Problem 32 - 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? $$ \sum_{n=1}^{\infty} \frac{(3 x+1)^{n+1}}{2 n+2} $$

Step-by-Step Solution

Answer

The series converges absolutely at \( x = -\frac{2}{3} \) and conditionally for \( x \in (-\frac{2}{3}, 0) \).

Step 1: Interpret the Series

The series given is \( \sum_{n=1}^{\infty} \frac{(3x+1)^{n+1}}{2n+2} \). Note that this can also be written as \( \sum_{n=1}^{\infty} \frac{(3x+1)\cdot(3x+1)^{n}}{2n+2} \). This is a power series in terms of \( (3x+1) \).

Step 2: Identify the Roots for Convergence Using Ratio Test

To find the radius of convergence, apply the Ratio Test: we need the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Compute the term \( \frac{a_{n+1}}{a_n} = \frac{\frac{(3x+1)^{n+2}}{2n+4}}{\frac{(3x+1)^{n+1}}{2n+2}} = \frac{(3x+1)(2n+2)}{(2n+4)} \). Simplified, \( \lim_{n \to \infty} \frac{(3x+1)(2n+2)}{2n+4} = |3x+1| \).

Step 3: Determine the Radius of Convergence

For the series to converge, \( |3x+1| < 1 \). Solving \( -1 < 3x+1 < 1 \) gives the interval \( -\frac{2}{3} < x < 0 \). Hence, the radius of convergence is \( R = \frac{1}{3} \).

Step 4: Verify Convergence at Endpoints

Check convergence at \( x = -\frac{2}{3} \) and \( x = 0 \).- For \( x = -\frac{2}{3} \), the series becomes \( \sum_{n=1}^{\infty} \frac{0}{2n+2} = 0 \), which converges absolutely.- For \( x = 0 \), the series becomes \( \sum_{n=1}^{\infty} \frac{1^{n+1}}{2n+2} = \sum_{n=1}^{\infty} \frac{1}{2n+2} \), which diverges by the harmonic series test.

Step 5: Conclusion on Convergence

The interval of convergence for \( x \) is \( [-\frac{2}{3}, 0) \).- The series converges absolutely at \( x = -\frac{2}{3} \).- The series converges conditionally for \( x \in (-\frac{2}{3}, 0) \).

Key Concepts

Radius of Convergence

In the realm of power series, the radius of convergence is a critical value that tells us how far from the center of the series we can move and still have the series sum to a finite number. In simpler terms, it's like the safe zone around a point where the series doesn't blow up.

• The radius of convergence, denoted usually by \( R \), is determined by finding within what bounds the variable \( x \) can change without making the series diverge.
• For our given series, we use the Ratio Test, a tool that helps us to find the individual's radius of convergence.
• We compute the limit of the absolute value of the ratio of successive terms as \( n \) approaches infinity. In this case, we found that \( \left|3x+1\right| < 1 \).

Solving \( -1 < 3x+1 < 1 \) gives us the interval \( -\frac{2}{3} < x < 0 \). Thus, the radius is \( \frac{1}{3} \).
This means values of \( x \) can only oscillate within this interval for the series to stay convergent.

Ratio Test

The Ratio Test is a powerful method in calculus, specifically used to determine the convergence or divergence of series. It anchors on the behavior of the series’ terms as they move towards infinity.

• The formula for the Ratio Test requires us to look at the limit of the sequence of successive term ratios: \( L = \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \).
• If \( L < 1 \), the series converges absolutely; if \( L > 1 \), it diverges; and if \( L = 1 \), the test is inconclusive.

In the given exercise, using the Ratio Test, we simplified the limit process and reached \( |3x+1| \). As we derived \( |3x+1| < 1 \) for convergence, indicating that for smaller values within this boundary, the series remains convergent.

Interval of Convergence

The interval of convergence is where the series finds a comfortable range for meeting the criteria of convergence. This is the set of all \( x \) values that make the series converge.

• After determining the radius of convergence, the next step is to check if the series converges at the boundary points of the interval.
• In our example, the interval generated by \( |3x+1| < 1 \) was \( -\frac{2}{3} < x < 0 \).
• Testing the endpoints \( x = -\frac{2}{3} \) and \( x = 0 \) clarifies whether to include them in the interval or not.

At \( x = -\frac{2}{3} \), the series is \( 0 \) at all points, and hence converges absolutely. Whereas at \( x = 0 \), it diverges, pushing the acceptable range a notch inside. Thus, the overall interval of convergence is \( [-\frac{2}{3}, 0) \).
Understanding the interval helps us know exactly where the series behaves well and what segments are assuredly finite.