Q. 66 - Calculus

Question

Determine whether the sequences in Exercises 63–74 are monotonic or not. Also determine whether the given sequence is bounded or unbounded.

$a_k=\frac{k^2}{k!}$

Step-by-Step Solution

Answer

The given sequence is monotonic and unbounded.

Step 1. Given Information.

The given sequence is $a_k=\frac{k^2}{k!}.$

Step 2. Determine whether the sequences are monotonic or not.

To determine whether the sequences are monotonic or not, we will use the ratio test.

Let the general term of the sequence is $a_k=\frac{k^2}{k!}.$

So, the $a_{k+1}$ term is

$a_{k+1}=\frac{{\left(k+1\right)}^2}{\left(k+1\right)!}.$

According to the ratio test,

$$\begin{gathered} \frac{a_{k+1}}{a_k} \\ =\frac{{\displaystyle \frac{{\left(k+1\right)}^2}{\left(k+1\right)!}}}{{\displaystyle \frac{k^2}{k!}}} \\ =\frac{k!{\left(k+1\right)}^2}{k^2\left(k+1\right)!} \\ =\frac{k!\left(k+1\right)\left(k+1\right)}{k^2\left(k+1\right)k!} \\ =\frac{k+1}{k^2} \end{gathered}$$

Now, $\frac{a_{k+1}}{a_k}<1 \mathrm{for} k\ge 2.$ 

Thus, the sequence is eventually decreasing for $k\ge 2,$ and it is a monotonic sequence.

Step 3. Determine whether the given sequence is bounded or unbounded.

As we have shown that the sequence is eventually decreasing for $k\ge 2.$Thus, it has an upper bound of 2 but it doesn't have a lower bound, so it is an unbounded sequence.