Proof that $p$-th total variation of a brownian motion is $0$ while $p>2$ The p-th total variation is defined as $$|f|_{p,TV}=\sup_{\Pi_n}\lim_{||\Pi_n||\to n}\sum^{n-1}_{i=0}|f(x_{i+1}-f(x_{i})|^p$$
And I know how to calculate the first total variation of the standard Brownian motion. But when dealing with high order TV, there are some problem. 
At first we assume that p is even.
First I define $$\xi_i=|B_{\frac{i+1}{n}}-B_{\frac{i}{n}}|^p$$ then we can get that $$E[\xi_i]=\left(\frac1n\right)^{\frac p2}(p-1)!!$$ and $$E[\xi_i^2]=\left(\frac1n\right)^{p}(2p-1)!!$$ 
Next, we define $V_n=\sum^{n-1}_{i=0}\xi_i$
Then we have$$E[V_n]=\sum^{n-1}_{i=0}\left(\frac 1n\right)^{\frac p2}(p-1)!!$$
But there's something wrong in the following step, when calculating $E[V_n^2]$
$$\begin{align} 
E[V_n^2] &= E\left[\sum^{n-1}_{i=0}\xi_i\sum^{n-1}_{j=0}\xi_j\right]\\
&=E\left[\sum^{n-1}_{i=0}\sum^{n-1}_{j=0}\xi_i\xi_j\right]\\ 
&=\sum^{n-1}_{i=0}\sum^{n-1}_{j=0}E\left[\xi_i\xi_j\right]\\
&=\sum^{n-1}_{i=0}E[\xi_i^2]+\sum_{i\neq j}E[\xi_i]E[\xi_j]\\
&=\left(\frac1n\right)^{p-1}(2p-1)!!+n(n-1)\left(\frac1n\right)^{p}\left[(p-1)!!\right]^2
\end{align}$$
And then the question is I have no idea that how to deal with this awesome equation. Do I need to brute it out or if there is any method more efficient to prove it?

Note that $2$-variation is not the same as quadratic variation. For the quadratic variation you take the limit as the partition gets finer, whereas for $p$-variation you take the supremum over all partitions.
In particular, the Brownian motion has finite quadratic variation on any finite interval but infinite $2$-variation (link).
Now coming to your question and assuming the interval $[0,1]$, we have that
$$\|B\|_{p-\text{var}} \geq |B_1|,$$
because the supremum is larger or equal to taking the partition that only contains $0$ and $1$.
So, the $p$-variation of Brownian motion for any $p\geq 1$ is clearly not converging to zero in any sense.
