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Chapter 9: Problem 68

Which series converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. $$\sum_{n=1}^{\infty} \frac{2^{n}+4^{n}}{3^{n}+4^{n}}$$

Short Answer

Expert verified
The series diverges because the nth-term test shows \( a_n \to 1 \neq 0 \). It doesn't converge, so there's no sum.

Step by step solution

01

Identify the Dominant Terms

In the given series, \( a_n = \frac{2^n + 4^n}{3^n + 4^n} \). As \( n \to \infty \), the dominant terms in both the numerator and the denominator are \( 4^n \). We can simplify the expression to \( a_n \approx \frac{4^n}{4^n} \). This simplifies to \( a_n \approx 1 \).
02

Determine Convergence with the Divergence Test

To apply the divergence test, check \( \lim_{n \to \infty} a_n \). Since we found that \( a_n \approx 1 \), \( \lim_{n \to \infty} a_n \) is 1. The divergence test states if \( \lim_{n \to \infty} a_n eq 0 \), then the series \( \sum_{n=1}^{\infty} a_n \) diverges.
03

Conclusion

Since \( \lim_{n \to \infty} a_n = 1 eq 0 \), the series \( \sum_{n=1}^{\infty} \frac{2^n+4^n}{3^n+4^n} \) diverges by the Divergence Test.

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Key Concepts

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

Understanding Dominant Terms in Series
When dealing with an infinite series, identifying the dominant terms is a fundamental step. These are the terms that grow the fastest as the number of terms increases.
In the expression \( a_n = \frac{2^n + 4^n}{3^n + 4^n} \), each component consists of terms raised to the power of \( n \).
As \( n \) becomes extremely large, both the numerator and the denominator are dominated by the terms \( 4^n \).
This is because, among the terms \( 2^n \, \text{and} \, 4^n \), the latter grows significantly faster due to its larger base.
  • The same logic applies to the denominator, where \( 4^n \) grows faster than \( 3^n \).
  • By simplifying, we see \( \frac{4^n}{4^n} = 1 \).
Using dominant terms simplifies the expression to \( a_n \approx 1 \), making it easier to determine convergence in the next steps.
Applying the Divergence Test
The divergence test is one of the simplest tools used to determine whether a series diverges.
This test evaluates \( \lim_{n \to \infty} a_n \) to check if it equals zero or not.
If the limit does not equal zero, then the series must diverge.
For the series \( \sum_{n=1}^{\infty} \frac{2^n + 4^n}{3^n + 4^n} \), we find \( a_n \approx 1 \).
  • The evaluation of the limit results in \( \lim_{n \to \infty} a_n = 1 \).
  • Since 1 is not zero, the series diverges by the divergence test.
The divergence test is particularly powerful when the limit of \( a_n \) is straightforward to compute, as seen in this example.
Exploring Infinite Series and Their Behavior
An infinite series is the sum of a sequence of numbers that goes on forever.
Understanding whether an infinite series converges or diverges is vital:
Convergence means that as you sum more terms, the total approaches a specific finite value, while divergence implies that the sum does not approach any particular number as more terms are added.
  • If \( \lim_{n \to \infty} a_n = 0 \), further tests are required to ascertain convergence, such as the comparison test or the ratio test.
  • If \( \lim_{n \to \infty} a_n eq 0 \), the series definitely diverges.
The series \( \sum_{n=1}^{\infty} \frac{2^n + 4^n}{3^n + 4^n} \) exemplifies divergence because its term limit does not vanish. Thus, understanding the behavior and form of the series terms is crucial for determining the series' nature.

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Most popular questions from this chapter

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=0}^{\infty} \frac{n+1}{(n+2) !}\left(\frac{3}{2}\right)^{n}$$

a. Find the interval of convergence of the power series $$\sum_{n=0}^{\infty} \frac{8}{4^{n+2}} x^{n}$$ b. Represent the power series in part (a) as a power series about \(x=3\) and identify the interval of convergence of the new series. (Later in the chapter you will understand why the new interval of convergence does not necessarily include all of the numbers in the original interval of convergence.)

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty}\left(1-\frac{2}{n}\right)^{n^{2}}$$

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.

Neither the Ratio Test nor the Root Test helps with \(p\) -series. Try them on $$ \sum_{n=1}^{\infty} \frac{1}{n^{p}} $$ and show that both tests fail to provide information about convergence.
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