Q. 31 - 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. 

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

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 $\left\{\frac{k-2^{-k}}{2^k}\right\}$ 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 $a_k=\frac{k-2^{-k}}{2^k}$.

The term 

$$\begin{gathered} a_{k+1}-a_k \\ =\frac{k+1-2^{-(k+1)}}{2^{(k+1)}}-\frac{k-2^{-k}}{2^k} \\ =\frac{k+1-2^{-(k+1)}}{2.2^k}-\frac{k-2^{-k}}{2^k} \\ =\frac{k+1-2^{-(k+1)}-2k+2.2^{-k}}{2.2^k} \\ =\frac{1-k-2^{-(k+1)}+2^{-k+1}}{2^{k+1}} \\ =\frac{1-k-2^{-k}\left({\displaystyle \frac{1}{2}}-2\right)}{2^{k+1}} \\ =\frac{1-k+2^{-k}.{\displaystyle \frac{3}{2}}}{2^{k+1}} \\ =\frac{1-k+3.2^{-k-1}}{2^{k+1}} \\ <0(k>1) \\ \therefore a_{k+1}<a_k \end{gathered}$$

The sequence is strictly decreasing so it is monotonic.

Step 3. Finding bounded

The sequence $\left\{\frac{k-2^{-k}}{2^k}\right\}$ 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.

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