/*! 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 16 Find a formula for the partial s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 16

Find a formula for the partial sums of the series. For each series, determine whether the partial sums have a limit. If so, find the sum of the series. $$ \sum_{n=1}^{\infty}\left(\frac{1}{4}\right)^{n} $$

Short Answer

Expert verified
The series converges to a sum of \( \frac{4}{3} \).

Step by step solution

01

Identify the Type of Series

The series given is \( \sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^n \). We can observe that this series is a geometric series of the form \( \sum_{n=1}^{\infty} ar^n \), where \( a = 0 \) (starting from \(n=1\)), and \( r = \frac{1}{4} \).
02

Formula for the Partial Sums

For a geometric series starting from \( n = 1 \), the formula for the \( n \)-th partial sum \( S_n \) is given by \( S_n = \frac{a(1 - r^n)}{1-r} \). Since \( a = 1 \) for our series, the partial sum formula is \( S_n = \frac{1(1 - \left(\frac{1}{4}\right)^n)}{1-\frac{1}{4}} \).
03

Simplify the Formula

Substitute \( a = 1 \) and \( r = \frac{1}{4} \) into the formula: \[ S_n = \frac{1 - \left(\frac{1}{4}\right)^n}{1 - \frac{1}{4}} = \frac{1 - \left(\frac{1}{4}\right)^n}{\frac{3}{4}} = \frac{4}{3}(1 - \left(\frac{1}{4}\right)^n) \]. This gives us the formula for the partial sums, \( S_n = \frac{4}{3}(1 - \left(\frac{1}{4}\right)^n) \).
04

Determine the Limit of the Partial Sums

To find if the partial sums have a limit, we examine \( \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{4}{3}(1 - \left(\frac{1}{4}\right)^n) \). As \( n \to \infty \), \( \left(\frac{1}{4}\right)^n \to 0 \).
05

Calculate the Limit

Apply the limit: \[ \lim_{n \to \infty} S_n = \frac{4}{3}(1 - 0) = \frac{4}{3} \]. Therefore, the series converges to this limit.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Sums
In mathematics, particularly in the study of series, the term **partial sums** is used to describe the sum of the first few terms of a sequence or series. When we speak of the partial sum, denoted as \( S_n \), we're essentially tallying up the first \( n \) elements. This helps us understand how the whole infinite series behaves.
In our task, we investigated the series \( \sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^n \), a classic example of a geometric series.
Here is how you calculate each partial sum:
  • Identify the series terms,
    like our example \( \left(\frac{1}{4}\right)^n \).
  • Use the formula for the nth partial sum of a geometric series:
    \( S_n = \frac{a(1 - r^n)}{1-r} \).
    • It's important to note that \( a \) is the first term of the series,
      and \( r \) is the common ratio.
    \( S_n = \frac{4}{3}(1 - \left(\frac{1}{4}\right)^n) \) tells us what the sum is up to the nth term.
    Using partial sums, we can analyze the trend of the entire series.
Series Convergence
The idea of **series convergence** is fundamental to understanding infinite series. In simple terms, convergence suggests that as you sum more and more terms of a series, you approach a specific value, known as the limit.
For a geometric series, the rule of thumb is:
  • If the absolute value of the common ratio \( r \) is less than 1
    (i.e., \( |r| < 1 \)), the series converges.
  • If \( |r| \geq 1 \), the series does not converge.
In our example, our common ratio \( r = \frac{1}{4} \). Because \( \left|\frac{1}{4}\right| < 1 \), we concluded that the series converges.
This holds true generally and means that as we add more terms, the partial sums approach a fixed value.
Limit of a Series
Understanding the notion of a **limit of a series** is crucial to grasping how infinite series operate. When we say a series has a limit, we mean that as the number of terms in the series (n) grows infinitely large, its partial sums focus in on a single, finite number.
For the series \( \sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^n \), we determined the limit as follows:
  • Start with the formula for the partial sums,
    \( S_n = \frac{4}{3}(1 - \left(\frac{1}{4}\right)^n) \).
  • Determine the behavior as \( n \to \infty \).
  • Recognize that \( \left(\frac{1}{4}\right)^n \to 0 \),
    because multiplying a number smaller than one by itself over and over
    makes it smaller and smaller.
Therefore, the limit of our series is:
\( \lim_{n \to \infty} S_n = \frac{4}{3} \).
This shows that as we consider more terms from the series, their sum gets closer to, and ultimately reaches, \( \frac{4}{3} \).

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 the repeating decimal as a fraction. $$ 0.453232232232 \ldots $$

Determine whether the series converges or diverges. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{n^{1 / 10}} $$

Determine whether the series converges or diverges. $$ \sum_{n=2}^{\infty}(-1)^{n+1} \frac{n+1}{4 n} $$

Let \(f(x)=e^{\left(x^{2}\right)} \int_{0}^{x} e^{-\left(t^{2}\right)} d t\) a. Show that \(f(0)=0\) and \(f^{\prime}(x)=2 x f(x)+1\). b. Find the Taylor series of \(f\) about 0 . (Hint: Note that \(f^{(n)}(x)=2(n-1) f^{(n-2)}(x)+2 x f^{(n-1)}(x)\) for \(\left.n \geq 2 .\right)\)

Suppose \(\sum_{n=1}^{\infty} a_{n}\) converges and is not necessarily positive. Give an example to show that \(\sum_{n=1}^{\infty} a_{n}^{2}\) need not converge. (Hint: Let \(\sum_{n=1}^{\infty} a_{n}\) be an alternating series whose terms approach 0 very slowly.)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Applied Mathematics

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.