Calculate mean square limit of Brownian motion - 4th moment of normal distribution

An example interpreting the mean square limit of a brownian motion.
Assumption: X(t) follows a brownian motion.

Examine the quantity:

$$E[(\sum_{j=1}^{n}(X(t_j)-X(t_{j-1}))-t)^2]$$

This can be expanded as:
$$E[\sum_{j=1}^{n}(X(t_j)-X(t_{j-1}))^4+\sum_{i=1}^{n}\sum_{i\neq j}(X(t_i)-X(t_{i-1}))^2(X(t_j)-X(t_{j-1}))^2$$ $$+\sum_{j=1}^{n}(X(t_j)-X(t_{j-1}))^2+t^2]$$ since $$E[(X(t_j)-X(t_{j-1}))^2]=\frac{t}{n}$$ and $$E[(X(t_j)-X(t_{j-1}))^4]=\frac{3t^2}{n^2}$$ which can be calculated as(Y~normal): $$E(Y^4)= \int_{-\infty}^{\infty}Y^4\frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{Y^2}{\sigma^2}}dY$$ $$= -\int_{-\infty}^{\infty}Y^3\frac{\sigma^2}{\sqrt{2\pi\sigma^2}}de^{\frac{Y^2}{\sigma^2}}=\int_{-\infty}^{\infty}3Y^2\frac{\sigma^2}{\sqrt{2\pi\sigma^2}}e^{\frac{Y^2}{\sigma^2}}dY=3\sigma^2\sigma^2=3\sigma^4$$ Thus the original equation becomes: $$n\frac{3t^2}{n^2}+n(n-1)\frac{t^2}{n^2}-2tn\frac{t}{n}+t^2=O(\frac{1}{n})$$ As n converts to infinity, this tends to zero. We therefore have that: $$\sum_{j=1}^{n}(X(t_j)-X(t_{j-1}))^2=t$$