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Converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. $$ \sum_{n=0}^{\infty} e^{-2 n} $$

Short Answer

Expert verified
The series converges with a sum of approximately 1.156.

Step by step solution

01

Recognize series type

The series given is \( \sum_{n=0}^{\infty} e^{-2n} \). This is an instance of a geometric series. A geometric series takes the form \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term, and \( r \) is the common ratio.
02

Identify elements of the geometric series

For the series \( \sum_{n=0}^{\infty} e^{-2n} \), we need to identify \( a \) and \( r \). Comparing with \( \sum_{n=0}^{\infty} ar^n \), it is evident that \( a = 1 \) and \( r = e^{-2} \).
03

Determine convergence or divergence

For a geometric series to converge, the absolute value of the common ratio must be less than 1, i.e., \( |r| < 1 \). Here, \( r = e^{-2} \). Since \( e^{-2} \approx 0.135 \), which is less than 1, the series converges.
04

Calculate the sum

The sum of an infinite geometric series is given by the formula \( S = \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio. Here, \( a = 1 \) and \( r = e^{-2} \). So, the sum is \( S = \frac{1}{1 - e^{-2}} \).
05

Simplify the sum expression

Calculate \( 1 - e^{-2} \approx 1 - 0.135 = 0.865 \) (using a calculator for approximation). Thus, the sum is \( S \approx \frac{1}{0.865} \approx 1.156 \) (round as necessary).

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Key Concepts

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

Convergence
In mathematics, particularly when dealing with series, the concept of convergence is pivotal. A series is said to converge if its terms approach a specific finite value as they extend to infinity. This is profound when evaluating infinite series, like our geometric series with the form \(\sum_{n=0}^{\infty} ar^n\).

When we talk about convergence in the context of a geometric series, what we are really checking is whether the absolute value of the common ratio \(r\) is less than 1. For the series \(\sum_{n=0}^{\infty} e^{-2n}\), we identified the common ratio \(r = e^{-2}\), which approximately equals 0.135. Since this value is less than 1, it indicates that the series converges.

It's important to note that convergence tells us that as we consider more and more terms in the series, the sum stabilizes or "settles down" to a finite number rather than going off to infinity.
Series Summation
The process of series summation is the act of adding terms in a series to find a total sum. For a finite series, this is straightforward; however, it becomes more intriguing with an infinite series, where we look at its convergence as well.

For a convergent geometric series, there's a neat formula to find its sum: \( S = \frac{a}{1 - r}\). Here, \(a\) is the first term, and \(r\) is the common ratio. Using this formula for the series \(\sum_{n=0}^{\infty} e^{-2n}\), we established that first term \(a = 1\) and \(r = e^{-2}\). Plugging these values into the formula gives us the series sum \(S = \frac{1}{1 - e^{-2}}\).

This formula enables us to quickly compute the sum of a convergent series without having to add up infinitely many terms directly. After calculating the value \(1 - e^{-2}\), we find \(S \approx 1.156\). This is the sum of all terms from the infinite series.
Infinite Series
Infinite series are sequences of numbers added indefinitely, often possessing particular properties that make them fascinating and useful in calculus and beyond. A crucial aspect of these series is their behavior as they stretch into infinity.

Infinite series can either converge, meaning they settle to a finite sum, or diverge, meaning they do not approach any set limit. The geometric series is a common type of infinite series, notable for its clear criteria for convergence.

For the series \(\sum_{n=0}^{\infty} e^{-2n}\), it is specifically a geometric infinite series. We discovered it converges due to its common ratio \(r = e^{-2}\) being less than 1. Such converging infinite series are quite valuable, as they allow mathematicians to simplify problems and understand behaviors at the limits of infinite processes.

Infinite series find applications in various fields—physics, engineering, and economics—to model and solve problems involving repeated processes or growth patterns.

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Most popular questions from this chapter

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ y^{\prime \prime}+y=x, \quad y^{\prime}(0)=1 \text { and } y(0)=2 $$

Compound interest, deposits, and withdrawals If you invest an amount of money \(A_{0}\) at a fixed annual interest rate \(r\) compounded \(m\) times per year, and if the constant amount \(b\) is added to the account at the end of each compounding period (or taken from the account if \(b<0 ),\) then the amount you have after \(n+1\) compounding periods is $$ A_{n+1}=\left(1+\frac{r}{m}\right) A_{n}+b $$ a. If \(A_{0}=1000, r=0.02015, m=12,\) and \(b=50\) , calculate and plot the first 100 points \(\left(n, A_{n}\right) .\) How much money is in your account at the end of 5 years? Does \(\left\\{A_{n}\right\\}\) converge? Is \(\left\\{A_{n}\right\\}\) bounded? b. Repeat part (a) with \(A_{0}=5000, r=0.0589, m=12,\) and \(b=-50 .\) c. If you invest 5000 dollars in a certificate of deposit (CD) that pays 4.5\(\%\) annually, compounded quarterly, and you make no further investments in the CD, approximately how many years will it take before you have \(20,000\) dollars? What if the CD earns 6.25\(\% ?\) d. It can be shown that for any \(k \geq 0\) , the sequence defined recursively by Equation \((1)\) satisfies the relation $$ A_{k}=\left(1+\frac{r}{m}\right)^{k}\left(A_{0}+\frac{m b}{r}\right)-\frac{m b}{r} $$ For the values of the constants \(A_{0}, r, m,\) and \(b\) given in part (a), validate this assertion by comparing the values of the first 50 terms of both sequences. Then show by direct substitution that the terms in Equation \((2)\) satisfy the recursion formula in Equation ( 1\()\) .

Replace \(x\) by \(-x\) in the Taylor series for \(\ln (1+x)\) to obtain a series for \(\ln (1-x) .\) Then subtract this from the Taylor series for \(\ln (1+x)\) to show that for \(|x|<1\) , $$ \ln \frac{1+x}{1-x}=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots\right) $$

Are there any values of \(x\) for which \(\sum_{n=1}^{\infty}(1 /(n x))\) converges? Give reasons for your answer.

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ y^{\prime \prime}+y=0, \quad y^{\prime}(0)=0 \text { and } y(0)=1 $$

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