-1}n\rho\sigma[/tex]
Then as [tex]n\to\infty[/tex], we find that [tex]\mathrm V[\bar X]\to\rho\sigma[/tex]. Simply put, the failure of the CLT/LLN has to do with the fact that the [tex]X_i[/tex] are not independent.
