/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 F眉r welche \(x \in \mathbb{R}\)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 7

F眉r welche \(x \in \mathbb{R}\) konvergieren die folgenden Potenzreihen? (a) \(\left(\sum_{n=1}^{\infty} \frac{(-1)^{n}\left(2^{n}+1\right)}{n}\left(x-\frac{1}{2}\right)^{n}\right)\) (b) \(\left(\sum_{n=0}^{\infty} \frac{1-(-2)^{-n-1} n !}{n !}(x-2)^{n}\right)\) (c) \(\left(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\left[\sqrt{n^{2}+n}-\sqrt{n^{2}+1}\right]^{n}(x+1)^{n}\right)\)

Short Answer

Expert verified
a) The power series converges for all x in 鈩�. b) The power series converges only for x = 2. c) The power series converges for all x in 鈩�.

Step by step solution

01

Identify the sequence

We have a power series in the form of: \(\sum_{n=1}^{\infty} a_n \left(x-\frac{1}{2}\right)^n\), where \(a_n = \frac{(-1)^{n}\left(2^{n}+1\right)}{n}\).
02

Apply the root test

Apply the root test to the sequence \(a_n\). The root test is used to determine convergence by examining the limit \(\lim_{n\to\infty} \sqrt[n]{|a_n|}\): $$\lim_{n\to\infty}\sqrt[n]{\left|\frac{(-1)^{n}\left(2^{n}+1\right)}{n}\right|} = \lim_{n\to\infty}\frac{2+\frac{1}{2^{n}}}{\sqrt[n]{n}}$$
03

Evaluate the limit

In order to evaluate the limit, divide the numerator and the denominator by \(2^n\): $$\lim_{n\to\infty} \frac{\frac{2^{n+1}+1}{2^n}}{\frac{n^{1/n}}{2^n}} = \lim_{n\to\infty} \frac{2+\frac{1}{2^n}}{\frac{n^{1/n}}{2^n}}$$ Since \(\lim_{n\to\infty} \frac{1}{2^n} = 0\) and \(\lim_{n\to\infty} \frac{n^{1/n}}{2^n} = 0\), the limit becomes: $$\lim_{n\to\infty} \frac{2+\frac{1}{2^n}}{\frac{n^{1/n}}{2^n}} = \lim_{n\to\infty} \frac{2}{0} = \infty$$
04

Determine convergence

As the limit from the root test is infinite, the power series converges for all \(x \in \mathbb{R}\). (b) \(\left(\sum_{n=0}^{\infty} \frac{1-(-2)^{-n-1} n !}{n !}(x-2)^{n}\right)\)
05

Identify the sequence

We have a power series in the form of: \(\sum_{n=0}^{\infty} b_n (x-2)^n\), where \(b_n = \frac{1-(-2)^{-n-1} n !}{n !}\).
06

Apply the ratio test

Apply the ratio test to the sequence \(b_n\). The ratio test is used to determine convergence by examining the limit \(\lim_{n\to\infty} |\frac{b_{n+1}}{b_n}|\): $$\lim_{n\to\infty} \left| \frac{\frac{1-(-2)^{-(n+1)-1} (n+1) !}{(n+1) !}}{\frac{1-(-2)^{-n-1} n !}{n !}} \right| = \lim_{n\to\infty} \left| \frac{1-(-2)^{-n} (n+1)}{1-(-2)^{-n-1} n} \right|$$
07

Evaluate the limit

In order to evaluate the limit, multiply the numerator and the denominator by \((-2)^n\): $$\lim_{n\to\infty} \left| \frac{-2^n + n+1}{-2^n n + n} \right|$$ Since the leading term on both the numerator and the denominator is \(-2^n\), we can simplify the limit as follows: $$\lim_{n\to\infty} \left| \frac{n+1}{n} \right| = 1$$
08

Determine convergence

As the limit from the ratio test is 1, the power series converges only for \(x = 2\) (where \(|x-2| < 1\)). (c) $\left(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\left[\sqrt{n^{2}+n}-\sqrt{n^{2}+1}\right]^{n}(x+1)^{n}\right)$
09

