I need to show that $$\frac{\sqrt{2n}}{\theta +1}\left(\frac{1}{\bar{X}_n}-1-\theta \right) \to^{d} N(0,1)$$ where $\bar{X}_n = \frac{1}{n} \sum_{i=1}^nX_i$ and iid random variables $X_i$, $X_1 \tilde{\;}\exp(\theta+1) $. However i onyl succeed at proving this without the $\sqrt{2}$. ($\to^{d}$ stands for convergence in distribution). $$\frac{\sqrt{2n}}{\theta +1}\left(\frac{1}{\bar{X}_n}-1-\theta \right) = - \frac{\sqrt{2}}{(\theta+1)\bar{X}_n} \cdot \sqrt{n}(\bar{X}_n*(\theta+1)-1) \quad \star$$ where $$\sqrt{n}(\bar{X}_n*(\theta+1)-1) \to^{d} N(0,1)$$ by the central limit theorem. $$\bar{X}_n \to \frac{1}{\theta+1}$$ by some strong limit theorem and so $$- \frac{\sqrt{2}}{(\theta+1)\bar{X}_n} \to^d -\sqrt{2} $$ by the theorem of slutsky for multiplication. The $''-''$ doesnt really matter, because if $Z \tilde{} N(0,1)$ then it is $-Z \tilde{} N(0,1)$ as well. I cannot find my mistake. Looking forward to any help. If you thik my mistake lies in the Line marked by the $\star$ i might write my steps as well.
Thanks.
The $\sqrt 2$ was a mistake in the exercise.