Q. 17 - Calculus

Question

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

Find a divergent sequence $\left\{a_k\right\}$ such that the sequence given by $\left\{\frac{k!}{a_k}\right\}$ converges.

Step-by-Step Solution

Answer

Examples of the sequence is $a_k={\left(k!\right)}^2$.

Step 1. Given information.

Consider the given question,

The condition is $\left\{\frac{k!}{a_k}\right\}$.

Step 2. Consider a divergent sequence.

Consider the divergent sequence $\left\{a_k\right\}=\left\{{\left(k!\right)}^2\right\}$.

The sequence $\left\{a_k\right\}=\left\{{\left(k!\right)}^2\right\}$ is an increasing sequence and is not bounded above.

Thus, the sequence is divergent.

The sequence $\left\{\frac{k!}{a_k}\right\}=\left\{\frac{k!}{{\left(k!\right)}^2}\right\}$ is written below,

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

Step 3. Find the ratio.

The general term of the sequence $\left\{A_k\right\}=\left\{\frac{k!}{a_k}\right\}$ is $A_k=\frac{1}{k!}$.

The ratio $\frac{A_{k+1}}{A_k}$ gives,

$$\begin{gathered} \frac{A_{k+1}}{A_k}=\frac{k!}{\left(k+1\right)!} \\ \frac{A_{k+1}}{A_k}=\frac{1}{k} \\ \frac{1}{k}\le 1 \end{gathered}$$

Thus, $A_{k+1}\le A_k$.

Step 4. Determine the boundedness of the sequence.

The sequence $\left\{A_k\right\}=\left\{\frac{1}{k!}\right\}$ is bounded below as,

$0<A_k$ for $k>0$.

The monotonically sequence which is decreasing and is bounded below is convergent.

Thus, the sequence is convergent.

The divergent sequence $\left\{a_k\right\}$ such that the sequence $\left\{\frac{k!}{a_k}\right\}$ becomes convergent is $\left\{a_k\right\}=\left\{{\left(k!\right)}^2\right\}$.