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?
 A: It is widely known that for $p=2$ the quadratic variation
$$S_{\Pi} := \sum_{t_i \in \Pi} |B_{t_{i+1}}-B_{t_i}|^2$$
converges to $T$ in $L^2$ as $|\Pi| \to 0$. Here $\Pi = \{0=t_0<\ldots< t_n \leq T\}$ denotes a partition of the interval $[0,T]$ with mesh size
$$|\Pi| = \max_i |t_{i+1}-t_i|.$$
For a proof of this statement see any book on Brownian motion (e.g. René L. Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 9). Since $L^2$-convergence implies almost sure convergence of a subsequence, we may assume that
$$S^{\Pi_n} \to T \qquad \text{almost surely} \tag{1}$$
for a (given) sequence of partitions $(\Pi_n)_n$ satisfying $|\Pi_n| \to 0$. Now, as $p>2$,
$$\begin{align*} \sum_{t_i \in \Pi_n} |B_{t_{i+1}}-B_{t_i}|^p &= \sum_{t_i \in \Pi_n} |B_{t_{i+1}}-B_{t_i}|^{p-2} |B_{t_{i+1}}-B_{t_i}|^2 \\ &\leq \sup_{s,t \leq T, |s-t| \leq |\Pi_n|} |B_t-B_s|^{p-2} S_{\Pi_n}. \tag{2}\end{align*}$$
Since $(B_t)_{t \geq 0}$ is a Brownian motion, we know that $t \mapsto B_t(\omega)$ is continuous almost surely; hence, $[0,T] \ni t \mapsto B_t(\omega)$ is uniformly continuous. This implies in particular
$$\sup_{s,t \leq T, |s-t| \leq |\Pi_n|} |B_t-B_s| \stackrel{n \to \infty}{\to} 0 \qquad \text{almost surely}$$
as the mesh size $|\Pi_n|$ tends to $0$. Combining $(1)$ and $(2)$ yields
$$\lim_{n \to \infty} \sum_{t_i \in \Pi_n} |B_{t_{i+1}}-B_{t_i}|^p = 0.$$
This finishes the proof.
Remark: For a more general statement see this question.
A: 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.
