Answer
$$Convergent,\frac{1}{ln2}$$
Work Step by Step
$$\int_{-\infty }^{0}2^{r}dr=\lim_{t\rightarrow -\infty}\int_{t}^{0}2^{r}dr$$
$$=\lim_{t\rightarrow -\infty}\left| \frac{2^{r}}{ln2} \right |_{t}^{0}$$
$$= \lim_{t\rightarrow -\infty} \frac{1-2^{t}}{ln2}$$
$$=\frac{1}{ln2}$$
Convergent.