Q. 82 - 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.

If $\left|r\right|<1,$then the sequence $r^k\to 0$

Step-by-Step Solution

Answer

The sequence $\left\{a_k\right\}=\left\{r^k\right\}$ for $\left|r\right|<1$ converges to 0

Step 1. Given information

The given sequence if $\left|r\right|<1$,then the sequence $r^k\to 0$

Step 2. To prove that,use the result of convergence of monotonic sequence

The general term of the sequence $\left\{a_k\right\}=\left\{r^k\right\} is a_k=r^4$

If $r=0$the geometric sequence $\left\{r^k\right\}$ with ration $r=0$ is a constant sequence with each term equal to 0.

The term of the sequence $\left\{r^k\right\}$is

$\left\{r^k\right\}=\left\{0,0,0......\right\}$

The sequence $\left\{r^k\right\}$ is a constant sequence and is bounded

The constant sequence is a always convergent and the sequence $\left\{r^k\right\}$ is converging to 0.

Therefore,$r=0 then r^k\to 0$holds

Now it is sufficient to prove the result for $0<r<1$

It is observed that 

$-\left|r^n\right|\le r^n\le \left|r^n\right|$

If $\left|r\right|<1$ then

$a_{k+1}=\left|r\right|a_k$

Also,$0\le a_{k+1}<a_k$

The sequence $\left\{a_k\right\}=\left\{r^k\right\}$ is decreasing sequence and is bounded below by 0.

The monitonic decreasing sequence which is bounded below is convergent

Therefore,sequence $\left\{a_k\right\}=\left\{r^k\right\}$ is convergent

Assume that $a_k\to l$

Therefore,$a_{k+1}=\left|r\right|a_k$

$\underset{k\to \infty }{\lim }\left(a_{k+1}\right)=\underset{k\to \infty }{\lim }\left(\left|r\right|a_k\right)$ (take limit)

$$\begin{gathered} l=\left|r\right|l \left(Because a_k\to 1,then a_{k+1}\to 1\right) \\ l\left(1-\left|r\right|\right)=0 \left(Transpose\right) \\ l=0,\left|r\right|=1 \\ l=0 (Because \left|r\right|<1) \end{gathered}$$

Therefore,the sequence $\left\{a_k\right\}=\left\{r^k\right\} for \left|r\right| <1$ converges to 0