Q. 33 - 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.  

$\left\{\frac{k}{2+k}\right\}$

Step-by-Step Solution

Answer

The given sequence is monotonic, bounded and convergent. 

The limit of the sequence is $1$.

Step 1. Given Information

We are given the sequence $\left\{\frac{k}{2+k}\right\}$ 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 $a_k=\frac{k}{2+k}$.

The term  
$$\begin{gathered} a_{k+1}-a_k \\ =\frac{\left(k+1\right)}{2+\left(k+1\right)}-\frac{k}{2+k} \\ =\frac{\left(k+1\right)}{3+k}-\frac{k}{2+k} \\ =\frac{\left(k+1\right)\left(k+2\right)-k\left(k+3\right)}{\left(3+k\right)\left(2+k\right)} \\ =\frac{k^2+k+2k+2-k^2-3k}{\left(3+k\right)\left(2+k\right)} \\ =\frac{2}{\left(3+k\right)\left(2+k\right)} \\ >0(k>0) \\ \therefore a_{k+1}>a_k \end{gathered}$$T

The sequence is strictly increasing so it is monotonic. 

Step 3. Finding bounded

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

Therefore, the given sequence is bounded. 

Step 4. Finding the limit

The monotonic decreasing sequence is bounded above and hence convergent. 

$$\begin{gathered} \underset{k\to \infty }{\lim }{a}_k \\ =\underset{k\to \infty }{\lim }\left(\frac{k}{2+k}\right) \\ =\underset{k\to \infty }{\lim }\left(\frac{1}{{\displaystyle \frac{2}{k}}+1}\right) \\ =1 \end{gathered}$$