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

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

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

$0<a_{k}fork>0$

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.