Problem 8 - Thomas Calculus

Question

In Exercises \(7-14,\) write out the first eight terms of each series to show how the series starts. Then find the sum of the series or show that it diverges. $$ \sum_{n=2}^{\infty} \frac{1}{4^{n}} $$

Step-by-Step Solution

Answer

The sum of the series is \( \frac{1}{12} \).

Step 1: Understanding the Series

The given series is \( \sum_{n=2}^{\infty} \frac{1}{4^{n}} \). This series is a geometric series where the first term starts at \( n = 2 \).

Step 2: Writing Out the First Eight Terms

To find the first eight terms, we substitute \( n = 2, 3, \ldots, 9 \) into the formula \( \frac{1}{4^n} \). This gives us: \( \frac{1}{16} \), \( \frac{1}{64} \), \( \frac{1}{256} \), \( \frac{1}{1024} \), \( \frac{1}{4096} \), \( \frac{1}{16384} \), \( \frac{1}{65536} \), and \( \frac{1}{262144} \).

Step 3: Identifying the Common Ratio

This is a geometric series with the first term \( a = \frac{1}{16} \) and a common ratio \( r = \frac{1}{4} \). Since \(|r| < 1\), the series converges.

Step 4: Finding the Sum of the Series

For a convergent geometric series, the sum is given by \( S = \frac{a}{1 - r} \). Here, \( a = \frac{1}{16} \) and \( r = \frac{1}{4} \), so \[ S = \frac{\frac{1}{16}}{1 - \frac{1}{4}} = \frac{\frac{1}{16}}{\frac{3}{4}} = \frac{1}{16} \cdot \frac{4}{3} = \frac{1}{12}. \]

Step 5: Conclusion of the Series Sum

The sum of the series \( \sum_{n=2}^{\infty} \frac{1}{4^n} \) is \( \frac{1}{12}. \) The infinite series converges to this sum.

Key Concepts

Series Convergence

When we talk about series convergence, we refer to the behavior of the sum of an infinite series as the number of terms goes to infinity. A series is said to converge if it approaches a specific value, known as the sum of the series. Conversely, a series diverges if it does not settle towards any value and either grows indefinitely or oscillates without approaching a limit. For example, in the geometric series given by \( \sum_{n=2}^{\infty} \frac{1}{4^n} \), we observed convergence because the sum converges to \( \frac{1}{12} \).
The convergence of a geometric series is determined by the common ratio \( r \). If the absolute value of \( r \) is less than 1, the series will converge. On the other hand, if \(|r|\) is equal to or greater than 1, the series will diverge.

Infinite Series

An infinite series is a sum of an unending sequence of terms, typically represented as \( a_1 + a_2 + a_3 + \dots \). Despite having innumerable terms, some infinite series can result in a finite sum. This happens especially when each subsequent term gets progressively smaller, reducing their impact on the total sum.
The series \( \sum_{n=2}^{\infty} \frac{1}{4^n} \) is an example, which despite extending to infinity, leads to a concise sum of \( \frac{1}{12} \). Infinite series are essential in calculus and mathematical analysis, helping us understand behaviors that oppose simple arithmetic. They appear in various fields from physics to financial calculations, illustrating how seemingly limitless processes can still converge to a finite result.

Common Ratio

In a geometric series, the common ratio \( r \) is a crucial factor. It defines the relationship between successive terms of the series. A geometric series takes the form \( a, ar, ar^2, ar^3, \ldots \), where \( a \) is the first term, and each term is obtained by multiplying the previous term by \( r \).
For the series \( \sum_{n=2}^{\infty} \frac{1}{4^n} \), the first term \( a \) is \( \frac{1}{16} \), and the common ratio \( r \) is \( \frac{1}{4} \). The series relies heavily on the common ratio to determine its characteristics of convergence or divergence. If the absolute value of \( r \) is less than 1, as with \( \frac{1}{4} \), it implies convergence, allowing the series to potentially add up to a finite sum. This makes calculating the series sum feasible and mathematically sound.