Does this series converges?

We apply the root test, so we must investigate the limit $\begin{aligned}\lim_{n\to\infty}\sqrt[n]{\left|e^{an^2}\left(1-\frac{a}{n}\right)^{n^3}\right|} &= \lim_{n\to\infty} e^{an}\left|1-\frac{a}{n}\right|^{n^2} \\ &= \lim_{h\to 0^+} e^{a/h}\left|1-ah\right|^{1/h^2} \\ &= \exp\left(\lim_{h\to 0^+} \frac{ah + \ln|1-ah|}{h^2}\right) \qquad & \text{by continuity of } e^x = \exp(x)\\ &= \exp\left(\lim_{h\to 0^+} \frac{a + \frac{-a}{1-ah}}{2h}\right) & \text{by L'Hopital}\\ &= \exp\left(\lim_{h\to 0^+} \frac{-\frac{a^2}{(1-ah)^2}}{2}\right) & \text{by L'Hopital again}\\ &= \exp\left(-\frac{a^2}{2}\right) = e^{-\frac{a^2}{2}} \end{aligned}$ Since this limit exists and $e^{-a^2/2} < 1$ for all nonzero $a$ , it follows that the series converges for all nonzero $a$ .

For $a=0$ , the root test is inconclusive, but luckily setting $a=0$ in the original series yields $\sum_{n=1}^\infty e^{0\cdot n^2}\left(1-\frac{0}{n}\right)^{n^3} = \sum_{n=1}^\infty 1$ which obviously diverges.

We conclude that the series converges if and only if $a\neq 0$ .