Convergence of var$(\bar{X}_n)$ $\newcommand{\var}{\operatorname{var}}$
Given $X_1,X_2,\ldots$ is a sequence of random variables with $\rho=0$ and finite second moments.
I am trying to show that
$$\var(\bar{X}_n)\rightarrow 0\text{ as }n\rightarrow\infty$$
The obvious bit is that $\bar{X}_n=\frac1n\sum_{i=1}^n X_i$ and the exercise allows for the assumption that 
$$\frac1n\var(X_n)\rightarrow 0\text{ as }n\rightarrow\infty$$
Since there is no mention on variance being bounded, how does this change things, if at all?
 A: If all the covariances are zero, $\mathrm{var}(\bar X_n)=\frac1{n^2}\sum\limits_{k=1}^n\mathrm{var}(X_k)$. If furthermore $\frac1n\mathrm{var}(X_n)\to0$, then $\mathrm{var}(\bar X_n)\to0$.
Edit: Assume that $\frac1n\mathrm{var}(X_n)\to0$ when $n\to\infty$. Let $\varepsilon\gt0$. There exists $n_\varepsilon$ such that $\mathrm{var}(X_n)\leqslant\varepsilon n$ for every $n\geqslant n_\varepsilon$. Let $C_\varepsilon=\sum\limits_{k=1}^{n_\varepsilon}\mathrm{var}(X_k)$. Then,  for every $n$,
$$
\sum\limits_{k=1}^n\mathrm{var}(X_k)\leqslant C_\varepsilon+\varepsilon\sum\limits_{k=1}^nk\leqslant C_\varepsilon+\varepsilon n^2. 
$$
Thus, $\mathrm{var}(\bar X_n)\leqslant\frac1{n^2}C_\varepsilon+\varepsilon$ for every $n$, which implies that $\limsup\limits_{n\to\infty}\mathrm{var}(\bar X_n)\leqslant\varepsilon$. This holds for every $\varepsilon\gt0$, and $\mathrm{var}(\bar X_n)\geqslant0$ for every $n$, hence $\lim\limits_{n\to\infty}\mathrm{var}(\bar X_n)=0$.
A: $\newcommand{\var}{\operatorname{var}}$
$$
\var\left( \frac{X_1+\cdots+X_n}{n} \right) = \frac{1}{n^2} \var(X_1+\cdots+X_n) = \frac{1}{n^2}\left(\var(X_1)+\cdots+\var(X_n)\right).
$$
I think it's easy to show this does not go to $0$ as $n\to\infty$ if the sequence of variances grows fast enough.  If they're all no bigger than $A$, then the whole thing is $\le nA/n^2 = A/n$, and that goes to $0$.  One could get by with a weaker hypothesis than such boundedness, but you can't just drop the assumption.
(That $\rho\ne0$ means you don't get a bunch of covariance terms in the last expression on the first line above.)
