Q. 38 - 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.  

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

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 $\left\{\frac{k^2-2}{k^2+2k+2}\right\}$ 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 $a_k=\frac{k^2-2}{k^2+2k+2}$.

The term

$$\begin{gathered} a_{k+1}-a_k \\ =\frac{{\left(k+1\right)}^2-2}{{\left(k+1\right)}^2+2\left(k+1\right)+2}-\frac{k^2-2}{k^2+2k+2} \\ =\frac{k^2+2k+1-2}{k^2+2k+1+2k+2+2}-\frac{k^2-2}{k^2+2k+2} \\ =\frac{k^2+2k-1}{k^2+4k+5}-\frac{k^2-2}{k^2+2k+2} \\ =\frac{\left(k^2+2k-1\right)\left(k^2+2k+2\right)-\left(k^2-2\right)\left(k^2+4k+5\right)}{\left(k^2+4k+5\right)\left(k^2+2k+2\right)} \\ =\frac{10k+8}{\left(k^2+4k+5\right)\left(k^2+2k+2\right)} \\ >0(k>0) \\ \therefore a_{k+1}>a_k \end{gathered}$$

The sequence is strictly increasing so it is monotonic.  

Step 3. Finding bounded

The sequence $\left\{\frac{k^2-2}{k^2+2k+2}\right\}$ is bounded below because $0<a_k$. As $k>1, a_k<1$. 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 increasing sequence is bounded above and hence convergent. 

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