I am currently trying to understand some lecture notes (which offer little depth) and the following exercise (without a solution) occurs:
Using the definition $$ \int_0^t f_s \hspace{1mm} dW_s := \text{l.i.m}_{\lambda \rightarrow 0} \left[ \sum_i f_{s_i} (W_{s_{i+1}} - W_{s_i}) \right] \hspace{10mm} (*) $$ where l.i.m = limit in (square) mean, show that $$ \int_0^t W_s \hspace{1mm} dW_s = \frac{W_t^2}{2} + \frac{t}{2} $$
Can someone please help me to understand how to do this? Plugging $W_s$ into the formula $(*)$ we get $$ \int_0^t W_s \hspace{1mm} dW_s = \text{l.i.m}_{\lambda \rightarrow 0} \left[ \sum_i W_{s_i} (W_{s_{i+1}} - W_{s_i}) \right] $$
Unfortunately, I am not sure where to go from here. I am also not certain what $\lambda$ is supposed to denote, since it is not defined in the lecture notes. I suspect (by comparing this formula to the formula for Riemann integration) that $$ \lambda = s_{i+1} - s_i $$
The integral value should be: $$\int_0^tW_s\;dW_s=\frac{W_t^2}{2}-\frac{t}{2}$$ Notice that $$\sum_iW_{s_i}(W_{s_{i+1}}-W_{s_i})=\frac{1}{2}\sum_i(W_{s_{i+1}}^2-W_{s_i}^2)-\frac{1}{2}\sum_i(W_{s_{i+1}}-W_{s_i})^2$$ The first sum is a telescoping sum that reduces to $\frac{W_t^2}{2}$ (since $W_0 = 0$). The second sum becomes half the quadratic variation of $W_s$. So your integral is...