Question

Prove the statements about the convergence or divergence of sequences in Exercises 78–83, referring to theorems in the section as necessary. For each of these statements, assume that r is a real number and p is a positive real number.

$k^p\to \infty$

Short answer

The given sequence$\left\{a_k\right\}=\left\{k^p\right\}$is divergence

Step-by-step solution

Step 1. Given information

The given sequence $k^p\to \infty$

Step 2. Find the given sequence is increasing or decreasing

The general term of the sequence$\left\{a_k\right\}=\left\{k^p\right\}is{a}_k=k^p$

The ratio $\frac{a_{k+1}}{a_k}$gives

$$\begin{gathered} \frac{a_{k+1}}{a_k}=\frac{{\left(k+1\right)}^p}{k^p}\left(substitution\right\} \\ ={\left(\frac{K+1}{k}\right)}^p \\ ={\left(1+\frac{1}{k}\right)}^p\left(simplify\right) \\ >1(Fork>0) \end{gathered}$$

Thus,$a_{k+1}>a_k$

The sequence $\left\{a_k\right\}=\left\{k^p\right\}$is strickly increasing sequence

Step 3. Find the given sequence is converges or divergent 

The sequence $\left\{a_k\right\}=\left\{k^p\right\}$is bounded below as for $p>0$

$0<a_k$

The sequence$\left\{a_k\right\}=\left\{k^p\right\}$is increasing and there is no upper bound

The monitoring increasing sequence which is bounded above is convergent

The sequence role="math" localid="1649300083524" $\left\{a_k\right\}=\left\{k^p\right\}$is strickly increasing but it is not bounded above.Hence the sequence is divergent

The sequence $\left\{a_k\right\}=\left\{k^p\right\}$is divergent.Therefore,

$$\begin{gathered} \underset{k\to \infty }{\lim }{a}_k=\underset{x\to \infty }{\lim }{k}^p \\ =\infty \end{gathered}$$

Hence,for$p>0$ it is proved that$k^p\to \infty$