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.

$\left\{\tan \left(\frac{k\pi }{2}\right)\right\}$

Short answer

Ans: The sequence$\left\{a_k\right\}=\left\{\tan \left(\frac{k\pi }{2}\right)\right\}$is not convergent.

Step-by-step solution

Step 1. Given information.

given,

$\left\{\tan \left(\frac{k\pi }{2}\right)\right\}$

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 $\left\{a_k\right\}=\left\{\tan \left(\frac{k\pi }{2}\right)\right\}$the general term is $a_k=\tan \left(\frac{k\pi }{2}\right)$.

Step 3. The general term of the sequence is ak=tan⁡kπ2.

The sequence$a_k=\tan \left(\frac{k\pi }{2}\right)$is not monotonic because the sign of $\tan \left(\frac{k\mathrm{\pi }}{2}\right)$varies as k increases. Therefore, the given sequence is not monotonic.

Step 4. Now,

The sequence $\left\{a_k\right\}=\left\{\tan \left(\frac{k\pi }{2}\right)\right\}$is bounded below because

$-\mathrm{\infty }<a_k<\mathrm{\infty }\text{for}k>0$

The given sequence has no lower and upper bounds, therefore, the sequence is not bounded.

Step 5. Thus,

The sequence $\left\{a_k\right\}=\left\{\tan \left(\frac{k\pi }{2}\right)\right\}$is not monotonic and not bounded. Therefore, the sequence $\left\{a_k\right\}=\left\{\tan \left(\frac{k\pi }{2}\right)\right\}$is not convergent.