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Chapter 2: Problem 10

a) Prove that \(\lim n^{1 / n}=1 .\) Hint: Write \(n^{1 / n}=1+b_{n}\) and note \(b_{n}>0 .\) Then show that \(\left(1+b_{n}\right)^{n} \geq \frac{n(n-1)}{2} b_{n}^{2}\) and use this to show that \(\lim b_{n}=0 .\) b) Use the result of part a) to show that if \(\sum a_{n} x^{n}\) is a convergent power series with radius of convergence \(R\), then \(\sum n a_{n} x^{n}\) is also convergent with the same radius of convergence.

Short Answer

Expert verified
a) \( \lim_{n \to \infty} n^{1/n} = 1. \) b) \( \sum n a_n x^n \) converges with radius \( R \).

Step by step solution

01

Understanding the Limit Statement

The exercise requires us to show that \( \lim_{n \to \infty} n^{1/n} = 1 \). The hint suggests rewriting \( n^{1/n} \) as \( 1 + b_n \) with \( b_n > 0 \), leading towards a proof structure that leverages inequalities.
02

Rearrange and Analyze \( n^{1/n} \)

Using \( n^{1/n} = 1 + b_n \), express it as \( (1 + b_n)^n = n \). To dive deeper, use the inequality given: \( (1 + b_n)^n \geq \frac{n(n-1)}{2}b_n^2 \).
03

Develop the Inequality \( \frac{n(n-1)}{2}b_n^2 \leq n \)

Since \((1 + b_n)^n = n \), and assuming \( (1 + b_n)^n \geq \frac{n(n-1)}{2}b_n^2 \), it follows that \( \frac{n(n-1)}{2}b_n^2 \leq n \). Simplifying, we get \( b_n^2 \leq \frac{2}{n-1} \).
04

Establish Limit \( \lim_{n \to \infty} b_n = 0 \)

From \( b_n^2 \leq \frac{2}{n-1} \), taking limits on both sides gives \( \lim_{n \to \infty} b_n^2 = 0 \), which implies \( \lim_{n \to \infty} b_n = 0 \). Thus, \( n^{1/n}=1 + b_n \to 1.\)
05

Apply Result to Power Series

Since part a) shows \( \lim_{n \to \infty} n^{1/n} = 1 \), apply this to the convergence condition of power series. For the series \( \sum a_n x^n \) converging with radius \( R \), the radius is determined by \( \limsup_{n \to \infty} |a_n|^{1/n} = 1/R \).
06

Conclusion for Power Series \( \sum n a_n x^n \)

The new series \( \sum n a_n x^n \) retains the same radius \( R \) if \( (n a_n)^{1/n} \to |a_n|^{1/n} \) as \( n^{1/n} \to 1 \). Therefore, it converges within the same radius.

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Key Concepts

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

Power series
A power series is an infinite series of the form \( \sum a_n x^n \), where \( a_n \) represents the coefficients and \( x \) is the variable. These series play a significant role in calculus and mathematical analysis, particularly in describing functions as infinite polynomials. The convergence of a power series depends on the value of \( x \) and the properties of its coefficients. Reliable convergence ensures we can use the power series to approximate functions accurately.

Understanding power series involves recognizing:
  • The general form: \( \sum a_n x^n \)
  • The significance of the coefficients \( a_n \)
  • The variable \( x \) determining where the series converges
By exploring these components, you get a clearer picture of how power series can represent complex functions in a unified and useful manner.
Convergence
Convergence of a power series refers to the series approaching a finite value as the number of terms increases. For a series like \( \sum a_n x^n \), it converges when the sum tends to a specific number as \( n \) becomes very large.

Critical factors in convergence are:
  • The limit of series terms: \( \lim_{n \to \infty} a_n x^n = 0 \)
  • The series' behavior at specific values of \( x \): determining its interval of convergence
In the context of the original exercise, using limits like \( \lim_{n \to \infty} n^{1/n} \) helps deduce if modifying a series still guarantees convergence. Understanding convergence not only aids in grasping series behavior but also provides a foundation for advanced concepts in calculus and function approximation.
Radius of convergence
The radius of convergence \( R \) is a critical value that determines the interval within which a power series \( \sum a_n x^n \) converges. This radius tells us how far along the x-axis we can move and the series will still converge. To find \( R \), one commonly uses the formula: \[R = \frac{1}{\limsup_{n \to \infty} |a_n|^{1/n}}\]
The radius of convergence sets the boundaries of the domain where our series is reliable.
  • If \(|x| < R\), the series converges absolutely.
  • If \(|x| > R\), the series diverges.
  • If \(|x| = R\), convergence relies on additional testing.
The exercise illustrates using the result \( n^{1/n} \to 1 \) to argue that modifying a series slightly, like multiplying by \( n \), retains the essential radius \( R \), ensuring convergence in the same domain.
Inequalities
Inequalities are crucial tools in mathematics, often used to estimate the bounds of functions or series. In the original exercise, inequalities play a key role in proving limits, such as showing \( (1+b_n)^n \geq \frac{n(n-1)}{2}b_n^2 \).

Here鈥檚 how inequalities contribute:
  • Show relationships: By comparing terms, inequalities help establish whether a series converges or a limit holds.
  • Provide bounds: They give you upper and lower constraints on behavior, as seen when \( b_n^2 \leq \frac{2}{n-1} \) helps us conclude about \( b_n \).
  • Validate limits: Inequalities can affirm that a calculated limit is sound, assisting in detailed analysis of series convergence.
Utilizing inequalities lets us dissect complex problems and assert claims like the convergence of a series or the behavior of its terms.

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Most popular questions from this chapter

Suppose \(x_{1}:=\frac{1}{2}\) and \(x_{n+1}:=x_{n}^{2}\). Show that \(\left\\{x_{n}\right\\}\) converges and find \(\lim x_{n}\). Hint: You cannot divide by zero!

Prove Abel's theorem: Theorem. Suppose \(\sum x_{n}\) is a series whose partial sums are a bounded sequence, \(\left\\{\lambda_{n}\right\\}\) is a sequence with \(\lim \lambda_{n}=0,\) and \(\sum\left|\lambda_{n+1}-\lambda_{n}\right|\) is convergent. Then \(\sum \lambda_{n} x_{n}\) is convergent.

True or false, prove or find a counterexample. If \(\left\\{x_{n}\right\\}\) is a sequence such that \(\left\\{x_{n}^{2}\right\\}\) converges, then \(\left\\{x_{n}\right\\}\) converges.

Suppose \(\sum a_{n} x^{n}\) converges for \(x=1\). a) What can you say about the radius of convergence? b) If you further know that at \(x=1\) the convergence is not absolute, what can you say?

Suppose the kth partial sum of \(\sum_{n=1} x_{n}\) is \(s_{k}=\frac{k}{k+1} .\) Find the series, that is find \(x_{n}\), prove that the series converges, and then find the limit.
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