Q. 41 - Calculus

Question

Find the least upper bound of the sequences in Exercises 37–42 

$\left\{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^k}\right\}$

Step-by-Step Solution

Answer

The least upper bound is $1.$

Step 1. Given information.

The given sequence is $\left\{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^k}\right\}$.

Step 2. Least upper bound.

We know, the sum of the series is,

$$\begin{gathered} S_1=\frac{\frac{1}{2}\left(1-{\left(\frac{1}{2}\right)}^k\right)}{1-\frac{1}{2}} \\ =1-{\left(\frac{1}{2}\right)}^k \\ \underset{k\to \infty }{\lim }{S}_k=\underset{k\to \infty }{\lim }\left[1-{\left(\frac{1}{2}\right)}^k\right] \\ =1 \end{gathered}$$