Brownian motion: Show $\lim \sum W_{i} (W_{i+1}-W_{i})=\frac12 W^2_t-\frac12 t$ in probability. Let $\{t_i\}_{i=1}^n$ be a partition of $[0,t]$ and $W$ a standard Brownian motion. Write $W_i$ for $W_{t_i}$.
Show
$$
\lim \sum W_{i} (W_{i+1}-W_i)=\frac12 W^2_t-\frac12 t
$$
where the limit is in probability.
The proof is in our textbook (Kurtz, Stochastic Analysis). It goes as follow
\begin{align}
\lim \sum_{i=1}^n W_{i} (W_{i+1}-W_i) &= \lim \sum_{i=1}^n \left( W_i W_{i+1} - \frac12 W^2_{i+1}-\frac12 W^2_i \right)+\sum_{i=1}^n \left( \frac12 W^2_{i+1} - \frac12 W^2_i \right) \\
&=\frac12 W_t^2 - \lim \frac12 \sum_{i=1}^n \left( W_{i+1}-W_{i} \right)^2 \\
&=\frac12 W_t^2 - \frac12 t^2
\end{align}
How does the second equality follows?
 A: The second sum in the RHS of the first equality is telescopic and gives the first term in the second line. For the first sum, note that 
$$W_iW_{i+1}-\frac 12W_{i+1}^2-\frac 12W_i^2=-\frac 12(W_{i+1}-W_i)^2,$$
hence a factor $\frac 12$ is missing.
A: Usually, for a Riemann integral, you say 
$$ \int f(t) dt = \lim_{|\delta t| \to 0} \sum f(t) \Delta t $$
Your textbook is showing that $\Delta W = W_{i+1}- W_i \approx \sqrt{\Delta t}$ since variance of random walk grows as the square-root of time.
First of all there's a telescoping sum:
$$ \sum_{i=1}^n \left( \frac12 W^2_{i+1} - \frac12 W^2_i \right) = \frac12 W_t^2 $$
and then binomial formula $(x+y)^2 = x^2 + 2xy + y^2$:
$$ \lim \sum_{i=1}^n \left( W_i W_{i+1} - \frac12 W^2_{i+1}-\frac12 W^2_i \right)
=  \lim \sum_{i=1}^n \left( W_{i+1}-W_{i} \right)^2  =  \frac12 t^2 $$
The word "lim" is very important.  We can approximate the random walk over small time interval as binomial random variable
$$ \mathbb{P}(\Delta W = \sqrt{\Delta t} )= \mathbb{P}(\Delta W = -\sqrt{\Delta t} )= \frac{1}{2}$$
and no matter what $(\Delta W)^2 = \Delta t$.
