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.

$\left\{\frac{k^2-1}{10000k+2}\right\}$

Short answer

Ans:$\left\{\frac{k^2-1}{10000k+2}\right\}$

Step-by-step solution

Step 1. Given information.

given,

$\left\{\frac{k^2-1}{10000k+2}\right\}$

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 $\left\{a_k\right\}=\left\{\frac{k^2-1}{10000k+2}\right\}$the gentral term is $a_k=\frac{k^2-1}{10000k+2}$.

Step 3.  The general term of the sequence is ak=k2−110000k+2.

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

$\begin{array}{r}{a}_{k+1}-a_k=\frac{(k+1{)}^2-1}{10000(k+1)+2}-\frac{k^2-1}{10000k+2}\text{(Substitution)}\\ =\frac{k^2+2k+1-1}{10000k+10000+2}-\frac{k^2-1}{10000k+2}\\ =\frac{k^2+2k}{10000k+10002}-\frac{k^2-1}{10000k+2}(\text{Simplify)}\\ =\frac{\left(k^2+2k\right)(10000k+2)-\left(k^2-1\right)(10000k+10002)}{(10000k+10002)(10000k+2)}\\ =\frac{10000k^2+10004k+10002}{(10000k+10002)(10000k+2)}\\ >0\text{(For}k>0)\\ \text{Thus,}{a}_{k+1}>a_k\end{array}$

The sequence $\left\{a_k\right\}=\left\{\frac{k^2-1}{10000k+2}\right\}$strictly increases. Therefore, the given sequence is monotonic.