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Chapter 9: Problem 64

Consider the following infinite series. a. Find the first four terms of the sequence of partial sums. b. Use the results of part (a) to find a formula for \(S_{n}\) c. Find the value of the series. $$\sum_{k=1}^{\infty} \frac{1}{2^{k}}$$

Short Answer

Expert verified
Based on the given infinite series, the first four terms of the sequence of partial sums are: Partial Sum 1: \(S_1 = \frac{1}{2}\) Partial Sum 2: \(S_2 = \frac{3}{4}\) Partial Sum 3: \(S_3 = \frac{7}{8}\) Partial Sum 4: \(S_4 = \frac{15}{16}\) The formula for the partial sum \(S_n\) is given by: \(S_n = \frac{2^n - 1}{2^n}\) The value of the series is 1 as n approaches infinity.

Step by step solution

01

Find the first four terms of the series

First, we need to find the first four terms of the given series. Let's plug in k=1, k=2, k=3, and k=4 into the given formula. Term 1: $$\frac{1}{2^1}=\frac{1}{2}$$ Term 2: $$\frac{1}{2^2}=\frac{1}{4}$$ Term 3: $$\frac{1}{2^3}=\frac{1}{8}$$ Term 4: $$\frac{1}{2^4}=\frac{1}{16}$$
02

Find the first four terms of the sequence of partial sums

Now, we sum the terms we just found to obtain the sequence of partial sums: Partial Sum 1: $$S_1 = \frac{1}{2}$$ Partial Sum 2: $$S_2 = \frac{1}{2}+\frac{1}{4} = \frac{3}{4}$$ Partial Sum 3: $$S_3 = \frac{1}{2}+\frac{1}{4}+\frac{1}{8} = \frac{7}{8}$$ Partial Sum 4: $$S_4 = \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16} = \frac{15}{16}$$
03

Find a formula for \(S_n\)

To find a formula for the partial sum \(S_n\), let's analyze the sequence of partial sums we've found. Notice the pattern that forms in the denominators and numerators: $$S_1 = \frac{1}{2} = \frac{2-1}{2}$$ $$S_2 = \frac{3}{4} = \frac{4-1}{4}$$ $$S_3 = \frac{7}{8} = \frac{8-1}{8}$$ $$S_4 = \frac{15}{16} = \frac{16-1}{16}$$ We can see a pattern in the partial sums, where the denominator is equal to \(2^n\) and the numerator is equal to \(2^n - 1\), so let's generalize the formula for \(S_n\): $$S_n = \frac{2^n - 1}{2^n}$$
04

Find the value of the series

To find the value of the series, we need to find the limit of the \(S_n\) formula as n approaches infinity: $$\lim_{n\to\infty} \frac{2^n - 1}{2^n}$$ To find the limit, we can divide both the numerator and denominator by \(2^n\): $$\lim_{n\to\infty} \frac{2^n - 1}{2^n} = \lim_{n\to\infty} \frac{2^n/2^n - 1/2^n}{2^n/2^n} = \lim_{n\to\infty} \frac{1 - \frac{1}{2^n}}{1}$$ The limit of \(\frac{1}{2^n}\) as n approaches infinity is 0, thus: $$\lim_{n\to\infty} \frac{1 - \frac{1}{2^n}}{1} = \frac{1}{1} = 1$$ Therefore, the value of the series is 1.

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