Problem 19
Question
In Exercises \(7-28,\) find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.) $$ \sum_{n=0}^{\infty} \frac{(-1)^{n+1}(x-1)^{n+1}}{n+1} $$
Step-by-Step Solution
Verified Answer
The interval of convergence of the given series is \([0, 2)\). The series converges on this interval, including the point \(x = 0\) but excluding the point \(x = 2\).
1Step 1: Apply the Ratio Test
We refresh the basic idea of the Ratio Test, which states that for a series of the form \(\sum a_{n}\), the series converges if \(L < 1\), diverges if \(L > 1\), and is inconclusive if \(L = 1\), where \(L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_{n}}\right|\). Therefore, we'll find the limit \(L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_{n}}\right|\) for our series \(\sum_{n=0}^\infty (-1)^{n+1} (x-1)^{n+1}/(n+1)\). This gives us \(L = \lim_{n \to \infty} \left|\frac{(-1)^{n+2} (x-1)^{n+2} / (n+2)}{(-1)^{n+1} (x-1)^{n+1} / (n+1)}\right|\) which simplifies to \(L = \lim_{n \to \infty} \left|\frac{n+1}{n+2}(x-1)\right|\).
2Step 2: Determine the Interval of Convergence
We know the series converges if \(L < 1\) and diverges if \(L > 1\). Thus, we solve for \(x\) in \(L < 1\) and \(L > 1\) and find the interval of \(x\) where \(L < 1\), excluding the points where \(L = 1\) because the Ratio Test is inconclusive. Solving \(\left|\frac{n+1}{n+2}(x-1)\right| < 1\) as \(n \to \infty\) gives us \(-1 < (x-1) < 1\), or \(0 < x < 2\), which is the interval of convergence excluding the endpoints.
3Step 3: Check for Convergence at the Endpoints
The endpoints \(x = 0\) and \(x = 2\) must be checked individually for convergence. If they converge, they are included in the interval. When \(x = 0\), the series becomes \(\sum_{n=0}^\infty (-1)^{n+1}/(n+1)\), which is the alternating harmonic series and is convergent by the Alternating Series Test. When \(x = 2\), the series is \(\sum_{n=0}^\infty 1/(n+1)\), which is the harmonic series and is divergent. Including the convergent endpoint, the final interval of convergence is \([0, 2)\).
Key Concepts
Power SeriesRatio TestAlternating Series Test
Power Series
A power series is a series of the form \(\sum_{n=0}^{\infty} a_n (x-c)^n\), where \(a_n\) represents the coefficient of the nth term, \(x\) is a variable, and \(c\) is a constant known as the center of the series. Power series are used to represent functions in a way that is easy to manipulate and understand. The power series' ability to approximate functions near the center is a cornerstone in the fields of calculus and mathematical analysis.
For a power series to be useful, it's crucial to determine where it converges, meaning the values of \(x\) for which the series sums to a finite number. This set of values is referred to as the 'interval of convergence'. Finding this interval helps us understand the behavior and applicability of the function represented by the series. In the exercise, the power series converges to a function within a certain interval around \(x=1\), which is the center in this case.
For a power series to be useful, it's crucial to determine where it converges, meaning the values of \(x\) for which the series sums to a finite number. This set of values is referred to as the 'interval of convergence'. Finding this interval helps us understand the behavior and applicability of the function represented by the series. In the exercise, the power series converges to a function within a certain interval around \(x=1\), which is the center in this case.
Ratio Test
The Ratio Test is a method used to determine the convergence or divergence of infinite series.
For a series \(\sum a_{n}\), where \(a_{n}\) represents the nth term of the series, the test involves evaluating the limit \(L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_{n}}\right|\). If the result \(L < 1\), the series converges absolutely. If \(L > 1\), or if the limit does not exist, the series diverges. However, if \(L = 1\), the test is inconclusive, and other methods must be sought to establish convergence or divergence.
In the context of the exercise, applying the Ratio Test involves expressing each term using the relationship \( (-1)^{n+1}(x-1)^{n+1}/(n+1)\) and then finding the limit of their ratio as \(n\) approaches infinity to determine the interval of convergence for a given \(x\).
For a series \(\sum a_{n}\), where \(a_{n}\) represents the nth term of the series, the test involves evaluating the limit \(L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_{n}}\right|\). If the result \(L < 1\), the series converges absolutely. If \(L > 1\), or if the limit does not exist, the series diverges. However, if \(L = 1\), the test is inconclusive, and other methods must be sought to establish convergence or divergence.
In the context of the exercise, applying the Ratio Test involves expressing each term using the relationship \( (-1)^{n+1}(x-1)^{n+1}/(n+1)\) and then finding the limit of their ratio as \(n\) approaches infinity to determine the interval of convergence for a given \(x\).
Alternating Series Test
The Alternating Series Test is specifically designed to determine the convergence of series whose terms alternate in sign.
An alternating series takes the form \(\sum (-1)^{n}b_{n}\), where \(b_n\) is a sequence of positive numbers. The test states that if the absolute value of the terms \(b_n\) decreases monotonically (each term is smaller than the one preceding it) and the limit of \(b_n\) as \(n\to\infty\) is zero, then the series converges.
This test is crucial in practice because many functions have alternating series in their power series expansions. In the exercise, when \(x=0\), the given power series reduces to an alternating series that passes the Alternating Series Test, indicating convergence at that endpoint. Endpoints need individual checking since the general conditions on the interval of convergence do not necessarily apply at these points.
An alternating series takes the form \(\sum (-1)^{n}b_{n}\), where \(b_n\) is a sequence of positive numbers. The test states that if the absolute value of the terms \(b_n\) decreases monotonically (each term is smaller than the one preceding it) and the limit of \(b_n\) as \(n\to\infty\) is zero, then the series converges.
This test is crucial in practice because many functions have alternating series in their power series expansions. In the exercise, when \(x=0\), the given power series reduces to an alternating series that passes the Alternating Series Test, indicating convergence at that endpoint. Endpoints need individual checking since the general conditions on the interval of convergence do not necessarily apply at these points.
Other exercises in this chapter
Problem 19
In Exercises \(19-24,\) verify that the infinite series converges. $$ \sum_{n=1}^{\infty} \frac{1}{n(n+1)}(\text { Use partial fractions. }) $$
View solution Problem 19
Write the next two apparent terms of the sequence. Describe the pattern you used to find these terms. \(3,-\frac{3}{2}, \frac{3}{4},-\frac{3}{8}, \ldots\)
View solution Problem 19
Use the Integral Test to determine the convergence or divergence of the \(p\) -series. $$ \sum_{n=1}^{\infty} \frac{1}{n^{3}} $$
View solution Problem 19
In Exercises \(19-24,\) find the \(n\) th Taylor polynomial centered at \(c\). $$ f(x)=\frac{1}{x}, \quad n=4, \quad c=1 $$
View solution