Showing $X_{(n)}$ is an unbiased and consistent estimator for $\theta$. Let $X_1,X_2,\ldots, X_n$ is distributed iid $\mathrm{Uniform}(0, \theta)$ with $\theta$ in being real positive. Show $\frac{n+1}{n}X_{(n)}$ is an unbiased and consistent estimator for $\theta$.
I have already found the distribution of $X_{(n)}$ and have shown it is unbiased, but I am stuck on showing it is consistent. I found $f(x)=\frac{n}{\theta}(\frac{x}{\theta})^{n-1}$. From there I found $V[X_{(n)}]=\frac{n\theta^2}{n+1}$. However when I try to show that $\lim_{n \rightarrow \infty} V[\frac{n+1}{n}X_{(n)}]=0$ it does not come out to $0$. What am I missing?

\begin{align}
& \int_0^\theta x^2 f(x) \, dx = \int_0^\theta x^2 \frac n \theta \left( \frac x \theta \right)^{n-1} \, dx = n\theta^2 \int_0^\theta \left( \frac x \theta \right)^2 \left( \frac x \theta \right)^{n-1} \left( \frac{dx}\theta \right) \\[10pt]
= {} &  n\theta^2 \int_0^1 u^2 \cdot u^{n-1} \, du = \frac{n\theta^2}{n+2}. \\[30pt]
& \operatorname{var}\left( \frac{n+1} n X_{(n)} \right) = \left(\frac{n+1} n\right)^2 \operatorname{var}(X_{(n)}) = \left(\frac{n+1} n\right)^2 \left( \operatorname{E} \left( X_{(n)}^2 \right) - \left( \operatorname{E}\left( X_{(n)} \right) \right)^2\right)  \\[10pt]
= {} & \left(\frac{n+1} n\right)^2 \left( \frac{n\theta^2}{n+2} - \left(\frac{n\theta}{n+1}\right)^2 \right) = \frac{\theta^2}{n+2}.
\end{align}
