Q. 56 - Calculus

Question

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

$\left\{\sqrt{1-\frac{1}{k}}\right\}$

Step-by-Step Solution

Answer

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

Step 1. Given Information.

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

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 $a_k=\sqrt{1-\frac{1}{k}}.$

So, the term $a_{k+1}$ is

$a_{k+1}=\sqrt{1-\frac{1}{1+k}}.$

According to the ratio test,

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

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

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