Answer
Converges
Work Step by Step
Here, we have $a_n= \dfrac{n}{e^{n^2}}$
Ratio Test states that when $\Sigma a_n$ is an infinite series with positive terms and, then $l=|\lim\limits_{n \to \infty}\dfrac{a_{n+1}}{a_n}|$
a) When $0 \leq l \lt 1$, the series converges. (b) When $l \gt 1$, or, $\infty$, so the series diverges. (c) When $l=1$, the ratio test is said to be inconclusive.
Now, $l=\lim\limits_{n \to \infty}|\dfrac{\dfrac{n+1}{e^{(n+1)^2}}}{\dfrac{n}{e^{n^2}}}|\\=\lim\limits_{n \to \infty}|\dfrac{n+1}{n e^{2n+1}}|\\=\lim\limits_{n \to \infty}|\dfrac{n+1}{n}| \times \lim\limits_{n \to \infty}|\dfrac{1}{e^{2n+1}}|\\=0 $
so, $l \lt 1$
Therefore, the series converges by the ratio test.
