Sample variance order Is it true that (and if so, how does one prove) the following.
 $$E\left|\hat{Var}_{n}(X)-Var(X)\right|^{2}=O(n^{-1})$$
  where:
• $X$ is a random variable with mean $\mu$ and variance $\sigma^{2}$
• $\hat{Var}_{n}(X)$ is the sample variance of $X$ from $n$
  i.i.d. random variables $X_{1},\cdots,X_{n}$ with mean $\mu$
  and variance $\sigma^{2}$.
Many thanks in advance. (Feel free to change my notation).
 A: I assume by the term, sample variance, you are referring to the unbiased estimator of population variance $\mu_2$, i.e. the 2nd h-statistic, namely:
$$ h_2 = \hat{Var}_{n}(X) = \frac{1}{n-1}\sum _{i=1}^n \left(X_i-\bar{X}\right){}^2$$
Either way, the expectation you seek is just the MSE of $\hat{Var}_{n}(X)$. Note that the absolute value is irrelevant due to the squaring. That should be enough to do a google search and find an answer somebody has worked out in a journal paper or book. 
More generally, these sorts of calculations are known as moments of moments and can be solved by working with power sum notation $s_r=\sum _{i=1}^n X_i^r$. First, express h2 in terms of power sums:

where I am using the HStatistic function in the mathStatica software (of which I am one of the authors). Next: find $E[(h_2-\mu_2)^2]$ ... which is just the 1st RawMoment of $(h_2-\mu_2)^2$, so we can find it with:

The ___ToCentral bit expresses the answer in terms of central moments $\mu_i$ of the population.
All done ... you can now work out what happens as $n$ gets large etc
