Q. 78 - Calculus

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.

$k^p\to \infty$

Step-by-Step Solution

Answer

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

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

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

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 $\left\{a_k\right\}=\left\{k^p\right\}$ 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 $k^p\to \infty$