91Ó°ÊÓ

given,

$\left\{\frac{1â‹\dots 3â‹\dots 5⋯(2k−1)}{{10}^k}\right\}$

The sequence $\left\{a_k\right\}=\left\{\frac{1â‹\dots 3â‹\dots 5⋯(2k−1)}{{10}^k}\right\}$the general term is$a_k=\frac{1â‹\dots 3â‹\dots 5⋯(2k−1)}{{10}^k}$

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

$\begin{array}{r}\frac{a_{k+1}}{a_k}=\frac{\frac{1â‹\dots 3â‹\dots 5⋯(2k−1)(2k+1)}{{10}^{k+1}}}{\frac{1â‹\dots 3â‹\dots 5⋯(2k−1)}{{10}^k}}\text{(Substitution)}\\ =\frac{1â‹\dots 3â‹\dots 5⋯(2k−1)(2k+1)}{{10}^{k+1}}â‹\dots \frac{{10}^k}{1â‹\dots 3â‹\dots 5⋯(2k−1)}\\ =\frac{2k+1}{10}\left(\text{Simplify}\right)\\ >1\text{(For}k>4)\end{array}$

Thus $a_{k+1}>a_k$when the value of $k>4$.

The sequence $\left\{a_k\right\}=\left\{\frac{1â‹\dots 3â‹\dots 5⋯(2k−1)}{{10}^k}\right\}$is eventually strictly increasing. The given sequence is monotonic.

The sequence $\left\{a_k\right\}=\left\{\frac{1â‹\dots 3â‹\dots 5⋯(2k−1)}{{10}^k}\right\}$is bounded below because

$0<a_k$for $k>1$

As the index k increases, the term role="math" localid="1649264167687" $\left\{a_k\right\}=\left\{\frac{1â‹\dots 3â‹\dots 5⋯(2k−1)}{{10}^k}\right\}$increases.

Thus the increasing sequence has no upper bound.

The given sequence has a lower bound, Therefore, the sequence is bounded below.

The eventually strictly increasing sequence $\left\{a_k\right\}=\left\{\frac{1â‹\dots 3â‹\dots 5⋯(2k−1)}{{10}^k}\right\}$is not bounded above and hence is not convergent. therefore the sequence is not convergent.