Question

Give examples of sequences satisfying the given conditions or explain why such an example cannot exist.

Complete Example 3 by showing that the limit of the sequence$\left\{\frac{k^3}{e^k}\right\}$is$0$.

Short answer

To prove that the limit of $\left\{a_k\right\}=\left\{\frac{k^3}{e^k}\right\}$is $0$, first apply the L'Hospital's Rule and simply then again apply the L'Hospital's Rule and simply the following.

Step-by-step solution

Step 1. Given information.

Consider the given question,

The limit of the sequence$\left\{\frac{k^3}{e^k}\right\}$is$0$.

Step 2. Find the limit.

Consider the sequence $\left\{a_k\right\}=\left\{\frac{k^3}{e^k}\right\}$.

The given sequence is bounded and eventually monotonic. Then the sequence is convergent.

The limit $\underset{k\to \infty }{\lim }{a}_k=\underset{k\to \infty }{\lim }\frac{k^3}{e^k}$.

The expression $\frac{k^3}{e^k}$is indeterminate from of type $\frac{\infty }{\infty }$.

Applying limits,

$$\begin{gathered} \underset{k\to \infty }{\lim }{a}_k=\underset{k\to \infty }{\lim }\frac{3k^2}{e^k} \\ =\underset{k\to \infty }{\lim }\frac{6k}{e^k} \\ =\underset{k\to \infty }{\lim }\frac{6}{e^k} \\ =0 \end{gathered}$$

Hence, it has been proven that the limit of the sequence$\left\{a_k\right\}=\left\{\frac{k^3}{e^k}\right\}$is$0$.