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.

$\left\{\frac{k+1}{k+7}\right\}$

Short answer

Ans: The sequence$\left\{\frac{k+1}{k+7}\right\}$is convergent and converges to$1$.

Step-by-step solution

Step 1. Given information.

given,

$\left\{\frac{k+1}{k+7}\right\}$

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. 

The sequence $\left\{a_k\right\}=\left\{\frac{k+1}{k+7}\right\}$the general term is $a_k=\frac{k+1}{k+7}$.

The term $a_{k+1}-a_k$gives

$\begin{array}{r}{a}_{k+1}-a_k=\frac{k+2}{k+8}-\frac{k+1}{k+7}\text{(Substitution)}\\ =\frac{(k+2)(k+7)-(k+1)(k+8)}{(k+8)(k+7)}\\ =\frac{k^2+9k+14-\left(k^2+9k+8\right)}{(k+8)(k+7)}\text{(Simplify)}\\ =\frac{6}{(k+8)(k+7)}\\ >0(\text{For}k>0)\end{array}$

Thus $a_{k+1}>a_k$

The sequence $\left\{a_k\right\}=\left\{\frac{k+1}{k+7}\right\}$is strictly increasing. The given sequence is monotonic.

Step 3. Now,

The sequence $\left\{a_k\right\}=\left\{\frac{k+1}{k+7}\right\}$is a bounded sequence because

$0<a_k<1$for $k>0$

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

Step 4. The monotonic increasing sequence is bounded above is convergent. 

The strictly increasing sequence $\left\{a_k\right\}=\left\{\frac{k+1}{k+7}\right\}$is bounded upper and hence is convergent. Therefore, the sequence is convergent.

Step 5. Find limit,

The limit of the sequence $\left\{a_k\right\}=\left\{\frac{k+1}{k+7}\right\}$

$\begin{array}{r}\underset{k\to \mathrm{\infty }}{lim} a_k=\underset{k\to \mathrm{\infty }}{lim} \left(\frac{k+1}{k+7}\right)\\ =1\text{(Simplify)}\end{array}$

Therefore the sequence $\left\{a_k\right\}=\left\{\frac{k+1}{k+7}\right\}$converges to $1$.