Answer
Convergent.
Work Step by Step
$$
\int_{-\infty}^{0} z e^{2 z} d z
$$
Observe that the given integral is improper
$$
\begin{aligned} \int_{-\infty}^{0} z e^{2 z} d z &=\lim _{t \rightarrow-\infty} \int_{t}^{0} z e^{2 z} d z \\
& \quad\quad\quad\left[\text { use integration by parts} \right] \\
& \quad\quad\quad\left[\begin{array}{c}{u=z, \quad d v=e^{2z} dz } \\ {d u=dz, \quad v=\frac{1}{2}e^{2z} \quad}\end{array}\right] \\
&= \lim _{t \rightarrow-\infty}\left[ \frac{1}{2} z e^{2 z} -\frac{1}{2}\int_{t}^{0} e^{2 z} d z \right]\\
&=\lim _{t \rightarrow-\infty}\left[\frac{1}{2} z e^{2 z}-\frac{1}{4} e^{2 z}\right]_{t}^{0} \\
&=\lim _{t \rightarrow-\infty}\left[\left(0-\frac{1}{4}\right)-\left(\frac{1}{2} t e^{2 t}-\frac{1}{4} e^{2 t}\right)\right] \\
&=\lim _{t \rightarrow-\infty}\left[\left(0-\frac{1}{4}\right)-\left(\frac{1}{2}\frac{t}{e^{-2 t}} -\frac{1}{4} e^{2 t}\right)\right] \\
& \quad\quad\quad\quad[\text { by L'Hospital's Rule }] \\
&=-\frac{1}{4}-0+0 \\
&=-\frac{1}{4} . \end{aligned}
$$
It follows that the integral
$$
\int_{-\infty}^{0} z e^{2 z} d z
$$
is convergent.
