Problem 20
Question
Find the radius of convergence and interval of convergence of the series. $$\sum_{n=1}^{\infty} \frac{(2 x-1)^{n}}{5^{n} \sqrt{n}}$$
Step-by-Step Solution
Verified Answer
Radius of convergence: 2.5. Interval of convergence: \((-2, 3)\).
1Step 1: Understanding the Given Series
We are given the series \( \sum_{n=1}^{\infty} \frac{(2x-1)^{n}}{5^{n} \sqrt{n}} \). This is a form of a power series where the general term is \( a_n = \frac{(2x-1)^n}{5^n \sqrt{n}} \). The power series is centered at 0 and involves the variable \( 2x-1 \).
2Step 2: Identifying the Test for Convergence
To find the radius and interval of convergence of the series, we will use the Ratio Test. The Ratio Test is particularly useful because our series has powers in the numerator and denominator. We will calculate \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
3Step 3: Applying the Ratio Test
Start by expressing \( a_{n+1} \) as \( \frac{(2x-1)^{n+1}}{5^{n+1} \sqrt{n+1}} \). Calculate the ratio:\[ \frac{a_{n+1}}{a_n} = \frac{(2x-1)^{n+1}}{5^{n+1} \sqrt{n+1}} \cdot \frac{5^n \sqrt{n}}{(2x-1)^n} = \frac{(2x-1) \cdot \sqrt{n}}{5 \cdot \sqrt{n+1}} \]Now, simplify and take the limit:\[ \lim_{n \to \infty} \left| \frac{(2x-1) \cdot \sqrt{n}}{5 \cdot \sqrt{n+1}} \right| = \frac{|2x-1|}{5} \].
4Step 4: Setting the Convergence Condition
According to the Ratio Test, the series converges if:\[ \frac{|2x-1|}{5} < 1 \]Thus, solving for \( x \), we have:\[ |2x-1| < 5 \]
5Step 5: Solving the Inequality for Interval of Convergence
Split the inequality \( |2x-1| < 5 \) into:\[ -5 < 2x-1 < 5 \]Add 1 to each part of the inequality:\[ -4 < 2x < 6 \]Divide everything by 2:\[ -2 < x < 3 \]Thus, the interval of convergence is \((-2, 3)\).
6Step 6: Finding the Radius of Convergence
The radius of convergence, \( R \), is half the length of this interval. Since the interval \((-2, 3)\) has a length of 5:\[ R = \frac{5}{2} = 2.5 \].
7Step 7: Checking Endpoints
Evaluate convergence at endpoints \( x = -2 \) and \( x = 3 \). Calculate the series separately by substituting these values in \( (2x-1) \) and determine whether each resulting series converges or diverges using tests such as the p-series or alternating series test. - At \( x = -2 \), \( 2x-1 = -5 \): the series becomes \( \sum \frac{(-5)^n}{5^n \sqrt{n}} \), diverges.- At \( x = 3 \), \( 2x-1 = 5 \): the series becomes \( \sum \frac{5^n}{5^n \sqrt{n}} \), diverges.
Key Concepts
Radius of ConvergenceInterval of ConvergenceRatio Test
Radius of Convergence
When we dive into the realm of power series, understanding the radius of convergence is fundamental. In essence, the radius of convergence (R) is the distance from the center of the series within which it converges. To determine the radius of convergence for a series, the Ratio Test is typically employed for its simplicity and effectiveness.
In the provided series \( \sum_{n=1}^{\infty} \frac{(2x-1)^n}{5^n \sqrt{n}} \), the application of the Ratio Test involves calculating the limit of the absolute value of the ratio between consecutive terms. For this example:\[R = \frac{5}{|2x-1|}\]Upon solving \( \frac{|2x-1|}{5} < 1 \), we find \( R = 2.5 \). This means that as long as \( x \) is within 2.5 units of the center, the series will continue to converge.
In layman's terms, just think of the radius of convergence as the safe zone around the center, where the power series behaves nicely and doesn't blow up.
