Problem 10

Question

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.

Step-by-Step Solution

Verified

Answer

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

1Step 1: 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.

2Step 2: 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 $.

3Step 3: 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} $.

4Step 4: 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.$

5Step 5: 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 $.

6Step 6: 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.

Key Concepts

Power seriesConvergenceRadius of convergenceInequalities

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’s 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.

Problem 10

Problem 11

Other exercises in this chapter

Problem 9

In the following exercises, feel free to use what you know from calculus to find the limit, if it exists. But you must prove that you found the correct limit, o

View solution

Problem 10

(Challenging): Let $\left\\{x_{n}\right\\}$ be a convergent sequence such that $x_{n} \geq 0$ and $k \in \mathbb{N}$. Then $$ \lim _{n \rightarrow \infty}

View solution

Problem 11

Let $r>0 .$ Show that starting with any $x_{1} \neq 0,$ the sequence defined by $$ x_{n+1}:=x_{n}-\frac{x_{n}^{2}-r}{2 x_{n}} $$ converges to $\sqrt{r}$ i

View solution

Problem 11

a) If $\sum a_{n}$ is convergent to a (in the usual sense), show that $\sum a_{n}$ is Cesàro summable (see above) to a. b) Show that in the sense of Cesàro

View solution