How do you calculate the following limit of a standard Brownian motion $lim_{n\rightarrow \infty} E\left[\sum_{k=1}^{n}\left(B(\frac{kt}{n})-B(\frac{(k-1)t}{n}) \right)^2\right]$, where $\{B(t); t\geq 0\}$ is a standard Brownian motion.
Here is my attempt to solve it. I'd appreciate any feedback on it: 
Recall that a standard Brownian motion has stationary increments and thus, $B\left( \frac{kt}{n} \right)-B\left( \frac{(k-1)t}{n} \right)$ ~$ N(0, t/n)$ . Further, by the CLT we get that $\frac{B\left( \frac{kt}{n} \right)-B\left( \frac{(k-1)t}{n} \right)}{\sqrt{\frac{t}{n}}}$ ~ $N(0,1)$. So, if we let for each n $Z_{n,k}= \frac{B\left( \frac{kt}{n} \right)-B\left( \frac{(k-1)t}{n} \right)}{\sqrt{\frac{t}{n}}}$, then $\{Z_{n,k}\}$  is a sequence of i.i.d N(0,1) random variables. Furthermore, by the WLLN we have that $E[\frac{1}{n}\sum_{k=1}^nZ^2_{n,k}]\rightarrow 1$ as $n\rightarrow \infty$. Therefore, $$lim_{n\rightarrow \infty} E\left[\sum_{k=1}^{n}\left(B(\frac{kt}{n})-B(\frac{(k-1)t}{n}) \right)^2\right] = lim_{n\rightarrow \infty} E \left[\sum_{k=1}^n \frac{t}{n}Z^2_{n,k} \right]=t\cdot lim_{n\rightarrow}E[\frac{1}{n}\sum_{k=1}^nZ^2_{n,k}]=t$$
 A: By linearity of expectation and expanding the quadratic term, we have
\begin{align}
\mathbb E\left[ \sum_{k=1}^n \left(B\left(\frac{kt}n\right) - B\left(\frac{(k-1)t}n\right)\right)^2\right] &= \sum_{k=1}^n \mathbb E\left[B\left(\frac{kt}n\right) - B\left(\frac{(k-1)t}n\right) \right]^2\\
&= \sum_{k=1}^n\mathbb E\left[B\left(\frac{kt}n\right)^2 - 2 B\left(\frac{kt}n\right)B\left(\frac{(k-1)t}n\right) + B\left(\frac{(k-1)t}n\right)^2 \right]\\
&= \sum_{k=1}^n\left(\mathrm{Var}\left(B\left(\frac{kt}n\right)\right) - 2\mathrm{Cov}\left((B\left(\frac{kt}n\right),\left(\frac{(k-1)t}n\right)\right) + \mathrm{Var}\left(B\left(\frac{(k-1)t}n\right)\right)\right)\\
&= \sum_{k=1}^n \mathrm{Var}\left(B\left(\frac{kt}n\right)\right) - 2\sum_{k=1}^n \mathrm{Cov}\left((B\left(\frac{kt}n\right),\left(\frac{(k-1)t}n\right)\right) + \sum_{k=1}^n \mathrm{Var}\left(B\left(\frac{(k-1)t}n\right)\right)\\
&= \sum_{k=1}^n \frac{kt}n - 2 \sum_{k=1}^n \frac{(k-1)t}n +\sum_{k=1}^n \frac{(k-1)t}n\\
&= \frac tn\left(\sum_{k=1}^n k - 2\sum_{k=1}^n (k-1) + \sum_{k=1}^n (k-1) \right)\\
&= \frac tn\left(\sum_{k=1}^n k - \sum_{k=1}^n (k-1) \right)\\
&= t.
\end{align}
