Problem 47
Question
Give examples that show that the convergence of a power series at an endpoint of its interval of convergence may be either conditional or absolute. Explain your reasoning.
Step-by-Step Solution
Verified Answer
The alternating harmonic series \(Σ((-1)^n/n)\) provides an example of conditional convergence at an end point, because it converges, but does not converge when all terms are replaced by their absolute values. On the other hand, the series \(Σx^n/n!\) is an example of absolute convergence because it does converge when all terms are replaced by their absolute values.
1Step 1: Provide an Example of Conditional Convergence
Consider the alternating harmonic series \(-1^n/n\) for \(x = -1\), also known as \(Σ((-1)^n/n)\). This series converges conditionally at \(x = -1\) because its terms do not converge to 0.
2Step 2: Explain Convergence Test
The alternating series test states that if the terms decrease to zero, which they do in the above series as \(n\) approaches infinity, the series converges.
3Step 3: Provide an Example of Absolute Convergence
Now consider the series \(Σx^n/n!\) for \(x = -1\) which equals \(Σ((-1)^n/n!)\). This series also converges at \(x = -1\) but unlike the above series, it converges absolutely.
4Step 4: Explain Absolute Convergence Test
The ratio test of convergence states if the absolute value of the ratio of the \(n+1\)th term to the nth term is less than 1, the series converges absolutely. This holds true for this series.
Other exercises in this chapter
Problem 47
In Exercises \(47-52,\) (a) write the repeating decimal as a geometric series and (b) write its sum as the ratio of two integers $$ 0 . \overline{4} $$
View solution Problem 47
Determine the convergence or divergence of the sequence with the given \(n\) th term. If the sequence converges, find its limit. \(a_{n}=\frac{n^{p}}{e^{n}}, p>
View solution Problem 47
Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} \frac{1}{n !} $$
View solution Problem 47
Verify the sum. Then use a graphing utility to approximate the sum with an error of less than 0.0001. $$ \sum_{n=0}^{\infty} \frac{2^{n}}{n !}=e^{2} $$
View solution