Convolution of discrete time signal

Question

I'm given a discrete time signal which is as follows:

$$ x[n] = \alpha^n u[n],\\$$ $$ h[n] = u[n],\\$$

Why is unit function neglected in the following step:

$$ y[n] = \sum_{k=0}^{n} \alpha^{k}\\$$

Answer 1

It isn't. The convolution sum is

$$y[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k]=\sum_{k=-\infty}^{\infty}\alpha^ku[k]u[n-k]\tag{1}$$

The step $u[k]$ in $(1)$ can be removed if the index $k$ starts at zero, and the step $u[n-k]$ can be removed by replacing the upper summation limit by $n$. For clarity, one should multiply the result by $u[n]$ because if $n<0$ the result of the convolution is zero:

$$y[n]=u[n]\sum_{k=0}^{n}\alpha^k\tag{2}$$