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Chapter 11: Problem 27

Determine whether the given infinite series converges or diverges. If it converges, find its sum. \(\sum_{n=0}^{x} \frac{7 \cdot 5^{n}+3 \cdot 11^{n}}{13^{n}}\)

Short Answer

Expert verified
The series converges with the sum \( \frac{247}{8} \).

Step by step solution

01

Identify the Series Terms

The given series is \( \sum_{n=0}^{\infty} \frac{7 \cdot 5^{n} + 3 \cdot 11^{n}}{13^{n}} \). This series can be separated into two series: \( \sum_{n=0}^{\infty} \frac{7 \cdot 5^{n}}{13^{n}} \) and \( \sum_{n=0}^{\infty} \frac{3 \cdot 11^{n}}{13^{n}} \). Each part needs to be checked for convergence separately.
02

Simplify the First Series

For the series \( \sum_{n=0}^{\infty} \frac{7 \cdot 5^{n}}{13^{n}} \), notice that it can be rewritten as \( 7 \sum_{n=0}^{\infty} \left( \frac{5}{13} \right)^{n} \). This is a geometric series with the common ratio \( r = \frac{5}{13} \).
03

Determine Convergence of the First Series

A geometric series \( \sum_{n=0}^{\infty} ar^n \) converges if \( |r| < 1 \). Here, \( |\frac{5}{13}| < 1 \), so the series converges. The sum \( S_1 = \frac{a}{1-r} \) where \( a = 7 \) is the first term. Thus, \( S_1 = \frac{7}{1-\frac{5}{13}} = \frac{7 \times 13}{13 - 5} = \frac{91}{8} \).
04

Simplify the Second Series

For the second series, we have \( \sum_{n=0}^{\infty} \frac{3 \cdot 11^{n}}{13^{n}} \) which simplifies to \( 3 \sum_{n=0}^{\infty} \left( \frac{11}{13} \right)^{n} \). This is another geometric series with the common ratio \( r = \frac{11}{13} \).
05

Determine Convergence of the Second Series

Similarly, since \( |\frac{11}{13}| < 1 \), this series converges. The sum \( S_2 = \frac{a}{1-r} \), where \( a = 3 \). Thus, \( S_2 = \frac{3}{1-\frac{11}{13}} = \frac{3 \times 13}{13 - 11} = \frac{39}{2} \).
06

Find the Total Sum

Since both parts of the series converge, the total sum is the sum of the individual sums: \( S = S_1 + S_2 = \frac{91}{8} + \frac{39}{2} = \frac{91}{8} + \frac{156}{8} = \frac{247}{8} \). Thus, the total sum of the series is \( \frac{247}{8} \).

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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 series where each term is a constant multiple of the previous term. This constant multiple is known as the common ratio. For example, in the series \(\sum_{n=0}^{\infty} ar^n\), each term is acquired by multiplying the previous term by the common ratio \(r\).Some key characteristics of geometric series include:
  • Every term except the first is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio.
  • If \(r\) (the common ratio) is between -1 and 1, the series converges, which means it sums to a finite number.
This type of series is extremely important in both mathematics and practical applications like financial calculations and physics.
Common Ratio
The common ratio in a geometric series determines how successive terms of the series relate to each other.In the context of a geometric series \(\sum_{n=0}^{\infty} ar^n\), the common ratio \(r\) profoundly affects whether the series converges or diverges.

Impact of Common Ratio

  • If \( |r| < 1\), the terms become progressively smaller, leading to a series that converges.
  • If \( |r| \geq 1\), the series diverges because the terms do not approach zero.
Understanding the role of the common ratio helps in making important decisions about the nature of the series.
Series Convergence
Series convergence refers to the behavior of a series as we sum its terms indefinitely. A convergent series has terms that approach a fixed finite value as the number of terms increases. For geometric series particularly, convergence is primarily determined by the absolute value of the common ratio \(r\).

Conditions for Convergence

For a geometric series \(\sum_{n=0}^{\infty} ar^n\), it converges if and only if:
  • \( |r| < 1 \)
If this condition is met, the series converges to a specific value, allowing us to use the formula for its sum. Geometric series are among the simplest to analyze because this straightforward condition makes it easy to determine convergence.
Sum of Series
The sum of a convergent geometric series can be found using a specific formula when the common ratio \(r\) satisfies \( |r| < 1\). The sum \(S\) of such a series \(\sum_{n=0}^{\infty} ar^n\) is calculated as follows:

Formula for the Sum

The sum \(S\) is given by:\[S = \frac{a}{1 - r}\]where \(a\) is the first term of the series.This formula allows us to quickly compute the sum of an infinite number of terms, provided the series converges. Understanding and applying this concept is crucial when working with infinite series in both theoretical and applied mathematics.

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Most popular questions from this chapter

Show that the indicated alternating series \(\sum(-1)^{n+1} a_{n}\) satisfies the condition that \(a_{n} \rightarrow 0\) as \(n \rightarrow \infty\), but nevertheless diverges. Tell why the alternating series test does not apply. It may be informative to graph the first 10 or 20 partial sums. \(a_{n}=\left\\{\begin{array}{ll}\frac{1}{\sqrt{n}} & \text { if } n \text { is odd. } \\ \frac{1}{n^{3}} & \text { if } n \text { is even. }\end{array}\right.\)

Find a power series representation for the given function \(f(x)\) by using termwise integration. \(\tanh ^{-1} x=\int_{0}^{1} \frac{1}{1-t^{2}} d t\)

Describe a way to rearrange the terms of the alternating harmonic series to obtain (a) a rearranged series that converges to \(-2 ;\) (b) a rearranged series that diverges to \(+\infty\).

Find the interval of convergence of each power series. \(\sum_{n=1}^{x} \frac{(-1)^{n} n x^{n}}{2^{n}(n+1)^{3}}\)

Suppose that \(\sum a_{n}\) is a conditionally convergent infinite series. For each \(n\), let $$a_{n}^{+}=\frac{a_{n}+\left|a_{n}\right|}{2} \text { and } a_{n}=\frac{a_{n}-\left|a_{n}\right|}{2} \text { . }$$ (a) Explain why \(\Sigma a_{n}^{+}\) consists of the positive terms of \(\Sigma a_{n}\) and why \(\Sigma a_{n}^{-}\) consists of the negative terms of \(\suma_{n}\). (b) Given a real number \(r\), show that some rearrangement of the conditionally convergent series \(\Sigma a_{n}\) converges to \(r\). (Snggestion: If \(r\) is positive, for instance, begin with the first partial sum of the positive series \(\Sigma a_{n}^{+}\) that exceeds \(r\). Then add just enough terms of the negative series \(\Sigma a_{n}\) so that the cumulative sum is less than \(r\). Next add just enough terms of the positive series that the cumulative sum is greater than \(r\), and continuc in this way to define the desired rearrangement.) Why does it follow that this rearranged infinite series converges to \(r\) ?
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