Q. 29 - Calculus

Question

For each of the sequences in Exercises 23–52 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.

29. $\left\{\left(3^{-k}-2\right)\right\}$

Step-by-Step Solution

Answer

The sequence is monotonic and bounded and convergent.

The limit of the sequence is $-2$.

Step 1. Given information

We have been given the sequence $\left\{\left(3^{-k}-2\right)\right\}$.

Step 2. Determine whether the sequence is monotonic or eventually monotonic and whether the sequence is bounded above and/or below.

$\left\{a_k\right\}=\left\{\left(3^{-k}-2\right)\right\}$

$$\begin{gathered} a_{k+1}-a_k \\ =\left(3^{-\left(k+1\right)}-2\right)-\left(3^{-k}-2\right) \\ =3^{-k-1}-3^{-k} \\ =3^{-k}{3}^{-1}-3^{-k} \\ =3^{-k}\left(\frac{1}{3}-1\right) \\ =-\frac{1}{3^k}\left(\frac{2}{3}\right) \\ =-\frac{2}{3^{k+1}} \\ <0 \end{gathered}$$

Thus, $a_{k+1}<a_k$.

The sequence is strictly decreasing. The given sequence is monotonic.

The sequence is bounded above because $a_k<0$ for $k>0$

$-2<3^{-k}-2<0$

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

The sequence is convergent.

Step 3. Determine the limit of the sequence.

$$\begin{gathered} \underset{k\to \infty }{\lim }{a}_k \\ =\underset{k\to \infty }{\lim }\left(3^{-k}-2\right) \\ =\underset{k\to \infty }{\lim }\left(\frac{1}{3^k}-2\right) \\ =0-2 \\ =-2 \end{gathered}$$

Thus, the limit of the sequence $\left\{a_k\right\}=\left\{\left(3^{-k}-2\right)\right\}$ is $-2$.