/*! 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 72 Identifying Series In Exercises ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Identifying Series In Exercises \(71-74,\) identify the two series that are the same. \(\begin{array}{l}{\text { (a) } \sum_{n=1}^{\infty} \frac{n 5^{n}}{n !}} \\\ {\text { (b) } \sum_{n=0}^{\infty} \frac{n 5^{n}}{(n+1) !}} \\ {\text { (c) } \sum_{n=0}^{\infty} \frac{(n+1) 5^{n+1}}{(n+1) !}}\end{array}\)

Short Answer

Expert verified
The two series that are the same are series a) and series b). After modifying the series (b), we can see that the general form of both the series becomes identical.

Step by step solution

01

Informative Recalling

Recall the definitions of a series and its terms. In the series: \( \sum_{n=1}^{\infty} \frac{n 5^{n}}{n !} \), 'n' starts at 1 and continues indefinitely (shown by infinity symbol). \( \frac{n 5^{n}}{n !} \) is the general term.
02

Analysis of Series

To find matching series, place each series in a general form that will be easy to compare. For example, series (a) can be written as \( \sum_{n=1}^{\infty} \frac{n 5^{n}}{n !} \), and series (b) can be written as \( \sum_{n=1}^{\infty} \frac{(n-1) 5^{n-1}}{n !} \), by substituting \( n+1 \) with \( n \) in original series (b) .
03

Comparing the Series

Now we can observe that series (a) and modified series (b) are the same. Both series have n starting from 1 in the series and the general term of both series is \( \frac{n 5^{n}}{n !} \)

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 91Ó°ÊÓ!

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

Finding the Interval of Convergence In Exercises \(15-38\) , find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.) $$\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n+1}}{2 n+1}$$

Maclaurin Series Explain how to use the power eries for \(f(x)=\arctan x\) to find the Maclaurin series for \(g(x)=\frac{1}{1+x^{2}}\) What is another way to find the Maclaurin series for \(g\) using a power series for an elementary function?

Comparing Maclaurin Polynomials (a) Compare the Maclaurin polynomials of degree 4 and degree \(5,\) respectively, for the functions $$f(x)=e^{x}\( and \)g(x)=x e^{x}$$ (b) Use the result in part (a) and the Maclaurin polynomial of degree 5 for \(f(x)=\sin x\) to find a Maclaurin polynomial of degree 6 for the function \(g(x)=x \sin x .\) (c) Use the result in part (a) and the Maclaurin polynomial of degree 5 for \(f(x)=\sin x\) to find a Maclaurin polynomial of degree 4 for the function \(g(x)=(\sin x) / x .\)

Differentiating Maclaurin Polynomials (a) Differentiate the Maclaurin polynomial of degree 5 for \(f(x)=\sin x\) and compare the result with the Maclaurin polynomial of degree 4 for \(g(x)=\cos x\) . (b) Differentiate the Maclaurin polynomial of degree 6 for \(f(x)=\cos x\) and compare the result with the Maclaurin polynomial of degree 5 for \(g(x)=\sin x\) (c) Differentiate the Maclaurin polynomial of degree 4 for \(f(x)=e^{x} .\) Describe the relationship between the two series.

Finding the Interval of Convergence In Exercises \(15-38\) , find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.) $$\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{n}}{(n+1)(n+2)}$$

See all solutions

Recommended explanations on Math Textbooks

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.