Question

Limits of sequences: Determine whether the sequences that follow are bounded, monotonic and/or eventually monotonic.

Determine whether each sequence converges or diverges. If the sequence converges, find its limit.

$\left\{\sin \frac{\mathrm{\pi }}{2k}\right\}$.

Short answer

The sequence is eventually monotonically decreasing sequence bounded between $-1$and 1.

The limit of the sequence is 0.

Step-by-step solution

Step 1. Given Information

The given sequence is$\left\{\sin \frac{\mathrm{\pi }}{2k}\right\}$.

Step 2. Check Boundedness and Monotonicity

• The sine function is bounded between $-1$and 1.
• So, the sequence is a bounded sequence.
• As $k\to \infty$, the expression $\frac{\mathrm{\pi }}{2k}\to 0$.
• So, the sequence is eventually monotonically decreasing sequence.

Step 3. Find the Limits

• The limit of the sequence is found as follows:

$\begin{array}{rcl}\underset{k\to \infty }{\lim }\sin \frac{\mathrm{\pi }}{2k}& =& \sin \left(0\right)\\ & =& 0\end{array}$