Determine whether the sequence converges or diverges. If it c DNE.)
$$
\begin{array}{c}
a_{n}=n^{2} e^{-3 n} \
\lim _{n \rightarrow \infty} a_{n}=\square
\end{array}
$$

Question

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ We start by considering the given sequence $ a_n = n^2 e^{-3n} $. ‖ ‖ step 2 ⋮ To find the limit of $ a_n $ as $ n $ approaches infinity, we need to evaluate $ \lim_{n \to \infty} n^2 e^{-3n} $. ‖ ‖ step 3 ⋮ Notice that as $ n $ approaches infinity, $ e^{-3n} $ approaches 0 very rapidly because the exponential function decays faster than any polynomial grows. ‖ ‖ step 4 ⋮ We can use L'Hôpital's Rule to evaluate the limit. Rewrite the expression as $ \frac{n^2}{e^{3n}} $. ‖ ‖ step 5 ⋮ Apply L'Hôpital's Rule, which states that if the limit of $ \frac{f(n)}{g(n)} $ as $ n $ approaches infinity is in the form $ \frac{\infty}{\infty} $ or $ \frac{0}{0} $, then $ \lim_{n \to \infty} \frac{f(n)}{g(n)} = \lim_{n \to \infty} \frac{f'(n)}{g'(n)} $. ‖ ‖ step 6 ⋮ Differentiate the numerator and the denominator: $ f(n) = n^2 $ and $ g(n) = e^{3n} $. So, $ f'(n) = 2n $ and $ g'(n) = 3e^{3n} $. ‖ ‖ step 7 ⋮ Apply L'Hôpital's Rule: $ \lim_{n \to \infty} \frac{n^2}{e^{3n}} = \lim_{n \to \infty} \frac{2n}{3e^{3n}} $. ‖ ‖ step 8 ⋮ Apply L'Hôpital's Rule again: Differentiate the numerator and the denominator: $ f'(n) = 2n $ and $ g'(n) = 3e^{3n} $. So, $ f''(n) = 2 $ and $ g''(n) = 9e^{3n} $. ‖ ‖ step 9 ⋮ Apply L'Hôpital's Rule again: $ \lim_{n \to \infty} \frac{2n}{3e^{3n}} = \lim_{n \to \infty} \frac{2}{9e^{3n}} $. ‖ ‖ step 10 ⋮ As $ n $ approaches infinity, $ e^{3n} $ grows exponentially, making $ \frac{2}{9e^{3n}} $ approach 0. ‖ ‖ step 11 ⋮ Therefore, $ \lim_{n \to \infty} n^2 e^{-3n} = 0 $. ‖ ∻Answer∻ ⚹ The sequence $ a_n = n^2 e^{-3n} $ converges to 0 as $ n $ approaches infinity. ⚹ ∻Key Concept∻ ⚹Exponential decay vs. polynomial growth⚹ ∻Explanation∻ ⚹The exponential function $ e^{-3n} $ decays much faster than the polynomial $ n^2 $ grows, leading the product $ n^2 e^{-3n} $ to approach 0 as $ n $ approaches infinity.⚹◊What is the mathematical definition of a sequence converging or diver⍭ Generate me a similar question◊

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