Q. 55 - Calculus

Question

In Exercises 55– 58 use the ratio test in Theorem 7.6 to analyze the monotonicity of the given sequence.

$\left\{\frac{k^2}{k!}\right\}$

Step-by-Step Solution

Answer

The given sequence is eventually decreasing for $k\ge 2.$

Step 1. Given Information.

The given sequence is $\left\{\frac{k^2}{k!}\right\}.$

Step 2. Use the ratio test.

To analyze the monotonicity of the given sequence we will use the ratio test.

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

So, the term $a_{k+1}$ is

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

According to the ratio test,

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

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

Thus, the given sequence is eventually decreasing for $k\ge 2.$