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In Exercises \(7-14,\) write out the first eight terms of each series to show how the series starts. Then find the sum of the series or show that it diverges. $$ \sum_{n=0}^{\infty}\left(\frac{5}{2^{n}}+\frac{1}{3^{n}}\right) $$

Short Answer

Expert verified
The series converges with a sum of 11.5.

Step by step solution

01

Identify the Series Components

The given series is \( \sum_{n=0}^{\infty}\left(\frac{5}{2^{n}}+\frac{1}{3^{n}}\right) \). It can be split into two separate geometric series: \( \sum_{n=0}^{\infty}\frac{5}{2^n} \) and \( \sum_{n=0}^{\infty}\frac{1}{3^n} \).
02

Write the First Eight Terms

To find how the series starts, calculate the first eight terms of both components:- For \( \sum_{n=0}^{\infty}\frac{5}{2^n} \): \( 5, \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \frac{5}{16}, \frac{5}{32}, \frac{5}{64}, \frac{5}{128} \).- For \( \sum_{n=0}^{\infty}\frac{1}{3^n} \): \( 1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \frac{1}{81}, \frac{1}{243}, \frac{1}{729}, \frac{1}{2187} \).Thus, the first eight terms of the entire series are the sums of these corresponding terms.
03

Add Corresponding Terms of the Series

Sum the corresponding terms of both geometric series to see the first eight terms:1. \( 5 + 1 = 6 \)2. \( \frac{5}{2} + \frac{1}{3} = \frac{15}{6} + \frac{2}{6} = \frac{17}{6} \)3. \( \frac{5}{4} + \frac{1}{9} = \frac{45}{36} + \frac{4}{36} = \frac{49}{36} \)4. \( \frac{5}{8} + \frac{1}{27} = \frac{135}{216} + \frac{8}{216} = \frac{143}{216} \)5. \( \frac{5}{16} + \frac{1}{81} = \frac{405}{1296} + \frac{16}{1296} = \frac{421}{1296} \)6. \( \frac{5}{32} + \frac{1}{243} = \frac{1215}{7776} + \frac{32}{7776} = \frac{1247}{7776} \)7. \( \frac{5}{64} + \frac{1}{729} = \frac{3645}{46656} + \frac{64}{46656} = \frac{3709}{46656} \)8. \( \frac{5}{128} + \frac{1}{2187} = \frac{10935}{280896} + \frac{128}{280896} = \frac{11063}{280896} \)
04

Determine If the Series Converges

For a geometric series \( \sum_{n=0}^{\infty} ar^n \), it converges if \( |r| < 1 \). In our series:- For \( \sum_{n=0}^{\infty}\frac{5}{2^n} \), \( r = \frac{1}{2} < 1 \), so it converges.- For \( \sum_{n=0}^{\infty}\frac{1}{3^n} \), \( r = \frac{1}{3} < 1 \), so it converges.Since both component series converge, the entire series converges.
05

Calculate the Sum of Each Series Component

Each geometric series converges to the sum \( \frac{a}{1-r} \).- For \( \sum_{n=0}^{\infty}\frac{5}{2^n} \), \( a = 5 \) and \( r = \frac{1}{2} \), so the sum is \( \frac{5}{1 - \frac{1}{2}} = 10 \).- For \( \sum_{n=0}^{\infty}\frac{1}{3^n} \), \( a = 1 \) and \( r = \frac{1}{3} \), so the sum is \( \frac{1}{1 - \frac{1}{3}} = \frac{3}{2} \).
06

Sum the Series Total

The sum of the entire series is the sum of the two component series:The total sum is \( 10 + \frac{3}{2} = 11.5 \).

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Key Concepts

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

Convergence of Series
Understanding the convergence of series is essential in mathematics, especially when dealing with infinite series. A series is said to converge if the sum of its infinite terms approaches a finite number. In simple terms, if you keep adding more and more terms, the total reaches a specific value rather than growing indefinitely.

For geometric series, convergence is determined by the common ratio, represented by \( r \). A geometric series of the form \( \sum_{n=0}^{\infty} ar^n \) converges if the absolute value of \( r \) is less than 1, \( |r| < 1 \). If this condition is met, we say the series converges, meaning the sum will settle at a particular value. Conversely, if \( |r| \geq 1 \), the series diverges, meaning it doesn't settle to a specific sum and instead grows endlessly.
Sum of Infinite Series
Calculating the sum of an infinite series can seem daunting at first. However, with geometric series, there's a straightforward approach. Once you confirm that the series converges, you can use the formula for the sum of a convergent geometric series: \( \frac{a}{1-r} \), where \( a \) is the first term, and \( r \) is the common ratio.

