Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 2

Find the sum for each of the series: a. \(\sum_{n=0}^{\infty} \frac{(-1)^{n} 3}{4^{n}}\) b. \(\sum_{n=2}^{\infty} \frac{2}{5^{n}}\). c. \(\sum_{n=0}^{\infty}\left(\frac{5}{2^{n}}+\frac{1}{3^{n}}\right)\). d. \(\sum_{n=1}^{\infty} \frac{3}{n(n+3)} .\)

Short Answer

Expert verified
The sums for the series are \(\frac{12}{5}\), \(\frac{1}{10}\), \(\frac{23}{2}\), and \(\frac{11}{6}\) respectively.

Step by step solution

01

Solve the first series

The first series is given by \( \sum_{n=0}^{\infty} \frac{(-1)^n 3}{4^n} \). The sum \( S \) can be found using the formula \( \frac{a}{1 - r} \), where \( a = 3 \) and \( r = -\frac{1}{4} \), i.e. \( S = \frac{3}{1 - (-\frac{1}{4})} = \frac{3}{1 + \frac{1}{4}} = \frac{3}{\frac{5}{4}} = \frac{12}{5} \).
02

Solve the second series

The second series is given by \( \sum_{n=2}^{\infty} \frac{2}{5^n} \). We can redefine \( n \) such that the summation starts from 0. So, it becomes \( \sum_{n=0}^{\infty} \frac{2}{5^{n+2}} = \sum_{n=0}^{\infty} \frac{2}{25} \times \frac{1}{5^n} \). This becomes a geometric series with \( a = \frac{2}{25} \) and \( r = \frac{1}{5} \). Using the sum formula, we get \( S = \frac{a}{1 - r} = \frac{2/25}{1 - \frac{1}{5}} = \frac{2/25}{\frac{4}{5}} = \frac{2/25}{\frac{5}{4}} = \frac{2}{20} = \frac{1}{10} \).
03

Solve the third series

The third series is \( \sum_{n=0}^{\infty} (\frac{5}{2^n} + \frac{1}{3^n}) \). Splitting the series, we get \( \sum_{n=0}^{\infty} \frac{5}{2^n} + \sum_{n=0}^{\infty} \frac{1}{3^n} \). Solving individually, we get \( S1 = \frac{5}{1 - \frac{1}{2}} = 10 \) and \( S2 = \frac{1}{1 - \frac{1}{3}} = \frac{3}{2} \). Summing these up, \( S = S1 + S2 = 10 + \frac{3}{2} = \frac{23}{2} \).
04

Solve the fourth series

The fourth series given is \( \sum_{n=1}^{\infty} \frac{3}{n(n+3)} \). This can be rewritten as \( \sum_{n=1}^{\infty} \frac{3}{n(n+3)} = \sum_{n=1}^{\infty} (\frac{1}{n} - \frac{1}{n+3}) \). The series telescopes and thus, the sum is \( S = 1 - \frac{1}{4} + \frac{1}{2} - \frac{1}{5} + \frac{1}{3} - \frac{1}{6} + ... = 1 + \frac{1}{2} + \frac{1}{3} - \frac{1}{4} - \frac{1}{5} - \frac{1}{6} = \frac{11}{6} \).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Consider Gregory's expansion $$ \tan ^{-1} x=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\cdots=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{2 k+1} x^{2 k+1} $$ a. Derive Gregory's expansion using the definition $$ \tan ^{-1} x=\int_{0}^{x} \frac{d t}{1+t^{2}} $$ expanding the integrand in a Maclaurin series, and integrating the resulting series term by term. b. From this result, derive Gregory's series for \(\pi\) by inserting an appropriate value for \(x\) in the series expansion for \(\tan ^{-1} x\).

For those sequences that converge, find the limit \(\lim _{n \rightarrow \infty} a_{n}\) a. \(a_{n}=\frac{n^{2}+1}{n^{3}+1} .\) b. \(a_{n}=\frac{3 n+1}{n+2}\). c. \(a_{n}=\left(\frac{3}{n}\right)^{1 / n}\). d. \(a_{n}=\frac{2 n^{2}+4 n^{3}}{n^{3}+5 \sqrt{2+n^{6}}} .\) e. \(a_{n}=n \ln \left(1+\frac{1}{n}\right)\). f. \(a_{n}=n \sin \left(\frac{1}{n}\right)\). g. \(a_{n}=\frac{(2 n+3) !}{(n+1) !} .\)

Consider the sum \(\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}\) a. Use an appropriate convergence test to show that this series converges. b. Verify that $$ \sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}=\sum_{n=1}^{\infty}\left(\frac{n+1}{n+2}-\frac{n}{n+1}\right) $$ c. Find the \(n\)th partial sum of the series \(\sum_{n=1}^{\infty}\left(\frac{n+1}{n+2}-\frac{n}{n+1}\right)\) and use it to determine the sum of the resulting telescoping series.

Evaluate the following expressions at the given point. Use your calculator and your computer (such as Maple). Then use series expansions to find an approximation to the value of the expression to as many places as you trust. a. \(\frac{1}{\sqrt{1+x^{3}}}-\cos x^{2}\) at \(x=0.015\). b. \(\ln \sqrt{\frac{1+x}{1-x}}-\tan x\) at \(x=0.0015\). c. \(f(x)=\frac{1}{\sqrt{1+2 x^{2}}}-1+x^{2}\) at \(x=5.00 \times 10^{-3}\). d. \(f(R, h)=R-\sqrt{R^{2}+h^{2}}\) for \(R=1.374 \times 10^{3} \mathrm{~km}\) and \(h=1.00 \mathrm{~m}\). e. \(f(x)=1-\frac{1}{\sqrt{1-x}}\) for \(x=2.5 \times 10^{-13}\)

Test for pointwise and uniform convergence on the given set. [The Weierstraß M-Test might be helpful.] a. \(f(x)=\sum_{n=1}^{\infty} \frac{\ln n x}{n^{2}}, x \in[1,2]\) b. \(f(x)=\sum_{n=1}^{\infty} \frac{1}{3^{n}} \cos \frac{x}{2^{n}}\) on \(R\).
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

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.

Sign-up for free