Q. 36 - 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.   

$\left\{{\left(1+\frac{1}{k}\right)}^k\right\}$

Step-by-Step Solution

Answer

The given sequence is monotonic, bounded and convergent. 

The limit of the sequence is $e$.

Step 1. Given Information

We are given the sequence $\left\{{\left(1+\frac{1}{k}\right)}^k\right\}$ 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 $a_k={\left(1+\frac{1}{k}\right)}^k$.

The ratio

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

The sequence is strictly increasing so it is monotonic. 

Step 3. Finding bounded

The sequence $\left\{{\left(1+\frac{1}{k}\right)}^k\right\}$ is bounded below because $2<a_k$. As $k>1, a_k\le 3$. The decreasing sequence has a lower bound and is $2$ and an upper bound $3$.

Therefore, the given sequence is bounded. 

Step 4. Finding the limit

The monotonic decreasing sequence is bounded below and hence convergent. 

$$\begin{gathered} \underset{k\to \infty }{\lim }{a}_k \\ =\underset{k\to \infty }{\lim }{\left(1+\frac{1}{k}\right)}^k \\ =e \end{gathered}$$