Problem 13 - 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? $$ \sum_{n=1}^{\infty} \frac{4^{n} x^{2 n}}{n} $$

Step-by-Step Solution

Answer

The radius of convergence is \( R = \frac{1}{2} \). The series converges absolutely for \(-\frac{1}{2} < x < \frac{1}{2}\) and conditionally at \( x = -\frac{1}{2} \).

Step 1: Identify the Power Series

The given series is \( \sum_{n=1}^{\infty} \frac{4^{n} x^{2n}}{n} \). It is a power series in terms of \( x^2 \), with the general term given by \( \frac{4^n (x^2)^n}{n} \).

Step 2: Apply the Ratio Test for Convergence

To find the interval of convergence, apply the ratio test. Consider the sequence of terms \( a_n = \frac{4^n x^{2n}}{n} \). We investigate \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \):\[\frac{a_{n+1}}{a_n} = \frac{\frac{4^{n+1} x^{2(n+1)}}{n+1}}{\frac{4^n x^{2n}}{n}} = \frac{4 \cdot x^2 \cdot n}{n+1}.\]Taking the limit as \( n \to \infty \), we get:\[|4x^2| \cdot \lim_{n \to \infty} \frac{n}{n+1} = |4x^2|.\]The series converges when \(|4x^2| < 1\), or equivalently \(|x^2| < \frac{1}{4}\), which implies \(-\frac{1}{2} < x < \frac{1}{2}\).

Step 3: Determine the Radius and Interval of Convergence

From the ratio test result, the radius of convergence is \( R = \frac{1}{2} \). The interval of convergence in terms of \( x \) is \(-\frac{1}{2} < x < \frac{1}{2}\).

Step 4: Test Convergence at Endpoints

Test the series at \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \):- At \( x = \frac{1}{2} \), the series becomes \( \sum_{n=1}^{\infty} \frac{1}{n} \), which diverges (harmonic series).- At \( x = -\frac{1}{2} \), the series becomes \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \), which converges (alternating harmonic series). Thus, the series converges conditionally at \( x = -\frac{1}{2} \).

Step 5: Determine Absolute and Conditional Convergence

- The series converges *absolutely* for \(-\frac{1}{2} < x < \frac{1}{2}\).- It converges *conditionally* at \( x = -\frac{1}{2} \).- It does not converge for \( x = \frac{1}{2} \).Therefore, the interval of absolute convergence is \(-\frac{1}{2} < x < \frac{1}{2}\), and the point of conditional convergence is \( x = -\frac{1}{2} \).

Key Concepts

Convergence Tests

Convergence tests are vital tools in mathematics when dealing with power series like the one given in your exercise. A convergence test helps determine the values for which the series converges. For power series, a popular method is the ratio test. This test involves examining the limit of the ratio of successive terms. For our series, \[ a_n = \frac{4^n x^{2n}}{n}, \] the ratio of the next term to the current one is calculated:\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = |4x^2|. \]The ratio test implies the series converges when this limit is less than 1. - This gives the condition: - \(|4x^2| < 1\), leading to the inequality: - \(|x^2| < \frac{1}{4}\).This simplifies to finding the interval of convergence for \(x\). Once the range is determined, it's also essential to check the endpoints because sometimes the test can be inconclusive for boundary values.

Radius of Convergence

The radius of convergence (\(R\)) indicates how far from the center (in this case, zero) on the \(x\)-axis the series converges. It's a measure derived from the condition â bar of the ratio test. If \||4x^2| < 1\| , then solving provides the radius \(R = \frac{1}{2}\).The radius tells us more than just how far the series converges. It defines a circle in the complex plane within which series converges absolutely. It’s computed using:- The formula \( R = \frac{1}{L} \), - where \(L\) is the limit found in the ratio test.For our series:- Use \(|x^2| < \frac{1}{4}\)The radius leads directly to the interval of convergence \(-\frac{1}{2} < x < \frac{1}{2}\) in this particular exercise.

Conditional Convergence

Conditional convergence occurs when a series converges, but not absolutely. This means the series converges when the terms are considered in the original form, not in absolute terms.In our series, testing at \(x = -\frac{1}{2}\), the series converts to an alternating harmonic series:- \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \).- This sequence converges according to the alternating series test.But- \( \sum_{n=1}^{\infty} \frac{1}{n} \) diverges.In simpler terms:- At \(x = -\frac{1}{2}\),the series still converges, but not if you take absolute values of terms: hence conditional convergence.Conditional convergence is crucial for understanding if, when, and how terms can reorder and still maintain convergence. It’s an essential distinction in series analysis, helping mathematicians ensure accuracy in convergence theories.