Q. 43 - Calculus

Question

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. 

     $\left\{{\left(\frac{2k^2}{k^2+2k+2}\right)}^3\right\}$

Step-by-Step Solution

Answer

Ans: The sequence is convergent and, the limit of the sequence $\left\{a_k\right\}=\left\{{\left(\frac{2k^2}{k^2+2k+2}\right)}^3\right\}$ is $8$

Step 1. Given information.

given,

    $\left\{{\left(\frac{2k^2}{k^2+2k+2}\right)}^3\right\}$

Step 2. The objective is to determine whether the sequence is monotonic, bounded above, or bounded below, and to find the limit of the sequence if the sequence is convergent.

   In the sequence $\left\{a_k\right\}=\left\{{\left(\frac{2k^2}{k^2+2k+2}\right)}^3\right\}$ in terms of $a_k={\left(\frac{2k^2}{k^2+2k+2}\right)}^3$

Step 3. The general term of the sequence is a k = 2 k 2 k 2 + 2 k + 2 3

The ratio  $\frac{a_{k+1}}{a_k}$ gives

    $\begin{array}{r}\frac{a_{k+1}}{a_k}=\frac{{\left(\frac{2(k+1{)}^2}{(k+1{)}^2+2(k+1)+2}\right)}^3}{{\left(\frac{2k^2}{k^2+2k+2}\right)}^3} \text{ (Substitution) }\\ ={\left(\frac{k^2+2k+1}{k^2+4k+5}\cdot \frac{k^2+2k+2}{k^2}\right)}^3 \\ >1(\text{ For }k>0) \end{array}$

Thus $a_{k+1}>a_k$

The sequence $a_k={\left(\frac{2k^2}{k^2+2k+2}\right)}^3$ is strictly increasing. Therefore, the given sequence is monotonic.

Step 4. Now,

The sequence $\left\{a_k\right\}=\left\{{\left(\frac{2k^2}{k^2+2k+2}\right)}^3\right\}$ is bounded below because

$0<a_k for k>0$

As the index k increases, the term $a_k={\left(\frac{2k^2}{k^2+2k+2}\right)}^3$ approaches $8$.

Thus, the decreasing sequence has an upper bound and is $8$.

Thus,

    $0<a_k<8$

The given sequence has lower and upper bounds, therefore, the sequence is bounded. 

Step 5. The monotonic increasing sequence is bounded above is convergent.

The monotonic decreasing sequence $\left\{a_k\right\}=\left\{{\left(\frac{2k^2}{k^2+2k+2}\right)}^3\right\}$ is bounded above and hence is convergent. Therefore, the sequence is convergent.

Step 6. Now,

The limit of the sequence is  $\left\{a_k\right\}=\left\{{\left(\frac{2k^2}{k^2+2k+2}\right)}^3\right\}$

   $\begin{array}{r}\underset{k\to \mathrm{\infty }}{lim} a_k=\underset{k\to \mathrm{\infty }}{lim} {\left(\frac{2k^2}{k^2+2k+2}\right)}^3\\ =\underset{k\to \mathrm{\infty }}{lim} {\left(\frac{\frac{2k^2}{k^2}}{\frac{k^2}{k^2}+\frac{2k}{k^2}+\frac{2}{k^2}}\right)}^3\\ =2^3(\text{ Simplify) }\\ =8\end{array}$

Thus the limit of the sequence $\left\{a_k\right\}=\left\{{\left(\frac{2k^2}{k^2+2k+2}\right)}^3\right\}$ is $8$.