Convolution and Change of Variables in Discrete-time

This isn't a complicated example by any means but I always look at it if I need a refresh on change of variables in sum math.

Let $ x[n] = u[n-3] $ and $ h[n] = u[n-1] $.

Find: $ y[n] = x[n] * h[n] $

$ y[n] = \sum_{k=-\infty}^{\infty}x[k]h[n-k] \, $

$ y[n] = \sum_{k=-\infty}^{\infty}u[k-3]u[n-k -1] \, $

Now $ u[n - 3] = 1 \, $ when $ n-3 \ge 0 $, so $ k > 3 \, $ is the new lower bound.

$ y[n] = \sum_{k=3}^{\infty}u[n-k - 1] \, $

Now $ n - k - 1 = 1 \, $ when $ n - k -1 \ge 0 $, so $ n - 1 \ge k $ is the new upper bound.

$ y[n]=\begin{cases} \sum_{k=3}^{n-1}1, & n -1 \ge 3 \\ 0, & else \end{cases} $

$ y[n]=\begin{cases} \sum_{k=3}^{n-1}1, & n \ge 4 \\ 0, & else \end{cases} $

$ y[n] = \sum_{k=3}^{n-1}u[n-4] \, $

$ y[n] = (\sum_{k=3}^{n-1}1)u[n-4] \, $

Now let $ r = k - 3 \, $. If $ k = 3 \, $ then $ r = 0 \, $.

And the following is then true: $ k = n - 1 \rightarrow r = n - 4 $

$ y[n] = (\sum_{r=0}^{n-4}1)u[n-4] = (n-3)u[n-4]\, $