In the provided series \( \sum_{n=1}^{\infty} \frac{(2x-1)^n}{5^n \sqrt{n}} \), the application of the Ratio Test involves calculating the limit of the absolute value of the ratio between consecutive terms. For this example:\[R = \frac{5}{|2x-1|}\]Upon solving \( \frac{|2x-1|}{5} < 1 \), we find \( R = 2.5 \). This means that as long as \( x \) is within 2.5 units of the center, the series will continue to converge.
In layman's terms, just think of the radius of convergence as the safe zone around the center, where the power series behaves nicely and doesn't blow up.
Interval of Convergence
The interval of convergence is a critical aspect that follows finding the radius. It specifies all the x values for which the series actually converges—not just within the radius, but at and around certain boundary values, too. From our series, solving the inequality \( |2x-1| < 5 \) yields the interval \(-2 < x < 3\).
This interval signifies that all real values of \( x \) between -2 and 3 allow the series to converge. However, it’s crucial to check what happens exactly at the endpoints; sometimes, the behavior changes, and convergence isn’t guaranteed there.
Upon testing endpoints, in this case at \( x = -2 \) and \( x = 3 \), both series diverge. This means that our interval of convergence is open, and the series excludes these boundary values, resulting in \((-2, 3)\).
Think of it as a stretchable piece of elastic band; within it, everything is snug and firm, but it doesn't quite attach at the very tips unless directly tested.
This interval signifies that all real values of \( x \) between -2 and 3 allow the series to converge. However, it’s crucial to check what happens exactly at the endpoints; sometimes, the behavior changes, and convergence isn’t guaranteed there.
Upon testing endpoints, in this case at \( x = -2 \) and \( x = 3 \), both series diverge. This means that our interval of convergence is open, and the series excludes these boundary values, resulting in \((-2, 3)\).
Think of it as a stretchable piece of elastic band; within it, everything is snug and firm, but it doesn't quite attach at the very tips unless directly tested.
Ratio Test
The Ratio Test is a pivotal tool to ascertain convergence of series, acting like a detective ensuring the series' terms shrink appropriately. To apply the Ratio Test effectively, consider any general term of a series \( a_n \).
For our series, by expressing the terms as \( a_n = \frac{(2x-1)^n}{5^n \sqrt{n}} \), we find:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{|2x-1|}{5}\]According to the Ratio Test, if this limit is less than 1, the series converges. Hence, rearranging gives us the condition \( |2x-1| < 5 \), leading us to the radius of convergence, demonstrated in previous sections.
Think of the Ratio Test as a universal sanity check. The test essentially evaluates if the series' terms shrink enough to zero as \( n \) grows infinitely large. If they do, using the established condition, the series passes the test and converges neatly within its radius!
It’s a method focusing on each step of the series to ensure stability and convergence like a conductor harmonizing an orchestra.
For our series, by expressing the terms as \( a_n = \frac{(2x-1)^n}{5^n \sqrt{n}} \), we find:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{|2x-1|}{5}\]According to the Ratio Test, if this limit is less than 1, the series converges. Hence, rearranging gives us the condition \( |2x-1| < 5 \), leading us to the radius of convergence, demonstrated in previous sections.
Think of the Ratio Test as a universal sanity check. The test essentially evaluates if the series' terms shrink enough to zero as \( n \) grows infinitely large. If they do, using the established condition, the series passes the test and converges neatly within its radius!
It’s a method focusing on each step of the series to ensure stability and convergence like a conductor harmonizing an orchestra.
Other exercises in this chapter
Problem 20
Suppose you know that $$f^{(n)}(4)=\frac{(-1)^{n} n !}{3^{n}(n+1)}$$ and the Taylor series of \(f\) centered at 4 converges to \(f(x)\) for all \(x\) in the int
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\(19-40=\) Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=1}^{\infty} \frac{(-2)^{n}}{n^{2}} $$
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\(9-32\) n Determine whether the sequence converges or diverges. If it converges, find the limit. $$a_{n}=\cos (2 / n)$$
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Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$\sum_{n=1}^{\infty} \arctan n$$
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