Proving that $\frac{n+1}{n} Y_n$ is consistent for $\theta$, where $X_i \sim U(0, \theta)$ Let $X_1, X_2,..., X_n$ be i.i.d $U(0, \theta)$ random variables.
I am attempting to prove that  $\theta_1=\frac{n+1}{n} Y_n$ is a consistent estimator for $\theta$, where $Y_n=\max(X_1, X_2,\ldots,X_n)$ using the definition of consistency directly.
I am familiar with other techniques but I'm not sure on how to proceed using the definition directly. 
I am aware that I have to prove that $P(|\theta_1 - \theta| > \delta)=P(|\frac{n+1}{n}Y_n - \theta| > \delta)< \epsilon$, for all $n>N(\epsilon)$, $\epsilon > 0$.
I'm unsure as to how I can proceed, any help would be greatly appreciated!
 A: \begin{align}
\mathbb{P}(|\theta_1(n) - \theta| > \delta) &= \mathbb{P}( \theta - \frac{n+1}{n} Y_n > \delta) + \mathbb{P}(\frac{n+1}{n} Y_n - \theta  > \delta) \\
&=\mathbb{P}(Y_n < \frac{ n }{ n+1 } ( \theta - \delta)) + \left(1 - \mathbb{P}(\frac{n+1}{n} Y_n \le \theta + \delta)\right)\\
&= F_{Y_n}\left(\frac{n}{ n  + 1 } (\theta-\delta)\right) + (1 - F_{Y_n}\left(\frac{n}{ n  + 1 } (\theta+\delta)\right) .
\end{align} 
Now, 
$$
F_{Y_n}(y)=(F_X(y))^n = ( 1/\theta ) ^ n,
$$
hence, 
\begin{align}
\lim_{n \to \infty} \mathbb{P}(|\theta_1(n) - \theta| > \delta)
 &= 
\lim_{n \to \infty} \left( \frac{n}{n+1}( 1 - \frac{ \delta }{ \theta } ) \right) ^ n 
+
(1 - \lim_{n \to \infty}  F_{Y_n}\left( \frac{n}{n+1}( \theta +  \delta ) \right) \\  
&= ( 1 - \delta/\theta)^\infty + ( 1 - F_{Y_n}( \theta + \delta))\\
&= 0 + (1-1) = 0.
\end{align}
The last line is abuse of notation, but the idea is that $n/(n+1)$ goes to $1$ and $\delta/\theta >0$, hence you have a fraction that goes to $0$ as the power goes to infinity. Moreover, $\theta + \delta > \theta$ while the support of $Y_n$ is $[0, \theta]$, thus the CDF is $1$. 
