Q. 53 - Calculus

Question

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

 

  $\left\{(k!{)}^{1/k}\right\}$

Step-by-Step Solution

Answer

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

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

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

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

Step 3. The sequence { a k } = ( 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.