Problem 35 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the convergence set for each series. (a) \(\sum_{n=1}^{\infty} \frac{(3 x+1)^{n}}{n \cdot 2^{n}}\) (b) \(\sum_{n=1}^{\infty}(-1)^{n} \frac{(2 x-3)^{n}}{4^{n} \sqrt{n}}\)

Step-by-Step Solution

Answer

The convergence set for (a) is \((-\frac{1}{3}, \frac{1}{3})\), and for (b) is \((-\frac{1}{2}, \frac{7}{2})\).

Step 1: Identify the series form for (a)

The given series for (a) is \(\sum_{n=1}^{\infty} \frac{(3x+1)^{n}}{n \cdot 2^{n}}\). This is a power series with the general term \(a_n = \frac{(3x+1)^{n}}{n \cdot 2^{n}}\).We can see it resembles the form \(\sum c_n (x - a)^n\), which is typical for power series with center \(a\).

Step 2: Apply the Ratio Test for (a)

The ratio test for convergence of a series \(\sum a_n\) considers \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1\). For our series, calculate:\[\lim_{n \to \infty} \left| \frac{(3x+1)^{n+1}}{(n+1)2^{n+1}} \cdot \frac{n2^n}{(3x+1)^n} \right| = \lim_{n \to \infty} \left| \frac{(3x+1)}{2} \cdot \frac{n}{n+1} \right|\]This simplifies to \(\left| \frac{3x+1}{2} \right| < 1\).

Step 3: Solve the inequality for (a)

Simplify \(|\frac{3x+1}{2}| < 1\) to find the convergence set:\[\frac{-1}{3} < x < \frac{1}{3}\]Thus, the series converges for \(x\) in the interval \(\left(\frac{-1}{3}, \frac{1}{3}\right)\).

Step 4: Identify the series form for (b)

The series for (b) is \(\sum_{n=1}^{\infty}(-1)^{n} \frac{(2x-3)^{n}}{4^{n} \sqrt{n}}\). This is an alternating series and also a power series with general term:\[a_n = (-1)^n \frac{(2x-3)^n}{4^n \sqrt{n}}\].

Step 5: Apply the Root Test for (b)

For the root test, consider \( \lim_{n \to \infty} \sqrt[n]{|a_n|} < 1\). In this series, apply:\[\lim_{n \to \infty} \left| \frac{(2x-3)^n}{4^n \sqrt{n}} \right|^{1/n} = \lim_{n \to \infty} \frac{|2x-3|^{1/n}}{4}\]The term \(|2x-3|^{1/n} \to 1\) as \(n \to \infty\), so it simplifies to \(\frac{|2x-3|}{4} < 1\).

Step 6: Solve the inequality for (b)

Solve \(|2x-3| < 4\):\[-4 < 2x - 3 < 4\]Adding 3 gives:\[-1 < 2x < 7\]Dividing by 2:\[-\frac{1}{2} < x < \frac{7}{2}\]Meaning the series converges for \(x\) in the interval \((-\frac{1}{2}, \frac{7}{2})\).

Key Concepts

Ratio Test

The Ratio Test is a powerful tool for determining the convergence of a series. It involves the limit of the absolute value of the ratio of successive terms in a series. For a series \( \sum a_n \), the Ratio Test considers the limit:

• \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \)

If this limit is less than 1, the series converges absolutely. If the limit is greater than 1, or if it is infinite, the series diverges. If the limit equals 1, the test is inconclusive.
The Ratio Test is particularly effective for series involving factorials or exponential terms. In our example, using the Ratio Test on the series \( \sum_{n=1}^{\infty} \frac{(3x+1)^n}{n \cdot 2^n} \), we found that:

• \( \left| \frac{3x+1}{2} \right| < 1 \)

This led us to solve the inequality and find the convergence set, \( \left( \frac{-1}{3}, \frac{1}{3} \right) \). This interval defines where the series converges based on the Ratio Test’s criteria.

Root Test

The Root Test is another method used to test the convergence of a series. It is similar to the Ratio Test but requires taking the nth root of the absolute value of the terms. For a series \( \sum a_n \), the Root Test evaluates:

• \( \lim_{n \to \infty} \sqrt[n]{|a_n|} \)

If this limit is less than 1, the series converges absolutely. If it's greater than 1, the series diverges. Like the Ratio Test, if the result is 1, the test gives no conclusion.
In the given exercise, we applied the Root Test to the series \( \sum_{n=1}^{\infty} (-1)^n \frac{(2x-3)^n}{4^n \sqrt{n}} \), leading to:

• \( \frac{|2x-3|}{4} < 1 \)

This inequality determines the set where the series converges and simplifies to: \( \left( -\frac{1}{2}, \frac{7}{2} \right) \). This interval denotes where the series is convergent according to the Root Test.

Power Series

A power series is a series of the form \( \sum_{n=0}^{\infty} c_n (x - a)^n \), where \( c_n \) are coefficients, and \( a \) is the center of the series. These are infinitely long polynomials and have a variable \( x \) which influences their convergence based on its value.
Power series can converge differently based on the value of \( x \), and usually require tests like the Ratio Test or Root Test to determine the convergence interval. The convergence set is the range of \( x \) where the power series converges. For example, the series \( \sum_{n=1}^{\infty} \frac{(3x+1)^n}{n \cdot 2^n} \) aligns with the definition of a power series centered at zero.
By using the Ratio Test, we deduced that it converges for \( x \) within \( \left( \frac{-1}{3}, \frac{1}{3} \right) \). Understanding power series is crucial, as they form the foundation of many functions in calculus and analysis.

Alternating Series

An alternating series is a series whose terms alternate in sign, typically expressed as \( \sum_{n=1}^{\infty} (-1)^n a_n \).
These series can converge even when the corresponding series of absolute values diverges, thanks to the Alternating Series Test. This test states that an alternating series converges if the absolute value of the terms decreases steadily and approaches zero.
In our example, the series \( \sum_{n=1}^{\infty} (-1)^n \frac{(2x-3)^n}{4^n \sqrt{n}} \) is alternating because of the term \((-1)^n\). The Root Test was used here, leading us to the conclusion that it converges for \( x \) values within \( \left( -\frac{1}{2}, \frac{7}{2} \right) \). Recognizing alternating series is important for solving complex problems involving conditional convergence and understanding the behavior of series expansions.