Question

Determine whether the sequence converges or diverges. If it converges, find the limit.

\({a_n} = {n^2}{e^{ - n}}\)

Short answer

Converges &\(\mathop {\lim }\limits_{n \to \infty } {a_n} = 0\)

Step-by-step solution

Definition

A sequence\(\left\{ {{a_n}} \right\}\)has the limit \(L\)and we write \(\mathop {\lim }\limits_{n \to \infty } {a_n} = L\;\;or\;\;{a_n} \to L\)as\(n \to \infty \)if we can make the terms\({a_n}\)as close to\(L\)as we like by taking\(n\)sufficiently large. If \(\mathop {\lim }\limits_{n \to \infty } {a_n}\)exists, we say the sequence converges (or is convergent). Otherwise, we say the sequence diverges (or is divergent).

Evaluate limit

Consider the sequence\({a_n} = {n^2}{e^{ - n}}\)

\(\begin{aligned}\mathop {\lim }\limits_{n \to \infty } {a_n} &= \mathop {\lim }\limits_{n \to \infty } {n^2}{e^{ - n}}\\ &= \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}}}{{{e^n}}}({\rm{ Form of }}(\infty /\infty ))\end{aligned}\)

Applying L- Hospital's rule

\(\begin{aligned}{c}\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}}}{{{e^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{2x}}{{{e^x}}}\\ = \mathop {\lim }\limits_{x \to \infty } \frac{2}{{{e^x}}}({\rm{AgainbyL - Hospital'sRule}})\\ = \frac{2}{\infty }\\ = 0\end{aligned}\)

\( \Rightarrow \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}}}{{{e^n}}} = 0\)