/*! 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 17 Evaluate \(\sum_{m=2}^{\infty} \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 17

Evaluate \(\sum_{m=2}^{\infty} \frac{5}{6^{m}}\).

Short Answer

Expert verified
The sum of the convergent geometric series \(\sum_{m=2}^{\infty} \frac{5}{6^{m}}\) is \(S = \frac{1}{6}\).

Step by step solution

01

Verify if the series is geometric

The given series is \(\sum_{m=2}^{\infty} \frac{5}{6^m}\). To check if the series is geometric, let's find the common ratio. We will divide the term with its previous term: \[\frac{\frac{5}{6^{m+1}}}{\frac{5}{6^m}} = \frac{5/6^{m+1}}{5/6^m} = \frac{6^m}{6^{m+1}} = \frac{1}{6}\] As the common ratio is constant and equals to \(\frac{1}{6}\), we can confirm that the series is geometric.
02

Check for convergence

To find out if the geometric series converges, we need to check if the absolute value of the common ratio is less than 1. Since the common ratio is \(\frac{1}{6}\) and its absolute value is also \(\frac{1}{6}\), which is less than 1, we can conclude that the series converges.
03

Compute the sum of the converging series

For a converging geometric series, the sum is given by the formula: \[S = \frac{a}{1 - r}\] where \(S\) is the sum of the series, \(a\) is the first term, and \(r\) is the common ratio. In the given series, the first term is \(a = \frac{5}{6^2} = \frac{5}{36}\) and the common ratio is \(r = \frac{1}{6}\). Now let's compute the sum: \[S = \frac{\frac{5}{36}}{1 - \frac{1}{6}} = \frac{\frac{5}{36}}{\frac{5}{6}}\] Simplify the expression: \[S = \frac{5}{36} \times \frac{6}{5} = \frac{1}{6}\] So, the sum of the series \(\sum_{m=2}^{\infty} \frac{5}{6^{m}}\) is \(\frac{1}{6}\).

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

Express $$ 5.1372647264 \ldots $$ as a fraction; here the digits 7264 repeat forever.

Write the series using summation notation (starting with \(k=1\) ). Each series is either an arithmetic series or a geometric series. $$ 2+4+6+\cdots+100 $$

Evaluate \(\lim _{n \rightarrow \infty} n\left(\ln \left(3+\frac{1}{n}\right)-\ln 3\right)\)

Show that $$ \sqrt{n^{2}+n}-n=\frac{1}{\sqrt{1+\frac{1}{n}}+1} $$ [Hint: Multiply the expression \(\sqrt{n^{2}+n}-n\) by \(\left(\sqrt{n^{2}+n}+n\right) /\left(\sqrt{n^{2}+n}+n\right) .\) Then factor \(n\) out of the numerator and denominator of the resulting expression.] [This identity was used in Example 1.]

Evaluate the geometric series. $$ \sum_{k=1}^{90} \frac{5}{7^{k}} $$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

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.