Q. 32 - 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. 

$\left\{\frac{2^k-2^{-k}}{2^k}\right\}$

Step-by-Step Solution

Answer

The given sequence is monotonic, bounded and convergent. 

The limit of the sequence is $0$.

Step 1. Given Information

We are given the sequence $\left\{\frac{2^k-2^{-k}}{2^k}\right\}$ and we need to find if the sequence is monotonic, bounded and the limit if it is convergent. 

Step 2. Finding monotonic

The general term is 
$$\begin{gathered} a_k=\left\{\frac{2^k-2^{-k}}{2^k}\right\} \\ \Rightarrow a_k=\left\{\frac{2^{2k}-1}{2^{2k}}\right\} \end{gathered}$$.

The term 

$$\begin{gathered} a_{k+1}-a_k \\ =\frac{2^{2\left(k+1\right)}-1}{2^{2k}.2^2}-\frac{2^{2k}-1}{2^{2k}} \\ =\frac{2^{2\left(k+1\right)}-1}{4.2^{2k}}-\frac{2^{2k}-1}{2^{2k}} \\ =\frac{2^{2(k+1)}-1-4.2^{2k}+4}{4.2^{2k}} \\ =\frac{2^{2(k+1)}-2^{2(k+1)}+3}{2^{2\left(k+1\right)}} \\ =\frac{3}{2^{2\left(k+1\right)}} \\ <0(k>0) \\ \therefore a_{k+1}<a_k \end{gathered}$$

The sequence is strictly decreasing so it is monotonic.

Step 3. Finding bounded

The sequence $\left\{\frac{2^k-2^{-k}}{2^k}\right\}$ is bounded below because $0<a_k$. As $k>0, a_k<1$. The decreasing sequence has a lower bound and is $0$.

Therefore, the given sequence is bounded. 

Step 4. Finding the limit

The monotonic decreasing sequence is bounded below and hence convergent.

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