Problem 39
Question
Let $$ f(x)=\sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}} $$ Find the intervals of convergence for \(f, f^{\prime},\) and \(f^{\prime \prime}\)
Step-by-Step Solution
Verified Answer
The interval of convergence for \(f(x)\) is \([-1, 1]\), for \(f'(x)\) is \([-1, 1)\), and for \(f''(x)\) is \((-1, 1)\).
1Step 1: Identify the Power Series
The given function is a power series: \( f(x) = \sum_{n=1}^{\infty} \frac{x^n}{n^2} \). It is centered at 0 with general term \( \frac{x^n}{n^2} \).
2Step 2: Apply the Ratio Test
To find the interval of convergence, apply the Ratio Test. For the general term \( a_n = \frac{x^n}{n^2} \), compute:\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{x^{n+1}}{(n+1)^2} \cdot \frac{n^2}{x^n} \right| = \lim_{n \to \infty} \left| x \cdot \frac{n^2}{(n+1)^2} \right|. \]This reduces to \( |x| \cdot \lim_{n \to \infty} \frac{n^2}{(n+1)^2} = |x| \).
3Step 3: Solve for Convergence
For convergence, we require \( |x| < 1 \). Hence, the series converges absolutely for \(-1 < x < 1\). Further testing is required for \(x = \pm 1\).
4Step 4: Test Endpoint x = 1
At \(x = 1\), the series becomes \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), which is a convergent p-series (p=2>1). Therefore, \(f(x)\) converges at \(x = 1\).
5Step 5: Test Endpoint x = -1
At \(x = -1\), the series becomes \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \), which converges by the Alternating Series Test since \( \frac{1}{n^2} \) is decreasing and approaches zero. Thus, \(f(x)\) converges at \(x = -1\).
6Step 6: Derivative Function and Convergence
The derivative series is \( f'(x) = \sum_{n=1}^{\infty} \frac{nx^{n-1}}{n^2} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{n} \). Apply the Ratio Test to find \( \lim_{n \to \infty} |x| = |x| < 1 \), using the same logic applied earlier, leading to \(-1 < x < 1\).
7Step 7: Test Derivative Endpoints
Test the boundary for \(x = 1\), you get \(\sum_{n=1}^{\infty} \frac{1}{n}\) which diverges (harmonic series). For \(x = -1\), \(\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\) converges (Alternating Series Test). So, \(f'(x)\) converges on \(-1 \leq x < 1\).
8Step 8: Second Derivative Function and Convergence
For the second derivative, the series is \( f''(x) = \sum_{n=1}^{\infty} x^{n-2} = \sum_{n=1}^{\infty} (n-1) x^{n-2} \). The Ratio Test applied gives \(-1 < x < 1\) and expands similar to \(f'(x)\).
9Step 9: Check Second Derivative Endpoints
For series expressions like this, convergence tests and comparison with similar findings suggest it converges within open interval \(-1 < x < 1\). No endpoints are included due to higher powers such as \(n-2\) in the denominator.
Key Concepts
Power SeriesRatio TestAlternating Series TestP-Series
Power Series
The given function, \[ f(x) = \sum_{n=1}^{\infty} \frac{x^n}{n^2} \], is an example of a power series. A power series is a mathematical expression that represents a function as an infinite sum of terms. Each term is in the form of \(c_n(x-a)^n\), where \(c_n\) are coefficients and \(a\) is the center of the series.
Our series is centered at 0 since the general term is \( \frac{x^n}{n^2} \). This means we consider how the series behaves as \( x \) approaches 0.
Power series are significant in mathematical analysis because they allow functions to be expressed in a way that is often easier to analyze. They are especially useful for approximating functions and solving differential equations. Understanding the interval of convergence, which specifies the set of \( x \) values for which the series converges, is crucial for using power series effectively. In this problem, determining these intervals helps understand where \( f(x) \), its first derivative, \( f'(x) \), and its second derivative, \( f''(x) \), converge.
Our series is centered at 0 since the general term is \( \frac{x^n}{n^2} \). This means we consider how the series behaves as \( x \) approaches 0.
Power series are significant in mathematical analysis because they allow functions to be expressed in a way that is often easier to analyze. They are especially useful for approximating functions and solving differential equations. Understanding the interval of convergence, which specifies the set of \( x \) values for which the series converges, is crucial for using power series effectively. In this problem, determining these intervals helps understand where \( f(x) \), its first derivative, \( f'(x) \), and its second derivative, \( f''(x) \), converge.
