Q. 54 - 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

  $$\begin{gathered} \frac{a_{k+1}}{a_k}=\frac{(k+1{)}^{1/(k+1)!}}{(k{)}^{1/k!}} \\ > 1 (for k>9) \\ Thus, a_{k+1}>a_k \end{gathered}$$

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

Step 3. Now,

The sequence $\left\{a_k\right\}=\left\{k^{1/k!}\right\}$ 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.