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Chapter 7: Problem 37

Find the sum of the convergent series. $$ \sum_{n=0}^{\infty}\left(-\frac{1}{2}\right)^{n} $$

Short Answer

Expert verified
The sum of the provided convergent series is \(\frac{2}{3}\).

Step by step solution

01

Identify the First Term and Common Ratio

In the given series \(\sum_{n=0}^{\infty}\left(-\frac{1}{2}\right)^{n}\), the first term \(a\) is when \(n = 0\) which gives \(a = (-\frac{1}{2})^{0} = 1\). The common ratio \(r\) is \(-\frac{1}{2}\).
02

Check if the Series is Convergent

For a series to be convergent, the absolute value of the common ratio must be less than 1. In this case: \(|-1/2| = 1/2 < 1\). Hence, the series is convergent.
03

Calculate the Sum of the Convergent Series

Using the formula for the sum of an infinite geometric series \(S = \frac{a}{1-r}\), substitute the found values for \(a\) and \(r\). The sum \(S\) of the series then is \(S = \frac{1}{1--1/2} = \frac{1}{1+1/2} = \frac{1}{1.5} = \frac{2}{3}\).

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Key Concepts

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

Geometric Series
A Geometric Series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In our case, the series is represented by \(\sum_{n=0}^{\infty}\left(-\frac{1}{2}\right)^{n}\). This means:
  • The first term is 1 when \(n = 0\).
  • Each subsequent term is obtained by multiplying the previous term by \(-\frac{1}{2}\).
The beauty of geometric series lies in their pattern and predictability. Each term quickly brings you to the next, offering a simple method to describe an endless series of numbers with just a couple of parameters, the starting term, and the common ratio.
Infinite Series
An Infinite Series is a sum of infinitely many terms. It starts adding terms from a sequence and continues indefinitely. These series can be daunting, but under certain conditions, they have finite sums. This depends on how the terms behave as they continue. In the given exercise:
  • The series \(\sum_{n=0}^{\infty}\left(-\frac{1}{2}\right)^{n}\) continues without end, aiming to find a sum.
  • To convert this to a finite sum, we need to determine if the series is convergent.
A convergent series approaches a specific value as more terms are added. If the terms decrease in a controlled manner, the series "settles down" to a sum. Geometric series like ours can often yield such results.
Common Ratio
The Common Ratio is a key component of a geometric series. It dictates the constant factor between any two consecutive terms. In the exercise, the common ratio is \(-\frac{1}{2}\). This tells us:
  • Each term is half the previous one, and negative, which alternates the sign.
  • The absolute value of the common ratio, \(\left| -\frac{1}{2} \right| = \frac{1}{2}\), is less than 1.
When the absolute value is less than 1, it ensures the series is convergent. A series with a common ratio like this will not diverge to infinity, but rather converge to a specific sum. The formula \(S = \frac{a}{1-r}\) uses the common ratio to calculate this finite sum efficiently.

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Most popular questions from this chapter

Conjecture Let \(x_{0}=1\) and consider the sequence \(x_{n}\) given by the formula \(x_{n}=\frac{1}{2} x_{n-1}+\frac{1}{x_{n-1}}, \quad n=1,2, \ldots .\) Use a graphing utility to compute the first 10 terms of the sequence and make a conjecture about the limit of the sequence.

Find the sum of the convergent series. $$ 3-1+\frac{1}{3}-\frac{1}{9}+\cdots $$

Compound Interest Consider the sequence \(\left\\{A_{n}\right\\}\) whose \(n\) th term is given by \(A_{n}=P\left(1+\frac{r}{12}\right)^{n}\) where \(P\) is the principal, \(A_{n}\) is the account balance after \(n\) months, and \(r\) is the interest rate compounded annually. (a) Is \(\left\\{A_{n}\right\\}\) a convergent sequence? Explain. (b) Find the first 10 terms of the sequence if \(P=\$ 9000\) and \(r=0.055\)

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(|r|<1,\) then \(\sum_{n=1}^{\infty} a r^{n}=\frac{a}{(1-r)} .\)

Find the sum of the convergent series. $$ 4-2+1-\frac{1}{2}+\cdots $$
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