Problem 6 - Thomas Calculus

Question

In Exercises \(1-36\) , (a) find the series' radius and interval of convergence. For what values of \(x\) does the series converge (b) absolutely, (c) conditionally? $$ \sum_{n=0}^{\infty}(2 x)^{n} $$

Step-by-Step Solution

Answer

The series converges for \(-\frac{1}{2} < x < \frac{1}{2}\), absolutely in the same interval; no conditional convergence exists.

Step 1: Identify the Series Type

The given series is \(\sum_{n=0}^{\infty} (2x)^n\), which is a geometric series of the form \(a + ar + ar^2 + \cdots\) where \(a=1\) and the common ratio \(r=(2x)\).

Step 2: Find the Radius of Convergence

For a geometric series \(\sum_{n=0}^{\infty} ar^n\) to converge, the absolute value of the common ratio must satisfy \(|r|<1\). Here, \(|2x|<1\) which simplifies to \(|x| < \frac{1}{2}\). Thus, the radius of convergence is \(\frac{1}{2}\).

Step 3: Determine the Interval of Convergence

Using the radius of convergence \(|x| < \frac{1}{2}\), the interval of convergence is \(-\frac{1}{2} < x < \frac{1}{2}\). In a geometric series, the endpoints are not included since the series converges only for \(|r|<1\). Thus, the interval is \((-\frac{1}{2}, \frac{1}{2})\).

Step 4: Check for Absolute Convergence

The series \(\sum_{n=0}^{\infty} (2x)^n\) absolutely converges wherever it converges since it is a geometric series and no additional terms are introduced to alter its convergence. Therefore, the series converges absolutely for \(-\frac{1}{2} < x < \frac{1}{2}\).

Step 5: Determine Conditional Convergence

For a series to converge conditionally, it must converge but not absolutely. However, since the series \(\sum_{n=0}^{\infty} (2x)^n\) converges absolutely wherever it converges, there are no values of \(x\) for which it converges conditionally.

Key Concepts

Geometric Series

A geometric series is a series that has a constant ratio between successive terms. For any geometric series, you begin with an initial term 'a' and multiply it by a constant ratio 'r' repeatedly. The series looks something like this: \(a, ar, ar^2, ar^3, \ldots\). In mathematical terms, it can be written as \(\sum_{n=0}^{\infty} ar^n\). In our case, the series \(\sum_{n=0}^{\infty} (2x)^n\) fits this definition with an initial term \(a = 1\) and a common ratio \(r = 2x\). The beauty of geometric series lies in their simplicity, especially when it comes to problems of convergence.

Absolute Convergence

A series is said to converge absolutely if the series of absolute values also converges. In simpler terms, if you take the absolute value of each term of the series and the resulting series converges, then the original series is absolutely convergent.For the geometric series \(\sum_{n=0}^{\infty} (2x)^n\), absolute convergence occurs within its interval of convergence, which is \(-\frac{1}{2} < x < \frac{1}{2}\). This is because the common ratio within this interval remains less than 1 in absolute value, ensuring all terms shrink effectively to zero as the series progresses. This implies that for these values of \(x\), the series sums to a finite value.

Interval of Convergence

The interval of convergence is the range of \(x\) values where a series converges. For a geometric series, we solve for \(x\) such that the absolute value of the common ratio is less than 1: \(|2x| < 1\).Solving this inequality, we get \(|x| < \frac{1}{2}\). Thus, the interval of convergence is \(( -\frac{1}{2}, \frac{1}{2} )\).It's crucial to note that for geometric series, unlike some other series, the endpoints of this interval are not included. At the endpoints, the series fails to meet the criteria necessary for convergence.

Conditional Convergence

Conditional convergence happens when a series converges, but it does not converge absolutely. In other words, the series \(\sum a_n\) converges, but \(\sum |a_n|\) does not converge.With our geometric series \(\sum_{n=0}^{\infty} (2x)^n\), you find that it converges absolutely within the interval \(-\frac{1}{2} < x < \frac{1}{2}\). This means there's no room left for conditional convergence since wherever the series converges, it converges absolutely.Hence, there are no values of \(x\) for which this geometric series converges conditionally.