Let $(P_\theta)_{\theta \in \Theta}$ a family of probability measures on $((0, \infty), \:(B_{(0,\infty)}))$ with the densities $f_{\theta}(t):= \theta^{-1}1_{(0,\theta)}(t)$. For $n \ge2$ we observe the statistical model $((0, \infty)^n, (B_{(0, \infty)^n}), (P_{\theta}^{\bigotimes n})_{\theta\in\Theta})$, where $X=(X_1, X_2, X_3,..., X_n)$ is a sample. Furthermore, we observe estimator of $\theta$ the geometrical mean $T_n := \sqrt[n]{X_1*...*X_n}$.
I need to determine the bias $b_{T_n}(\theta)$ and get unbiased estimator, also I should calculate the limit $\lim_{n \rightarrow\infty}b_{T_n}(\theta)$
MY TRY:
$$ E(\sqrt[n]{X_1}) = \int_{0}^{\theta}\theta^{-1}t^{\frac{1}{n}}dt = \theta^{-1}\frac{\theta^{\frac{1}{n}+1}}{\frac{1}{n}+1}=\theta^{-1}\frac{\theta^{\frac{n+1}{n}}}{\frac{n+1}{n}}=\frac{n}{n+1}\theta^{\frac{1}{n}}$$
$$ E(\sqrt[n]{X_1...X_n})=(\frac{n}{n+1}\theta^{\frac{1}{n}})^n=(\frac{n}{n+1})^n\theta $$
$$ b_{T_n}(\theta) = (\frac{n}{n+1})^n\theta-\theta $$ for $n\rightarrow \infty$ equals to $\frac{1}{e}\theta - \theta$
MY PROBLEM: I tried by myself and I stuck at the one part, moreover, I am not sure, whether this, what I have done so far, is correct at all. But there is something wrong about this since it should be unbiased.