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$r^k\to 0$

Short answer

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

Step-by-step solution

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,$r=0then{r}^k\to 0$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