Output for h[n] = u[n], x[n] = 2ⁿu[−n]

Given: $h[n] = u[n]$ (accumulator) and $x[n] = 2^n u[-n]$, which is non-zero only for $n \le 0$: $x[n] = \{\dots, \tfrac{1}{8}, \tfrac{1}{4}, \tfrac{1}{2}, \underset{\uparrow}{1}\}$.

$$y[n] = \sum_{k=-\infty}^{\infty} x[k]\,h[n-k] = \sum_{k=-\infty}^{\infty} 2^k u[-k]\,u[n-k]$$

Limits: $u[-k] = 1$ for $k \le 0$; $u[n-k] = 1$ for $k \le n$. So $k$ runs from $-\infty$ to $\min(n, 0)$.

Case 1: $n \le 0$ (here $\min = n$)

$$\begin{aligned} y[n] &= \sum_{k=-\infty}^{n} 2^k \quad \text{(put } m = n - k \text{, } m = 0 \text{ to } \infty\text{)} \\ &= 2^n\sum_{m=0}^{\infty}\left(\tfrac{1}{2}\right)^m = 2^n\cdot\frac{1}{1 - \frac{1}{2}} \\ &= 2^{n+1} \end{aligned}$$

Case 2: $n > 0$ (here $\min = 0$)

$$y[n] = \sum_{k=-\infty}^{0} 2^k = 1 + \tfrac{1}{2} + \tfrac{1}{4} + \dots = \frac{1}{1 - \frac{1}{2}} = 2$$

Answer:

$$y[n] = \begin{cases} 2^{n+1}, & n \le 0 \\ 2, & n > 0 \end{cases} \quad\text{i.e.}\quad y[n] = 2^{n+1}u[-n-1] + 2\,u[n]$$

(At $n = 0$ both forms give 2, so the pieces join.)

 y[n]
  2 |           o  o  o  o ...
  1 |        o  |  |  |  |
0.5 |     o  |  |  |  |  |
    |  o  |  |  |  |  |  |
  0 +--+--+--+--+--+--+--+--> n
      -3 -2 -1  0  1  2  3

This makes sense: $h[n] = u[n]$ is an accumulator, so $y[n] = \sum_{k=-\infty}^{n} x[k]$, the running sum of the input. The sum keeps growing until $n = 0$, then stays at the total $\sum x[k] = 2$ because the input is zero for $n > 0$.