Answer
First five terms are: $0.36787944; 0.54134113; 0. 44808362; 0.29305022; 0.16844867$
Converges to $0$.
Work Step by Step
Plugging $n = {1,2,3,4,5}$ in $n^2 e^{-n}$.
$\implies n=1$: $(1)^2 e^{-1}=0.36787944$
$\implies n=2$: $(2)^2 e^{-2}=0.54134113$
$\implies n=3$: $(3)^2 e^{-3}=0. 44808362$
$\implies n=4$: $(4)^2 e^{-4}=0.29305022$
$\implies n=5$: $(5)^2 e^{-5}=0.16844867$
We see that $\lim\limits_{n \to \infty} n^2 e^{-n}=\lim\limits_{n \to \infty} \dfrac{2n}{e^n}\\=\lim\limits_{n \to \infty} \dfrac{2}{e^n}\\=\dfrac{2}{\infty}\\=0$
Therefore, the given series converges to $0$.
