Question

Determine whether the sequence is monotonic or eventually monotonic and whether the sequence is bounded above and/or below. If the sequence converges, give the limit.

$\left\{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{{10}^k}\right\}$

Short answer

Ans: The sequence$\left\{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{{10}^k}\right\}$is not convergent.

Step-by-step solution

Step 1. Given information.

given,

$\left\{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{{10}^k}\right\}$

Step 2. The objective is to determine whether the sequence is monotonic, bounded above, or bounded below, and to find the limit of the sequence if the sequence is convergent.   

The sequence $\left\{a_k\right\}=\left\{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{{10}^k}\right\}$the general term is$a_k=\frac{1\cdot 3\cdot 5\cdots (2k-1)}{{10}^k}$

Step 3.  The general term of the sequence is ak=1⋅3⋅5⋯(2k−1)10k 

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

$\begin{array}{r}\frac{a_{k+1}}{a_k}=\frac{\frac{1\cdot 3\cdot 5\cdots (2k-1)(2k+1)}{{10}^{k+1}}}{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{{10}^k}}\text{(Substitution)}\\ =\frac{1\cdot 3\cdot 5\cdots (2k-1)(2k+1)}{{10}^{k+1}}\cdot \frac{{10}^k}{1\cdot 3\cdot 5\cdots (2k-1)}\\ =\frac{2k+1}{10}\left(\text{Simplify}\right)\\ >1\text{(For}k>4)\end{array}$

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

The sequence $\left\{a_k\right\}=\left\{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{{10}^k}\right\}$is eventually strictly increasing. The given sequence is monotonic.

Step 4. Now,

The sequence $\left\{a_k\right\}=\left\{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{{10}^k}\right\}$is bounded below because

$0<a_k$for $k>1$

As the index k increases, the term role="math" localid="1649264167687" $\left\{a_k\right\}=\left\{\frac{1\cdot 3\cdot 5\cdots (2k-1)}{{10}^k}\right\}$increases.

Thus the increasing sequence has no upper bound.

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

Step 5. The monotonic increasing sequence is bounded above is convergent 

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