The Formula for the Sum of a Geometric Series

Oct 13, 2025 by talynt.org 114 views

Introduction

In mathematics, a series is a sequence of numbers that are added together. The series we are considering in this article is $\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\ldots$ . This series is a classic example of a geometric series, where each term is obtained by multiplying the previous term by a fixed constant. In this case, the common ratio is $\frac{1}{2}$ . The sum of the first $n$ terms of this series is denoted by $S_n$ . In this article, we will explore the expression that defines $S_n$ .

What is a Geometric Series?

A geometric series is a series of the form $a + ar + ar^2 + ar^3 + \ldots$ , where $a$ is the first term and $r$ is the common ratio. In our series, $a = \frac{1}{8}$ and $r = \frac{1}{2}$ . The series can be written as:

\frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \frac{1}{128} + \ldots

The Formula for the Sum of a Geometric Series

The sum of the first $n$ terms of a geometric series is given by the formula:

S_n = \frac{a(1 - r^n)}{1 - r}

where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.

Applying the Formula to Our Series

In our series, $a = \frac{1}{8}$ and $r = \frac{1}{2}$ . Plugging these values into the formula, we get:

S_n = \frac{\frac{1}{8}\left(1 - \left(\frac{1}{2}\right)^n\right)}{1 - \frac{1}{2}}

Simplifying the expression, we get:

S_n = \frac{\frac{1}{8}\left(1 - \left(\frac{1}{2}\right)^n\right)}{\frac{1}{2}}

S_n = \frac{1}{4}\left(1 - \left(\frac{1}{2}\right)^n\right)

The Limit of the Series

As $n$ approaches infinity, the term $\left(\frac{1}{2}\right)^n$ approaches zero. Therefore, the limit of the series is:

\lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{1}{4}\left(1 - \left(\frac{1}{2}\right)^n\right)

\lim_{n \to \infty} S_n = \frac{1}{4}\left(1 - 0\right)

\lim_{n \to \infty} S_n = \frac{1}{4}

Conclusion

In conclusion, the expression that defines $S_n$ is $\frac{1}{4}\left(1 - \left(\frac{1}{2}\right)^n\right)$ . As $n$ approaches infinity, the limit of the series is $\frac{1}{4}$ .

Answer

The correct answer is:

A. $\lim _{n \rightarrow \infty} \frac{1}{2^n}$

However, this is not the correct expression that defines $S_n$ . The correct expression is $\frac{1}{4}\left(1 - \left(\frac{1}{2}\right)^n\right)$ .

Discussion

The series we are considering in this article is a classic example of a geometric series. The sum of the first $n$ terms of a geometric series is given by the formula $S_n = \frac{a(1 - r^n)}{1 - r}$ . In our series, $a = \frac{1}{8}$ and $r = \frac{1}{2}$ . Plugging these values into the formula, we get $S_n = \frac{1}{4}\left(1 - \left(\frac{1}{2}\right)^n\right)$ . As $n$ approaches infinity, the term $\left(\frac{1}{2}\right)^n$ approaches zero, and the limit of the series is $\frac{1}{4}$ .

References

[1] "Geometric Series" by Math Open Reference
[2] "Sum of a Geometric Series" by Wolfram MathWorld

Note

$The expression that defines $S_n$ is $\frac{1}{4}\left(1 - \left(\frac{1}{2}\right)^n\right)$. This is not the same as the expression $\lim _{n \rightarrow \infty} \frac{1}{2^n}$, which is the correct answer.<br/> **Q&A: Understanding the Series and Its Limit**$

Q: What is a geometric series?

A: A geometric series is a series of the form $a + ar + ar^2 + ar^3 + \ldots$ , where $a$ is the first term and $r$ is the common ratio. In our series, $a = \frac{1}{8}$ and $r = \frac{1}{2}$ .

Q: What is the formula for the sum of a geometric series?

A: The formula for the sum of a geometric series is $S_n = \frac{a(1 - r^n)}{1 - r}$ , where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.

Q: How do I apply the formula to our series?

A: To apply the formula to our series, we need to plug in the values of $a$ and $r$ . In our series, $a = \frac{1}{8}$ and $r = \frac{1}{2}$ . Plugging these values into the formula, we get:

S_n = \frac{\frac{1}{8}\left(1 - \left(\frac{1}{2}\right)^n\right)}{1 - \frac{1}{2}}

Simplifying the expression, we get:

S_n = \frac{1}{4}\left(1 - \left(\frac{1}{2}\right)^n\right)

Q: What is the limit of the series as $n$ approaches infinity?

A: As $n$ approaches infinity, the term $\left(\frac{1}{2}\right)^n$ approaches zero. Therefore, the limit of the series is:

\lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{1}{4}\left(1 - \left(\frac{1}{2}\right)^n\right)

\lim_{n \to \infty} S_n = \frac{1}{4}\left(1 - 0\right)

\lim_{n \to \infty} S_n = \frac{1}{4}

Q: Why is the limit of the series $\frac{1}{4}$ ?

A: The limit of the series is $\frac{1}{4}$ because as $n$ approaches infinity, the term $\left(\frac{1}{2}\right)^n$ approaches zero. This is because the common ratio $r = \frac{1}{2}$ is less than 1, and as we multiply a number less than 1 by itself repeatedly, the result approaches zero.

Q: What is the relationship between the series and the expression $\lim _{n \rightarrow \infty} \frac{1}{2^n}$ ?

A: The expression $\lim _{n \rightarrow \infty} \frac{1}{2^n}$ is not the same as the expression that defines $S_n$ . However, it is related to the series in that it is the limit of the series as $n$ approaches infinity.

Q: Why is the expression $\lim _{n \rightarrow \infty} \frac{1}{2^n}$ not the correct answer?

A: The expression $\lim _{n \rightarrow \infty} \frac{1}{2^n}$ is not the correct answer because it does not accurately represent the sum of the series. The correct expression that defines $S_n$ is $\frac{1}{4}\left(1 - \left(\frac{1}{2}\right)^n\right)$ .

Q: What is the significance of the series and its limit?

A: The series and its limit are significant because they demonstrate the concept of a geometric series and its properties. The series is a classic example of a geometric series, and its limit is a fundamental concept in mathematics.

Q: How can I apply the concepts learned from this article to real-world problems?

A: The concepts learned from this article can be applied to real-world problems in various fields, such as finance, economics, and engineering. For example, the concept of a geometric series can be used to model population growth, compound interest, and other phenomena that involve exponential growth or decay.

References

[1] "Geometric Series" by Math Open Reference
[2] "Sum of a Geometric Series" by Wolfram MathWorld
[3] "Limit of a Geometric Series" by Khan Academy