Q. 37 - Calculus

Question

For each of the sequences in Exercises 23–52 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.  

$\left\{\frac{2^k}{k!}\right\}$

Step-by-Step Solution

Answer

The given sequence is monotonic, bounded and convergent. 

The limit of the sequence is $0$.

Step 1. Given Information

We are given the sequence $\left\{\frac{2^k}{k!}\right\}$ and we need to find if the sequence is monotonic, bounded and the limit if it is convergent.     

Step 2. Finding monotonic

The general term is $a_k=\frac{2^k}{k!}$.

The ratio

$$\begin{gathered} \frac{a_{k+1}}{a_k} \\ =\frac{\left({\displaystyle \frac{2^{k+1}}{\left(k+1\right)!}}\right)}{\left({\displaystyle \frac{2^k}{k!}}\right)} \\ =\left(\frac{2^{k+1}}{\left(k+1\right)!}\right).\left(\frac{k!}{2^k}\right) \\ =\left(\frac{2.2^k}{\left(k+1\right)k!}\right).\left(\frac{k!}{2^k}\right) \\ =\frac{2}{k+1} \\ \le 1(k>0) \\ \therefore a_{k+1}\le a_k \end{gathered}$$

The sequence is strictly decreasing so it is monotonic.  

Step 3. Finding bounded

The sequence $\left\{\frac{2^k}{k!}\right\}$ is bounded below because $0<a_k$. As $k>1, a_k\le 2$. The decreasing sequence has a lower bound and is $0$ and an upper bound $2$.

Therefore, the given sequence is bounded. 

Step 4. Finding the limit

The monotonic decreasing sequence is bounded below and hence convergent. 

$$\begin{gathered} \underset{k\to \infty }{\lim }{a}_k \\ =\underset{k\to \infty }{\lim }\frac{2^k}{k!} \\ =0 \end{gathered}$$