Ratio Test
The ratio test is a crucial technique for determining the convergence of an infinite series, especially when dealing with power series. In this particular exercise, it's applied to find where the power series converges.
The idea is to examine the limit of the absolute value of the ratio of successive terms. Given a series with general term \( a_n \), the test analyses 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, which indicates that it converges for all real numbers in that interval.
In our exercise, for \[ a_n = \frac{x^n}{n^2} \], applying the ratio test ultimately gives us \[ |x| \cdot \lim_{n \to \infty} \frac{n^2}{(n+1)^2} = |x|.\] This simplifies to \( |x| < 1 \) for convergence. Therefore, by using this test, we conclude that the series converges absolutely in the interval \(-1 < x < 1\). After using the ratio test, further checks for the endpoints \( x = \pm 1 \) are necessary to know if the series converges or diverges at these points.
The idea is to examine the limit of the absolute value of the ratio of successive terms. Given a series with general term \( a_n \), the test analyses 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, which indicates that it converges for all real numbers in that interval.
In our exercise, for \[ a_n = \frac{x^n}{n^2} \], applying the ratio test ultimately gives us \[ |x| \cdot \lim_{n \to \infty} \frac{n^2}{(n+1)^2} = |x|.\] This simplifies to \( |x| < 1 \) for convergence. Therefore, by using this test, we conclude that the series converges absolutely in the interval \(-1 < x < 1\). After using the ratio test, further checks for the endpoints \( x = \pm 1 \) are necessary to know if the series converges or diverges at these points.
Alternating Series Test
Some series contain terms that alternate in sign, and these are evaluated using the alternating series test. This test is particularly useful for series of the form \[ \sum (-1)^n a_n, \]where \( a_n \) decreases to 0 as \( n \) increases.
It ensures convergence if two conditions are met:
In our series \[ \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}, \]we quickly establish it converges by checking both conditions. Each term \( \frac{1}{n^2} \) does indeed become smaller as \( n \) increases, and they tend towards zero. Therefore, the series converges when \( x = -1 \) using the alternating series test.
It ensures convergence if two conditions are met:
- The absolute value of the terms decreases monotonically (i.e., each term is smaller than its predecessor).
- The terms tend to 0 as the number of terms goes to infinity.
In our series \[ \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}, \]we quickly establish it converges by checking both conditions. Each term \( \frac{1}{n^2} \) does indeed become smaller as \( n \) increases, and they tend towards zero. Therefore, the series converges when \( x = -1 \) using the alternating series test.
P-Series
A \( p \)-series is a type of series that helps in analyzing convergence. It has the general form \[ \sum_{n=1}^{\infty} \frac{1}{n^p}. \]
For the series to converge, the power \( p \) must be greater than 1. The larger the value of \( p \), the faster the terms approach zero, and thus, more quickly it converges.
In this exercise, when testing \( x = 1 \), the series \[ \sum_{n=1}^{\infty} \frac{1}{n^2} \] appears—a classic example of a \( p \)-series with \( p = 2 > 1 \).
As expected, the series converges since it meets the criterion for \( p \)-series convergence. Therefore, in our situation, the series is convergent at \( x = 1 \), confirming that \( p \)-series rules are crucial in determining the endpoints' behavior in convergence solutions.
For the series to converge, the power \( p \) must be greater than 1. The larger the value of \( p \), the faster the terms approach zero, and thus, more quickly it converges.
In this exercise, when testing \( x = 1 \), the series \[ \sum_{n=1}^{\infty} \frac{1}{n^2} \] appears—a classic example of a \( p \)-series with \( p = 2 > 1 \).
As expected, the series converges since it meets the criterion for \( p \)-series convergence. Therefore, in our situation, the series is convergent at \( x = 1 \), confirming that \( p \)-series rules are crucial in determining the endpoints' behavior in convergence solutions.
Other exercises in this chapter
Problem 38
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View solution Problem 39
The meaning of the decimal representation of a number $$0 . d_{1} d_{2} d_{3} \ldots\( (where the digit \)d_{i}\( is one of the numbers \)0,1\( \)2, \ldots, 9 )
View solution