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Determine the radius of convergence of the given power series. $$ \sum_{n=1}^{\infty} \frac{(2 x+1)^{n}}{n^{2}} $$

Short Answer

Expert verified
The radius of convergence of the given power series is 1. This means that the power series converges absolutely for all values of x within the interval (-1, 0).

Step by step solution

01

Write down the Ratio Test formula

For a power series \(\sum_{n=1}^{\infty} a_n\), the Ratio Test states that the limit: $$ L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| $$ helps determine the convergence of the series. If \(L < 1\), the series converges. If \(L > 1\), the series diverges. If \(L = 1\), the test is inconclusive.
02

Apply the Ratio Test to the given power series

For the given power series, the general term is \(a_n = \frac{(2x+1)^n}{n^2}\). We want to find the ratio \(\frac{a_{n+1}}{a_n}\) and take the limit as n approaches infinity. $$ \frac{a_{n+1}}{a_n} = \frac{\frac{(2x+1)^{n+1}}{(n+1)^2}}{\frac{(2x+1)^n}{n^2}} $$
03

Simplify the ratio

Now we can simplify the ratio, canceling out some terms: $$ \frac{a_{n+1}}{a_n} = \frac{(2x+1)^{n+1} n^2}{(n+1)^2 (2x+1)^n} $$ Since \((2x+1)^n\) appears in both the numerator and the denominator, we can cancel it out: $$ \frac{a_{n+1}}{a_n} = \frac{(2x+1) n^2}{(n+1)^2} $$
04

Take the limit as n approaches infinity

Now we need to take the limit of the simplified ratio as n approaches infinity: $$ L = \lim_{n \to \infty} \left|\frac{(2x+1) n^2}{(n+1)^2}\right| $$ To evaluate this limit, we can divide both the numerator and denominator by \(n^2\): $$ L = \lim_{n \to \infty} \left|\frac{(2x+1)}{\left(1+\frac{1}{n}\right)^2}\right| $$ As n approaches infinity, the term \(\frac{1}{n}\) becomes 0, giving us: $$ L = \left|\frac{(2x+1)}{1^2}\right| = |2x+1| $$
05

Find the radius of convergence

For the series to converge, we need the limit \(L < 1\). We have \(L = |2x+1|\), so: $$ |2x+1| < 1 $$ To find the radius of convergence, we need to solve this inequality for the absolute value of x: $$ -1 < 2x+1 < 1 $$ Subtracting 1 from all sides gives: $$ -2 < 2x < 0 $$ Dividing by 2: $$ -1 < x < 0 $$ Thus, the radius of convergence is: $$ R = |x| = 1 $$ So the radius of convergence of the given power series is 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ratio Test
When it comes to analyzing the convergence of power series, the Ratio Test is an essential tool. It provides a way to determine whether a series converges absolutely, diverges, or remains inconclusive by using the limit of the ratio of successive terms. Here's a simplified explanation:

In the Ratio Test, we consider a series \( \sum_{n=1}^\infty a_n \) and calculate the limit \( L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| \) as n approaches infinity. If this limit, \( L \), is less than 1, the series is said to converge absolutely. If it's greater than 1, the series diverges. And if the limit equals 1, the test does not provide a conclusive answer on the convergence of the series.

To ensure students grasp this concept effectively:
  • Highlight the significance of the absolute value in this test, ensuring that the ratio considered is non-negative.
  • Provide examples where the Ratio Test clearly shows convergence or divergence.
  • Discuss how the Ratio Test can be inconclusive, requiring other convergence tests to make a determination.
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 coefficients, \( x \) is the variable, and \( c \) is the center of the series. The series expands in powers of \( (x-c) \).

Power series are used extensively in fields like physics and engineering to approximate functions, solve differential equations, and model complex phenomena. To further enhance comprehension for students:
  • Clarify the role of the center, \( c \), and how it affects the series.
  • Discuss examples where the power series converges to well-known functions.
  • Show the connection between Taylor series and power series for function approximation.

The application of the Ratio Test to a power series helps determine the radius of convergence, which indicates the interval around the center \( c \) where the series converges.
Limit of a Sequence
Understanding the limit of a sequence is fundamental in calculus and analysis. The limit describes the value that the terms of a sequence get closer to as the index \( n \) grows indefinitely. It is denoted as \( \lim_{n \to \infty} a_n \) for a sequence \( \{a_n\} \) and can be a finite number, infinity, or it may not exist at all.

For students to understand this concept standout points are:
  • Explain the epsilon-N definition of limits, which provides an intuitive grasp of how close terms must get to the limit as \( n \) increases.
  • Use graphical illustrations of sequences approaching their limits to visually demonstrate the concept.
  • Highlight differences between limits that exist (finite or infinite) and those that do not, using diverse examples.

The calculation of the limit is crucial for the Ratio Test during the investigation of the radius of convergence for power series.

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Most popular questions from this chapter

Consider the differential equation $$ x^{3} y^{\prime \prime}+\alpha x y^{\prime}+\beta y=0 $$ where \(\alpha\) and \(\beta\) are real constants and \(\alpha \neq 0\). (a) Show that \(x=0\) is an irregular singular point. (b) By attempting to determine a solution of the form \(\sum_{n=0}^{\infty} a_{n} x^{r+n},\) show that the indicial equation for \(r\) is linear, and consequently there is only one formal solution of the assumed form. (c) Show that if \(\beta / \alpha=-1,0,1,2, \ldots,\) then the formal series solution terminates and therefore is an actual solution. For other values of \(\beta / \alpha\) show that the formal series solution has a zero radius of convergence, and so does not represent an actual solution in any interval.

First Order Equations. The series methods discussed in this section are directly applicable to the first order linear differential equation \(P(x) y^{\prime}+Q(x) y=0\) at a point \(x_{0}\), if the function \(p=Q / P\) has a Taylor series expansion about that point. Such a point is called an ordinary point, and further, the radius of convergence of the series \(y=\sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\) is at least as large as the radius of convergence of the series for \(Q / P .\) In each of Problems 16 through 21 solve the given differential equation by a series in powers of \(x\) and verify that \(a_{0}\) is arbitrary in each case. Problems 20 and 21 involve nonhomogeneous differential equations to which series methods can be easily extended. Where possible, compare the series solution with the solution obtained by using the methods of Chapter 2 . $$ y^{\prime}-y=x^{2} $$

Referring to Problem \(14,\) use the method of reduction of order to show that the second solution of the Bessel equation of order zero contains a logarithmic term. Hint: If \(y_{2}(x)=J_{0}(x) v(x)\), then $$ y_{2}(x)=J_{0}(x) \int \frac{d x}{x\left[J_{0}(x)\right]^{2}} $$ Find the first term in the series expansion of \(1 / x\left[J_{0}(x)\right]^{2}\)

Show that the Bessel equation of order one-half, $$ x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-\frac{1}{4}\right) y=0, \quad x>0 $$ can be reduced to the equation $$ v^{\prime \prime}+v=0 $$ by the change of dependent variable \(y=x^{-1 / 2} v(x)\). From this conclude that \(y_{1}(x)=\) \(x^{-1 / 2} \cos x\) and \(y_{2}(x)=x^{-1 / 2} \sin x\) are solutions of the Bessel equation of order one-half.

Find all the regular singular points of the given differential equation. Determine the indicial equation and the exponents at the singularity for each regular singular point. \((x+1)^{2} y^{\prime \prime}+3\left(x^{2}-1\right) y^{\prime}+3 y=0\)

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