For $V_n$ = $\sum_{i=0}^{n-1}|W(t_{i+1})-W(t_i)|^2$ where $W(t)$ is a Brownian motion, I know the following:
$$ E[V_n] = \sum_{i=0}^{n-1} E[(W(t_{i+1}) - W(t_i))^2] = \sum_{i=0}^{n-1} (t_{i+1} - t_i) = T $$
Now, suppose I need to find the expectation of $Y_n$ which is defined as follows: $$ Y_n = \sum_{i=0}^{n-1} |(W(t_{i+1}) - W(t_i))^3| $$
How can I solve this? I have done the following and would like to know if I am on the right track:
$$ \sum_{i=0}^{n-1} |(W(t_{i+1}) - W(t_i))^3| = \sum_{i=0}^{n-1} |(W(t_{i+1}-W(t_i))(W(t_{i+1}) - W(t_i))^2| \\[2ex] \leq \sum_{i=0}^{n-1} |(W(t_{i+1}-W(t_i))| |(W(t_{i+1}) - W(t_i))^2| \\[2ex] \leq max_{i=0,1,...,n-1} |(W(t_{i+1}-W(t_i))| \sum_{i=0}^{n-1} |(W(t_{i+1}) - W(t_i))^2| $$
So:
$$ E[Y_n] = E\bigg[max_{i=0,1,...,n-1} |(W(t_{i+1}-W(t_i))| \sum_{i=0}^{n-1} |(W(t_{i+1}) - W(t_i))^2|\bigg] $$
Now, I am a bit confused because I do not know how I can take the expectation of the terms. Does the following make sense? (are the terms independent so I can do this):
$$ E[Y_n] = E\bigg[max_{i=0,1,...,n-1} |(W(t_{i+1}-W(t_i))|\bigg] \sum_{i=0}^{n-1} E[|(W(t_{i+1}) - W(t_i))^2|] \\[2ex] E[Y_n] = 0 \times T $$
You're trying to bound the random variable $Y_n$ and then take the expectation of the bound. It is easier to take the expectation first and then bound it.
We know that $W_{t_{i+1}} - W_{t_i} \sim N(0, t_{i+1}-t_i)$. Now one can see by scaling that the absolute third moment of $N(0, \sigma^2)$ must be of the form $c \sigma^{3}$. (In fact $c = 2^{3/2}/\sqrt{\pi}$ according to Wikipedia). So we have $E[|W_{t_{i+1}} - W_{t_i}|^3] = c (t_{i+1}-t_i)^{3/2}$. Hence $E[Y_n] = c \sum_i (t_{i+1}-t_i)^{3/2}$.
Now we can bound the latter by $$E[Y_n] \le c\max(t_{i+1} - t_i)^{1/2} \sum (t_{i+1} - t_i) = cT \max(t_{i+1} - t_i)^{1/2}.$$ I presume that when you say "as $n \to \infty$" you mean "as the mesh size $\max(t_{i+1} - t_i) \to 0$". Obviously just having more points in our partition won't ensure convergence unless they are actually all getting close to each other. Thus we indeed have $E[Y_n] \to 0$.
The same idea works for any $p > 2$, using instead that for $Z \sim N(0, \sigma^2)$ we have $E[|Z|^p] = c_p \sigma^{p}$, again by scaling, where the value of the constant $c_p$ is irrelevant here. You will then bound the $p$-variation by $c_p T \max(t_{i+1}-t_i)^{(p-1)/2}$ which likewise goes to zero.