Question

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

Answer: The value of the infinite series is 1.

Step-by-step solution

Find the first four partial sums of the series

To find the first four partial sums, substitute k=1,2,3,4 into the series one at a time. - For k=1: \(S_1 = \frac{1}{2^1} = \frac{1}{2}\) - For k=2: \(S_2 = S_1 + \frac{1}{2^2} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}\) - For k=3: \(S_3 = S_2 + \frac{1}{2^3} = \frac{3}{4} + \frac{1}{8} = \frac{7}{8}\) - For k=4: \(S_4 = S_3 + \frac{1}{2^4} = \frac{7}{8} + \frac{1}{16} = \frac{15}{16}\) The first four partial sums are \(\frac{1}{2}, \frac{3}{4}, \frac{7}{8},\) and \(\frac{15}{16}\).

Find a formula for \(S_n\)

Looking at the partial sums found in step 1, we notice a pattern: - \(S_1 = \frac{1}{2} = \frac{1}{2^1}\) - \(S_2 = \frac{3}{4} = \frac{2^2 - 1}{2^2}\) - \(S_3 = \frac{7}{8} = \frac{2^3 - 1}{2^3}\) - \(S_4 = \frac{15}{16} = \frac{2^4 - 1}{2^4}\) Following this pattern, we can express the general formula for the nth partial sum as: $$S_n = \frac{2^n - 1}{2^n}$$

Find the value of the series

To find the value of the series, we need to find the limit of \(S_n\) as n approaches infinity. We can do this by finding the limit of the formula: $$\lim_{n\to\infty} \frac{2^n - 1}{2^n}$$ We can simplify this expression by dividing both the numerator and the denominator by \(2^n\): $$\lim_{n\to\infty} \frac{1 - \frac{1}{2^n}}{1}$$ As n approaches infinity, the term \(\frac{1}{2^n}\) approaches 0: $$\lim_{n\to\infty} \left(1 - \frac{1}{2^n}\right) = 1$$ So, the value of the series is 1.

Key concepts

Partial Sums

Partial sums play a crucial role in the study of infinite series. To understand them fully, imagine attempting to "sum up" infinitely many numbers. This task seems daunting!
However, by calculating partial sums, we break down the problem into manageable pieces. A partial sum is the sum of the first few terms of a series.
With each new term added, a fresh partial sum is created, building towards the total, eventually helping us understand the series' behavior. For example, consider the series \(\sum_{k=1}^{\infty} \frac{1}{2^k}\). If we only take the first term \(\frac{1}{2}\), it's \(S_1\). Add the next term \(\frac{1}{4}\) to get \(S_2 = \frac{3}{4}\), then adding \(\frac{1}{8}\) gives \(S_3 = \frac{7}{8}\), and so on.
The resulting sequence \(\frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16}\) shows how each partial sum approaches something interesting, hinting at the series' final value.

Sequence

A sequence in mathematics refers to a list of numbers arranged in a definite order. Each number in the sequence is called a term, and sequences can be finite or infinite.
For infinite series like ours, each term \(a_k = \frac{1}{2^k}\), forms a part of the sequence. Sequences help set up the problem, offering an orderly roadmap for calculation.
As you write out more terms, you'll identify patterns and relationships essential for analyzing an infinite series. The progression \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots\) makes up a sequence that reveals more with each additional term.

• Term by term, you can see how each affects the whole.
• Studying sequences is crucial for building foundational understanding. This lays the groundwork for finding partial sums and exploring the series in full.

Recognizing and analyzing sequences allows students to look beyond individual numbers and see the patterns forming the structure of the infinite series.

Formula for Sn

Finding a formula for \(S_n\) creates a bridge between individual partial sums and the entire series. Observing your earlier partial sums \(\frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16} \), you might notice a pattern: \(\frac{2^n - 1}{2^n}\).
This formula allows us to compute any partial sum without having to add each term one by one.
The expression \(S_n = \frac{2^n - 1}{2^n}\) summarizes this; it's a shortcut!

• \(n\) represents how many terms we sum up.
• The numerator \(2^n - 1\) arises from adding powers of two.
• The denominator \(2^n\) sets a standard for normalization, bringing transparency to each cumulative sum.

This formula brings efficiency, letting students forecast the behavior of partial sums even as \(n\) grows large. It smoothly transitions us to the concept of limits.

Limit of a Series

The limit of a series is the ultimate goal when computing infinite series. This concept helps determine what happens as we add infinitely many terms.
Using the formula for \(S_n\), as \(n\) grows, the fraction \(\frac{1}{2^n}\) in \(1 - \frac{1}{2^n}\) diminishes because \(\frac{1}{2^n}\) approaches 0.
Thus, the limit becomes 1, indicated by \(\lim_{n \to \infty} S_n = 1\).

• The closer the terms \(n\) include, the series approaches a set value.
• It shows learners what the series truly converges to over the infinite horizon.

Understanding limits clarifies an infinite series' behavior, proving essential for gathering insights into series convergence or divergence. This interpretation unlocks answers to broader questions, showcasing mathematics' power to transform abstract ideas into concrete understanding.