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.

$\left\{\frac{\sin k}{k}\right\}$

Short answer

Ans: The sequence$\left\{\frac{\sin k}{k}\right\}$is convergent and converges to$0$.

Step-by-step solution

Step 1. Given information.

given,

$\left\{\frac{\sin k}{k}\right\}$

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 $\left\{a_k\right\}=\left\{\frac{\sin k}{k}\right\}$the general term is $a_k=\frac{\sin k}{k}$.

The sequence $\left\{a_k\right\}=\left\{\frac{\sin k}{k}\right\}$is not monotonic because the sign of $\left\{\frac{\sin k}{k}\right\}$varies as k increases.

Therefore, the given sequence is not a monotonic sequence,

Step 3.  Now,

The sequence $\left\{a_k\right\}=\left\{\frac{\sin k}{k}\right\}$is bounded because

$-1\le \frac{\sin k}{k}\le 1$

The sequence role="math" localid="1649276444082" $\left\{a_k\right\}=\left\{\frac{\sin k}{k}\right\}$is bounded.

Step 4. The sequence {ak}=sin⁡kk is bounded as

$\begin{array}{r}-1\le \sin k\le 1\\ -\frac{1}{k}\le \frac{\sin k}{k}\le \frac{1}{k}(\text{For}k>0)\end{array}$

The limit of the function$\underset{k\to \mathrm{\infty }}{lim} \left(\frac{\sin k}{k}\right)$ is obtained by the Squeeze Theorem.

$$\begin{gathered} \underset{k\to \mathrm{\infty }}{lim} \left(\pm \frac{1}{k}\right)=0 \\ \underset{k\to \mathrm{\infty }}{lim} \left(\frac{\sin k}{k}\right)=0 \end{gathered}$$

Thus, the sequence$\left\{a_k\right\}=\left\{\frac{\sin k}{k}\right\}$ is convergent and converges to$0$.