Question

A series is given. (a) Find a formula for \(S_{n},\) the \(n^{\text {th }}\) partial sum of the series. (b) Determine whether the series converges or diverges. If it converges, state what it converges to. $$\sum_{n=0}^{\infty} \frac{1}{4^{n}}$$

Short answer

The series converges to \( \frac{4}{3} \).

Step-by-step solution

Understand the Series Type

The series given is \( \sum_{n=0}^{\infty} \frac{1}{4^{n}} \). Recognize it as a geometric series where the common ratio \( r = \frac{1}{4} \).

Derive the Formula for the Partial Sum

For a geometric series, the \( n^{\text{th}} \) partial sum \( S_n \) is given by \( S_n = a \frac{1-r^{n+1}}{1-r} \), where \( a \) is the first term of the series. In this case, \( a = 1 \) and \( r = \frac{1}{4} \). Thus, \( S_n = \frac{1-\left(\frac{1}{4}\right)^{n+1}}{1-\frac{1}{4}} = \frac{1-\left(\frac{1}{4}\right)^{n+1}}{\frac{3}{4}} \). Further simplification gives \( S_n = \frac{4}{3}\left(1-\left(\frac{1}{4}\right)^{n+1}\right) \).

Determine Convergence or Divergence

Since \( |r| = \left|\frac{1}{4}\right| < 1 \), the geometric series converges. The sum of the infinite series is given by \( S = \frac{a}{1-r} = \frac{1}{1-\frac{1}{4}} = \frac{4}{3} \).

State the Conclusion

The series \( \sum_{n=0}^{\infty} \frac{1}{4^{n}} \) is a convergent geometric series and it converges to \( \frac{4}{3} \).

Key concepts

Partial Sum

When dealing with a geometric series like \( \sum_{n=0}^{\infty} \frac{1}{4^{n}} \), one important concept to understand is the partial sum. A partial sum, denoted as \( S_n \), represents the sum of the first \( n+1 \) terms of a series. It's essentially a way to measure how a series builds up over its first few terms.

For any geometric series, the formula to calculate the \( n^{\text{th}} \) partial sum is:

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

where \( a \) is the first term, and \( r \) is the common ratio.

In our case, \( a = 1 \) and \( r = \frac{1}{4} \), so the partial sum can be found using the formula:

\( S_n = \frac{4}{3}\left(1-\left(\frac{1}{4}\right)^{n+1}\right) \)

Using this, we can compute the sum of the first \( n+1 \) terms easily. Each calculation gives us insight into how the series grows and behaves as more terms are included.

Convergence and Divergence

The concepts of convergence and divergence are crucial when evaluating infinite series. A series is said to converge if the sum of its infinite terms approaches a specific value, and it diverges if the sum continues to grow without approaching a fixed limit.

For geometric series, determining convergence is straightforward:

If the absolute value of the common ratio \( |r| \) is less than 1, the series converges.

If \( |r| \) is equal to or greater than 1, the series diverges.

In our example, with \( r = \frac{1}{4} \), the series converges because \(|r| = \left|\frac{1}{4}\right| = 0.25 < 1\). This means that as we add more terms, the overall sum approaches a limit. By recognizing whether a series converges or diverges, we can predict its long-term behavior and determine if it is meaningful for practical applications.

Infinite Series

An infinite series is the summation of an infinite sequence of terms. This concept is essential in many areas of mathematics because it allows us to represent complex functions and solve problems in calculus and analysis.

The notation \( \sum_{n=0}^{\infty} \frac{1}{4^{n}} \) expresses the sum of all terms from \( n = 0 \) to infinity. For geometric series like this, if the series converges, we can find its sum by using the formula:

\( S = \frac{a}{1-r} \)

where \( a \) is the first term, and \( r \) is the common ratio.

In this series, because it converges, its sum becomes \( \frac{4}{3} \). This is an example of how an infinite series, if convergent, results in a finite sum even though it contains an infinite number of terms. Understanding how infinite series work helps us model and solve complex real-world phenomena by providing a finite value from an infinite process.