Q. 42 - 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. 

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

Step-by-Step Solution

Answer

Ans: The sequence is convergent and, the limit of sequence$\left\{\frac{k}{k^2+2k+2}\right\}$ is $0$.

Step 1. Given information.

given,

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

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 $\left\{a_k\right\}=\left\{\frac{k}{k^2+2k+2}\right\}$ in terms of $a_k=\frac{k}{k^2+2k+2}$

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

The terms $a_{k+1}-a_k$ gives

   $\begin{array}{r}{a}_{k+1}-a_k=\frac{k+1}{(k+1{)}^2+2(k+1)+2}-\frac{k}{k^2+2k+2} \text{ (Substitution) }\\ =\frac{k+1}{k^2+2k+1+2k+2+2}-\frac{k}{k^2+2k+2} \\ =\frac{k+1}{k^2+4k+5}-\frac{k}{k^2+2k+2} (\text{ Simplify) }\\ =\frac{(k+1)\left(k^2+2k+2\right)-k\left(k^2+4k+5\right) }{\left(k^2+4k+5\right)\left(k^2+2k+2\right)}\\ =\frac{-k^2+9k+2}{\left(k^2+4k+5\right)\left(k^2+2k+2\right)} \\ <0\text{ (For }k>9) \end{array}$

Thus, $a_{k+1}<a_k$ 

The sequence $\left\{a_k\right\}=\left\{\frac{k}{k^2+2k+2}\right\}$ is eventually decreasing. The given sequence is monotonic.

Step 4. Now,

The sequence $\left\{a_k\right\}=\left\{\frac{k}{k^2+2k+2}\right\}$ is bounded below because $0<a_k$ for $k>1$

As the index k increases, the term $\left\{a_k\right\}=\left\{\frac{k}{k^2+2k+2}\right\}$ approaches $0$.

Thus, the decreasing sequence has a lower bound and is $0$

Thus,

   $0<a_k\le 0.2$

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

 

Step 5. The monotonic decreasing sequence with a lower bound is convergent.

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

Step 6. Now,

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

   $\begin{array}{r}\underset{k\to \mathrm{\infty }}{lim} a_k=\underset{k\to \mathrm{\infty }}{lim} \left(\frac{k}{k^2+2k+2}\right)\\ =\underset{k\to \mathrm{\infty }}{lim} \left(\frac{\frac{k}{k^2}}{1+\frac{2k}{k^2}+\frac{2}{k^2}}\right)\\ =0 \text{ (Simplify) }\end{array}$

Thus the limit of the sequence $\left\{a_k\right\}=\left\{\frac{k}{k^2+2k+2}\right\}$ is $0$.