<s>[INST] Show that $\lim_{n\rightarrow \infty } Var(Y_{n}) = 0$. Given $Var(Y_{n}) = (\theta_{2}-\theta_{1})^2\frac{n}{(n+2)(n+1)^2}$.
My work: $$lim_{n\rightarrow \infty } Var(Y_{n}) = \lim_{n\rightarrow \infty } (\theta_{2}-\theta_{1})^2\frac{n}{(n+2)(n+1)^2} = \lim_{n\rightarrow \infty } \frac{(\theta_{2}-\theta_{1})^2}{(1 + \frac{2}{n}) (n+1)^{2}} = \lim_{n\rightarrow \infty } \frac{\frac{(\theta_{2}-\theta_{1})^2}{n^2}}{(1 + \frac{2}{n}) (1 + \frac{1}{n})^{2}} = 0.$$
Is this correct? 
 [/INST] Hint:
$\lim_{n\rightarrow \infty } (\theta_{2}-\theta_{1})^2\frac{n}{(n+2)(n+1)^2} = (\theta_{2}-\theta_{1})^2\lim_{n\rightarrow \infty }\frac{n}{(n+2)(n+1)^2} $
However, $\lim_{n\rightarrow \infty }\frac{n}{(n+2)(n+1)^2}=0$  (Do you know why?). 
ADDED part:
$\lim_{n\rightarrow \infty }\frac{n}{(n+2)(n+1)^2}=$\lim_{n\rightarrow \infty }\frac{1}{(1+\frac2n)(n+1)^2}=$\lim_{n\rightarrow \infty }\frac{1}{(n+1)^2}=0$
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