91Ó°ÊÓ

The geometric sequence ${\left\{c{r}^k\right\}}_{k=0}^{∞}$with $c>0$and $r>0.$

The purpose is to discuss the monotonicity and boundedness of the sequence. ${\left\{c{r}^k\right\}}_{k=0}^{∞}$with $c>0$and $r>0.$

The sequence $\left\{a_k\right\}={\left\{c{r}^k\right\}}_{k=0}^{∞}$has the general term $a_k=c{r}^k.$

The geometric sequence ${\left\{c{r}^k\right\}}_{k=0}^{∞}$with ratio $r=1$is a constant sequence with each term equal to c.

The terms of the sequence ${\left\{c{r}^k\right\}}_{k=0}^{∞}$is ${\left\{c{r}^k\right\}}_{k=0}^{∞}=\{c,c,c…\}$.

The sequence ${\left\{c{r}^k\right\}}_{k=0}^{∞}$is a constant sequence and is bounded.

The sequence ${\left\{c{r}^k\right\}}_{k=0}^{∞}$is convergent to c, and the constant sequence is always convergent.

Thus, the sequence${\left\{c{r}^k\right\}}_{k=0}^{∞}$with $c>0$is convergent for $r=1.$

The geometric sequence ${\left\{c{r}^k\right\}}_{k=0}^{∞}$with ratio $0<r<1:$

It has been noted that

$-\left|c{r}^k\right|≤c{r}^k≤\left|c{r}^k\right|$

If $\left|r\right|<1,$then

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

(or)

$0≤a_{k+1}<a_k$

The decreasing sequence $\left\{ak\right\}=\left\{c{r}^k\right\}{k=0}^{∞}$is constrained below by 0.

Convergence occurs in the monotonically decreasing sequence that is bound below.

As a result, the sequence$\left\{a_k\right\}={\left\{c{r}^k\right\}}_{k=0}^{∞}$ is convergent.

Assume that $a_k→l$.

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

$\underset{k→∞}{\lim }\left(a_{k+1}\right)=\underset{k→∞}{\lim }\left(\left|r\right|a_k\right)$(Take limit)

$l=\left|r\right|l$(Because $a_k→l$, then $(\left.a_{k+1}→l\right)$)

$l(1-|r\left|\right)=0$( Transpose )

$l=0$,$\left|r\right|=1$

$l=0$(Because $\left|r\right|<1$)

As a result, for $\left|r\right|<1$the sequence $\left\{a_k\right\}={\left\{c{r}^k\right\}}_{k=0}^{∞}$converges to $0$.

The geometric sequence ${\left\{c{r}^k\right\}}_{k=0}^{∞}$with ratio$\left|r\right|>1$.

Since $\left|r\right|>1$

Therefore,

$\frac{1}{r}<1$

The geometric sequence $\left\{\frac{1}{r^k}\right\}$with ratio$\frac{1}{r}<1$is convergent and converges to $0.$

Also, if $\underset{k→∞}{\lim }{a}_k=∞,$then$\frac{1}{a_k}→0$

The sequence $\left\{\frac{1}{c{r}^k}\right\}$is converging to 0, therefore,

$\underset{k→∞}{\lim }\frac{1}{c{r}^k}=0$

Consequently$\underset{k→∞}{\lim }c{r}^k=∞$for $\left|r\right|>1$

(Because if $\underset{k→∞}{\lim }{a}_k=∞$, then $\frac{1}{a_k}→0$)

As a result, if $\left|r\right|>1$then ${\left\{c{r}^k\right\}}_{k=0}^{∞}$is divergent.