Question

Determine whether the sequence converges or diverges. If the sequence converges, give the limit.

$\left\{(k!{)}^{1/k}\right\}$

Short answer

Ans: The sequence$\left\{a_k\right\}=\left\{(k!{)}^{1/k}\right\}$is divergent.

Step-by-step solution

Step 1. Given information.

given,

$\left\{(k!{)}^{1/k}\right\}$

Step 2. The objective is to determine whether the sequence is convergent or divergent and to find the limit of the sequence if the sequence is convergent.

In the sequence $\left\{a_k\right\}=\left\{(k!{)}^{1/k}\right\}$the general term is $a_k=(k!{)}^{1/k}$

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

role="math" localid="1649302149750" $\begin{array}{r}\frac{a_{k+1}}{a_k}=\frac{\left(\right(k+1)!{)}^{1/k+1}}{(k!{)}^{1/k}}\\ >1(\text{For}k>0)\\ \text{Thus,}{a}_{k+1}>a_k\end{array}$

The sequence $\left\{a_k\right\}=\left\{(k!{)}^{1/k}\right\}$is strictly increasing. The given sequence is monotonic.

Step 3. The sequence {ak}=(k!)1/k is bounded below because

$0<a_k$for $k>0$

The sequence is an increasing sequence and doesn't have any upper bound.

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

Step 4. The monotonic increasing sequence is bounded above is convergent. 

The monotonic decreasing sequence $\left\{a_k\right\}=\left\{(k!{)}^{1/k}\right\}$is not bounded above and hence is not convergent. Therefore, the given sequence is divergent.