Science:Math Exam Resources/Courses/MATH101 B/April 2025/Question 08 (a)/Solution 1

For any number $a$, consider the general term, $e^{n^2}{e}^{-an}$ and examine its limits,

$$\underset{n\to \mathrm{\infty }}{\lim }{e}^{n^2}{e}^{-an}=\underset{n\to \mathrm{\infty }}{\lim }{e}^{n^2-an}.$$

We note that: $n^2-an=n(n-a)$. Thus, as $n\to \mathrm{\infty }$, $(n-a)\to \mathrm{\infty }$, and so $e^{n^2-an}\to +\mathrm{\infty }$, and in particular,

$$\underset{n\to \mathrm{\infty }}{\lim }{e}^{n^2-an}=+\mathrm{\infty }\ne 0.$$ Since the terms of the series do not tend to zero, the series $\underset{n\to \mathrm{\infty }}{\lim }{e}^{n^2}{e}^{-an}$ diverges for all $a$ by the divergence test.

Alternatively, one can use a ratio test.