Identify the sequence

We have a power series in the form of: \(\sum_{n=1}^{\infty} c_n (x+1)^n\), where \(c_n = \frac{1}{n^2}\left[\sqrt{n^2+n}-\sqrt{n^2+1}\right]^n\).
10

Apply the root test

Apply the root test to the sequence \(c_n\). The root test is used to determine convergence by examining the limit \(\lim_{n\to\infty} \sqrt[n]{|c_n|}\): $$\lim_{n\to\infty}\sqrt[n]{\left|\frac{1}{n^2}\left[\sqrt{n^2+n}-\sqrt{n^2+1}\right]^n\right|}$$
11

Evaluate the limit

Observe that the expression \(\frac{1}{n^2}\left[\sqrt{n^2+n}-\sqrt{n^2+1}\right]^n\) is positive. We can rewrite the limit as: $$\lim_{n\to\infty}\frac{\sqrt[n]{\left[\sqrt{n^2+n}-\sqrt{n^2+1}\right]^n}}{\sqrt[n]{n^2}} = \lim_{n\to\infty}\frac{\sqrt{n^2+n}-\sqrt{n^2+1}}{\sqrt[ n]{n^2}}$$
12

Simplify the limit with L'H么pital's Rule

Apply L'H么pital's rule on limit, we have $$\lim_{n\to\infty}\frac{\frac{n}{\sqrt{n^2+n}}-\frac{n}{\sqrt{n^2+1}}}{\frac{2n}{\sqrt[n]{n^2}}}$$ Consider the limit of the numerator $$\lim_{n\to\infty} \frac{n}{\sqrt{n^2+n}}-\frac{n}{\sqrt{n^2+1}} = \lim_{n\to\infty} \frac{n\sqrt{n^2+1}-n\sqrt{n^2+n}}{\sqrt{n^2+n}\cdot\sqrt{n^2+1}}$$ Apply L'H么pital's rule on the numerator $$\lim_{n\to\infty}-\frac{n}{\sqrt{n^2+n}\cdot\sqrt{n^2+1}} = 0$$ So, we have $$\lim_{n\to\infty}\frac{\sqrt{n^2+n}-\sqrt{n^2+1}}{\sqrt[ n]{n^2}} = \frac{0}{\infty} = 0$$
13

Determine convergence

As the limit from the root test is 0, the power series converges for all \(x \in \mathbb{R}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Bestimmen Sie alle \(z \in \mathbb{C}\), die der folgenden Gleichung gen眉gen. (a) \(\cosh (z)=-1\), (b) \(\cosh z-\frac{1}{2}(1-8 \mathrm{i}) \mathrm{e}^{-z}=2+2 \mathrm{i}\)

Berechnen Sie mithilfe der Potenzreihendarstellung der Exponentialfunktion die ersten 5 Stellen der Dezimaldarstellung von \(\mathrm{e}^{2}\). 脺berlegen Sie sich dazu eine Absch盲tzung, die Ihnen die Richtigkeit Ihres Ergebnisses garantiert.

Berechnen Sie eine Potenzreihendarstellung der rationalen Funktion $$ f(z)=\frac{1+z^{3}}{2-z}, \quad z \in \mathbb{C} \backslash\\{2\\} $$ indem Sie die geometrische Reihe verwenden.

Bestimmen Sie jeweils alle \(z \in \mathbb{C}\), die L枚sungen der folgenden Gleichung sind. (a) \(\cos \bar{z}=\overline{\cos z}\), (b) \(\mathrm{e}^{\mathrm{i} \bar{z}}=\overline{\mathrm{e}^{\mathrm{i} z}}\)

Bestimmen Sie mithilfe der zugeh枚rigen Potenzreihen die folgenden Grenzwerte, (a) \(\lim _{x \rightarrow 0} \frac{1-\cos x}{x \sin x}\), (b) \(\lim _{x \rightarrow 0} \frac{\mathrm{e}^{\sin \left(x^{4}\right)}-1}{x^{2}(1-\cos (x))}\).
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.