Problem 2
Question
Suppose that the series \(\sum_{k=0}^{\infty} a_{k} x^{k}\) converges at \(x=-3\) and diverges at \(x=5 .\) What can you conclude about the convergence or divergence of the following series? (a) \sum_{k=0}^{\infty} a_{k} 2^{k} (b) \(\sum_{k=0}^{\infty} a_{k}(-6)^{k}\) (c) \(\sum_{k=0}^{\infty} a_{k} 4^{k}\) (d) \(\sum_{k=0}^{\infty}(-1)^{k} a_{k} 3^{k}\)
Step-by-Step Solution
Verified Answer
In conclusion, based on the radius of convergence, the following series converge:
(a) \(\sum_{k=0}^{\infty} a_{k} 2^{k}\)
(b) \(\sum_{k=0}^{\infty} a_{k}(-6)^{k}\)
(c) \(\sum_{k=0}^{\infty} a_{k} 4^{k}\)
(d) \(\sum_{k=0}^{\infty}(-1)^{k} a_{k} 3^{k}\)
1Step 1: Determine the radius of convergence
To determine the radius of convergence, we can use the information given. Since the series converges at \(x=-3\) and diverges at \(x=5\), this means that the radius of convergence, \(R\), must be less than the distance between these two points, which is \(|5 - (-3)| = 8\). Therefore, we have \(R < 8\).
2Step 2: Check the convergence of each series
Now we will check if the series converge or diverge for the given values of \(x\):
(a) For the series \(\sum_{k=0}^{\infty} a_{k} 2^{k}\), the value of \(x\) is 2. Since \(|2| < 8\), this series converges.
(b) For the series \(\sum_{k=0}^{\infty} a_{k}(-6)^{k}\), the value of \(x\) is -6. The distance from the center of convergence (\(x = -3\)) to \(x = -6\) is \(|-6 - (-3)| = 3\). Since \(3 < 8\), this series converges.
(c) For the series \(\sum_{k=0}^{\infty} a_{k} 4^{k}\), the value of \(x\) is 4. The distance from the center of convergence (\(x = -3\)) to \(x = 4\) is \(|4 - (-3)| = 7\). Since \(7 < 8\), this series converges.
(d) For the series \(\sum_{k=0}^{\infty}(-1)^{k} a_{k} 3^{k}\), the value of \(x\) is 3. The distance from the center of convergence (\(x = -3\)) to \(x = 3\) is \(|3 - (-3)| = 6\). Since \(6 < 8\), this series converges.
In conclusion, all the given series (a), (b), (c), and (d) converge.
Key Concepts
Power SeriesConvergence and DivergenceDistance Between PointsConvergent Series
Power Series
A power series is like an infinite polynomial. You might be familiar with polynomials having a finite number of terms, but a power series continues forever. It is expressed in the general form:
- \( \sum_{k=0}^{\infty} a_k x^k \)
- \(a_k\) are the coefficients,
- \(x\) is the variable,
- and \(k\) is the term number starting from zero.
Convergence and Divergence
Understanding when a series converges or diverges is crucial. A series converges if the sum of its terms approaches a finite limit as you add more terms. It diverges if the sum doesn't settle on a limit. In simple terms, a convergent series behaves and leads to a predictable outcome, while a divergent series doesn't.
The convergence of a power series depends on its radius of convergence. The radius is the distance between the center of the series and the boundary within which the series converges.
If you find that the power series converges at a particular point, any point closer to the center will also converge. However, if a series diverges at a point, any point further away will also diverge. This is essential when you determine the series behavior around different points.
Distance Between Points
The distance between points is a simple yet powerful concept for understanding the behavior of power series. In the context of power series, it helps determine whether a certain point will lead to convergence or divergence.
Consider a power series with a center point. By calculating the absolute distance between this center and any other point (where you may want to evaluate the series), you essentially check if the point is within the radius for convergence.
In practice, if the distance from the center to the point in question is less than the radius of convergence, the series will converge at that point. Conversely, if the distance is greater, the series will diverge.
Convergent Series
A convergent series adds up to a specific, finite value. Think of it as a recipe for success; all ingredients (or terms) work together to create something complete and well-defined.With a power series, once you've determined the radius of convergence, you can easily predict which values of \(x\) result in a convergent series. Any value of \(x\) that falls within the radius from the center ensures convergence.This concept is very helpful in various fields like mathematics, physics, and engineering. Knowing where a series converges allows scientists and engineers to use these series reliably for calculations, approximations, and even solving complex differential equations.
Other exercises in this chapter
Problem 1
Evaluate. $$\sum_{k=0}^{2}(3 k+1)$$
View solution Problem 2
Expand \(f(x)\) in powers of \(x,\) basing your calculations on the geometric series $$\frac{1}{1-x}=1+x+x^{2}+\cdots+x^{n}+\cdots$$ $$f(x)=\frac{1}{(1-x)^{3}}$
View solution Problem 2
Find the Taylor polynomial \(P_{4}\) for the function \(f\) $$f(x)=\sqrt{1+x}$$
View solution Problem 2
Find the Taylor polynomial of the function \(f\) for the given values of \(a\) and \(n\) and give the Lagrange form of the remainder. $$f(x)=\cos x ; \quad a=\p
View solution