Let's apply this to the original series from the exercise. The series splits into two components: \( \sum_{n=0}^{\infty}\frac{5}{2^n} \) and \( \sum_{n=0}^{\infty}\frac{1}{3^n} \). For the first component, \( a = 5 \) and \( r = \frac{1}{2} \). Using the formula, we find the sum:
  • \( \frac{5}{1 - \frac{1}{2}} = 10 \)
For the second component, \( a = 1 \) and \( r = \frac{1}{3} \), leading to:
  • \( \frac{1}{1 - \frac{1}{3}} = \frac{3}{2} \)
Adding these sums gives the total sum of the series: \( 10 + \frac{3}{2} = 11.5 \). This method is especially useful for series with multiple components, like the one in our exercise.
Mathematical Series
A mathematical series is essentially a way to add up a sequence of terms to find a sum. The terms in a series might follow a certain rule or pattern, and the series can be either finite or infinite.

In the context of infinite series, we often explore whether they converge or diverge. Geometric series, which have terms based on a constant ratio, are a classic example. They highlight not just how mathematicians think about sums but how we can manage calculations even when dealing with infinite numbers of terms.

Series like \( \sum_{n=0}^{\infty}\left(\frac{5}{2^n}+\frac{1}{3^n}\right) \) include both addition and multiplication throughout their structure. They present a beautiful intersection of algebra and calculus, showing us how mathematical principles govern patterns in numbers and their sums. Learning about the properties of different series, such as geometric or arithmetic, provides a deeper insight into the nature of these sums, equipping students with an understanding that extends beyond mere calculation.

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Most popular questions from this chapter

Taylor's Theorem and the Mean Value Theorem Explain how the Mean Value Theorem (Section \(4.2,\) Theorem 4 ) is a special case of Taylor's Theorem.

In the alternating harmonic series, suppose the goal is to arrange the terms to get a new series that converges to \(-1 / 2 .\) Start the new arrangement with the first negative term, which is \(-1 / 2 .\) Whenever you have a sum that is less than or equal to \(-1 / 2,\) start introducing positive terms, taken in order, until the new total is greater than \(-1 / 2 .\) Then add negative terms until the total is less than or equal to \(-1 / 2\) again. Continue this process until your partial sums have been above the target at least three times and finish at or below it. If \(s_{n}\) is the sum of the first \(n\) terms of your new series, plot the points \(\left(n, s_{n}\right)\) to illustrate how the sums are behaving.

Euler's constant Graphs like those in Figure 10.11 suggest that as \(n\) increases there is little change in the difference between the sum $$ 1+\frac{1}{2}+\dots+\frac{1}{n} $$ and the integral $$ \ln n=\int_{1}^{n} \frac{1}{x} d x $$ To explore this idea, carry out the following steps. a. By taking \(f(x)=1 / x\) in the proof of Theorem \(9,\) show that $$ \ln (n+1) \leq 1+\frac{1}{2}+\cdots+\frac{1}{n} \leq 1+\ln n $$ or $$ 0<\ln (n+1)-\ln n \leq 1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \leq 1 $$ Thus, the sequence $$ a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n $$ is bounded from below and from above. b. Show that $$ \frac{1}{n+1}<\int_{n}^{n+1} \frac{1}{x} d x=\ln (n+1)-\ln n $$ and use this result to show that the sequence \(\left\\{a_{n}\right\\}\) in part (a) is decreasing. since a decreasing sequence that is bounded from below converges, the numbers \(a_{n}\) defined in part (a) converge: $$ 1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n \rightarrow \gamma $$ The number \(\gamma,\) whose value is 0.5772\(\ldots\) is called Euler's constant.

Suppose that \(a_{n}>0\) and \(b_{n}>0\) for \(n \geq N(N\) an integer). If \(\lim _{n \rightarrow \infty}\left(a_{n} / b_{n}\right)=\infty\) and \(\sum a_{n}\) converges, can anything be said about \(\sum b_{n}^{?}\) Give reasons for your answer.

Show that if \(\sum_{n=1}^{\infty} a_{n}\) and \(\sum_{n=1}^{\infty} b_{n}\) both converge absolutely, then so do the following. $$\begin{array}{ll}{\text { a. }} & {\sum_{n=1}^{\infty}\left(a_{n}+b_{n}\right)} & {\text { b. } \sum_{n=1}^{\infty}\left(a_{n}-b_{n}\right)} \\ {\text { c. }} & {\sum_{n=1}^{\infty} k a_{n}(k \text { any number })}\end{array}$$

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