Question

In Exercises 51–54 use the difference test in Theorem 7.6 to analyze the monotonicity of the given sequence.

$\left\{\frac{1}{k!}\right\}$

Short answer

The given sequence is strictly decreasing for$k\ge 1.$

Step-by-step solution

Step 1. Given Information.

The given sequence is$\left\{\frac{1}{k!}\right\}.$

Step 2. Use the difference test. 

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

Let the general term of the sequence is$a_k=\frac{1}{k!}.$

So, the term $a_{k+1}$is

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

According to the test,

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

Now, $a_{k+1}-a_k\le 0\mathrm{for}\mathrm{all}k\ge 1.$

Thus, the sequence is strictly decreasing for$k\ge 1.$